3.13 HW 12

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3.13.1 chapter 15, problem 4.12

Problem Find the exponential Fourier transform of the given $f\left(x\right)$ and write $f(x)$ as a fourier integral.

$$f(x)=\{\begin{array}{lll}\sin x& & \left|x\right|<\frac{\pi }{2}\\ & & \\ 0,& & \left|x\right|>\frac{\pi }{2}\end{array}$$

Solution

Let $ϝ\left(\alpha \right)$ be the Fourier transform of $f\left(x\right)$ defined as $ϝ\left(\alpha \right)=\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}f(x)e^{-i\alpha x}dx$

$$\begin{array}{rl}ϝ\left(\alpha \right)& =\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}f(x)e^{-i\alpha x}dx\\ 2\pi ϝ\left(\alpha \right)& ={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}}\sin x{e}^{-i\alpha x}dx\end{array}$$

Integration by parts, $u=\sin x,du=\cos x,v=\frac{e^{-i\alpha x}}{-i\alpha }$, hence $\int udv=$ $uv-\int duv$

$$\begin{array}{rl}I& ={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}}\sin x{e}^{-i\alpha x}dx\\ & ={[\sin x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}}\cos x\frac{e^{-i\alpha x}}{-i\alpha }dx\\ & ={[\sin x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}+\frac{1}{i\alpha }{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}}\cos x{e}^{-i\alpha x}dx\end{array}$$

Integration by parts the second integral again. $u=\cos x,du=-\sin x$

$$\begin{array}{rl}I& ={[\sin x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}+\frac{1}{i\alpha }\{{[\cos x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}}(-\sin x)\frac{e^{-i\alpha x}}{-i\alpha }dx\}\\ I& ={[\sin x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}+\frac{1}{i\alpha }\{{[\cos x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-\left(\frac{1}{i\alpha }\right){\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}}\sin x{e}^{-i\alpha x}dx\}\\ I& ={[\sin x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}+\frac{1}{i\alpha }{[\cos x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-\frac{1}{i^2{\alpha }^2}{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}}\sin x{e}^{-i\alpha x}dx\\ I& ={[\sin x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}+\frac{1}{i\alpha }{[\cos x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}+\frac{1}{{\alpha }^2}{\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}}\sin x{e}^{-i\alpha x}dx\end{array}$$

But the last integral on the right above is the same as the integral we start with, so

$$\begin{array}{rl}I& ={[\sin x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}+\frac{1}{i\alpha }{[\cos x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}+\frac{1}{{\alpha }^2}I\\ I-\frac{1}{{\alpha }^2}I& ={[\sin x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}-\frac{i}{\alpha }{[\cos x\frac{e^{-i\alpha x}}{-i\alpha }]}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\\ I(1-\frac{1}{{\alpha }^2})& =[\sin \left(\frac{\pi }{2}\right)\frac{e^{-i\alpha \frac{\pi }{2}}}{-i\alpha }-\sin (-\frac{\pi }{2})\frac{e^{-i\alpha (-\frac{\pi }{2})}}{-i\alpha }]-\frac{i}{\alpha }[\cos \left(\frac{\pi }{2}\right)\frac{e^{-i\alpha \frac{\pi }{2}}}{-i\alpha }-\cos (-\frac{\pi }{2})\frac{e^{-i\alpha (-\frac{\pi }{2})}}{-i\alpha }]\\ I\left(\frac{{\alpha }^2-1}{{\alpha }^2}\right)& =[\frac{e^{-i\alpha \frac{\pi }{2}}}{-i\alpha }+\frac{e^{i\alpha \left(\frac{\pi }{2}\right)}}{-i\alpha }]-\frac{i}{\alpha }(0)\\ I& =\left(\frac{{\alpha }^2}{{\alpha }^2-1}\right)[\frac{-e^{-i\alpha \frac{\pi }{2}}}{i\alpha }-\frac{e^{i\alpha \frac{\pi }{2}}}{i\alpha }]\\ I& =\left(\frac{\alpha }{{\alpha }^2-1}\right)\frac{-1}{i}[e^{i\alpha \frac{\pi }{2}}+e^{-i\alpha \frac{\pi }{2}}]\\ I& =\left(\frac{\alpha i}{{\alpha }^2-1}\right)[e^{i\alpha \frac{\pi }{2}}+e^{-i\alpha \frac{\pi }{2}}]\\ I& =\left(\frac{\alpha i}{{\alpha }^2-1}\right)2\cos \left(\alpha \frac{\pi }{2}\right)\end{array}$$

Hence the Fourier transform of $f(x)$ is

$$\begin{array}{rl}2\pi ϝ\left(\alpha \right)& ={\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}}\sin x{e}^{-i\alpha x}dx\\ & =\left(\frac{\alpha i}{{\alpha }^2-1}\right)2\cos \left(\alpha \frac{\pi }{2}\right)\end{array}$$

Therefore

$$\begin{array}{rl}g\left(\alpha \right)& =\frac{1}{\pi }\left(\frac{\alpha i}{{\alpha }^2-1}\right)\cos \left(\alpha \frac{\pi }{2}\right)\\ & =\frac{\alpha i\cos \left(\alpha \frac{\pi }{2}\right)}{({\alpha }^2-1)\pi }\end{array}$$

To obtain $f(x)$ given its fourier transform $ϝ\left(\alpha \right),$ then we apply the inverse fourier transform

$$\begin{array}{rl}f(x)& ={\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}ϝ\left(\alpha \right)e^{i\alpha x}d\alpha \\ & ={\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}\frac{\alpha i\cos \left(\alpha \frac{\pi }{2}\right)}{({\alpha }^2-1)\pi }{e}^{i\alpha x}d\alpha \\ & =\frac{1}{\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}\frac{\alpha i\cos \left(\alpha \frac{\pi }{2}\right)}{{\alpha }^2-1}{e}^{i\alpha x}d\alpha \end{array}$$

3.13.2 chapter 15, problem 4.18

Problem Find the fourier sin transform of the given $f\left(x\right)$ and write $f(x)$ as a fourier integral. Verify the answer is the same as the exponential fourier transform.

$$f(x)=\{\begin{array}{lll}x& & \left|x\right|<1\\ & & \\ 0,& & \left|x\right|>1\end{array}$$

Solution

Let ${\mathcal{F}}_{s}f(x)$ be the Fourier sin transform of $f(x)$ defined as $g_s\left(\alpha \right)={\mathcal{F}}_{s}f(x)=\sqrt{\frac{2}{\pi }}{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}f(x)\sin \left(\alpha x\right)dx$. Hence, for the function above we get

$$g_s\left(\alpha \right)=\sqrt{\frac{2}{\pi }}{\displaystyle \underset{0}{\overset{1}{\int }}}x\sin \left(\alpha x\right)dx$$

Notice, we integrate from zero, not from -1, since the sin transform is defined only for positive x. Integrating by parts, $u=x,v=\frac{-\cos \left(\alpha x\right)}{\alpha }$, hence $\int udv=$ $uv-\int duv$

$$\begin{array}{rl}{g}_s\left(\alpha \right)& =\sqrt{\frac{2}{\pi }}\{{\left[x\left(\frac{-\cos \left(\alpha x\right)}{\alpha }\right)\right]}_0^1-{\displaystyle \underset{0}{\overset{1}{\int }}}\frac{-\cos \left(\alpha x\right)}{\alpha }dx\}\\ & =\sqrt{\frac{2}{\pi }}\{\frac{-1}{\alpha }{[x\cos \left(\alpha x\right)]}_0^1+\frac{1}{\alpha }{\displaystyle \underset{0}{\overset{1}{\int }}}\cos \left(\alpha x\right)dx\}\\ & =\sqrt{\frac{2}{\pi }}\{\frac{-1}{\alpha }{[x\cos \left(\alpha x\right)]}_0^1+\frac{1}{\alpha }{\left[\frac{\sin \left(\alpha x\right)}{\alpha }\right]}_0^1\}\\ & =\sqrt{\frac{2}{\pi }}\frac{1}{\alpha }\{-[\cos \left(\alpha \right)-0]+\frac{1}{\alpha }{[\sin \left(\alpha x\right)]}_0^1\}\\ & =\sqrt{\frac{2}{\pi }}\frac{1}{\alpha }\{-\cos \left(\alpha \right)+\frac{1}{\alpha }{[\sin \left(\alpha \right)-0]}^{}\}\\ & =\sqrt{\frac{2}{\pi }}\frac{1}{\alpha }(-\cos \left(\alpha \right)+\frac{1}{\alpha }\sin \left(\alpha \right))\\ & =\sqrt{\frac{2}{\pi }}\frac{1}{\alpha }(\frac{1}{\alpha }\sin \left(\alpha \right)-\cos \left(\alpha \right){}^{})\end{array}$$

Hence the Sin Fourier transform of $f(x)$ is

$$g_s\left(\alpha \right)=\sqrt{\frac{2}{\pi }}\frac{1}{\alpha }(\frac{1}{\alpha }\sin \left(\alpha \right)-\cos \left(\alpha \right){}^{})$$

To obtain $f(x)$ given its sin fourier transform $g_s\left(\alpha \right),$ then we apply the inverse sin fourier transform

$$\begin{array}{rl}{f}_s(x)& =\sqrt{\frac{2}{\pi }}{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}{g}_s\left(\alpha \right)\sin \alpha xd\alpha \\ & =\sqrt{\frac{2}{\pi }}{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}\sqrt{\frac{2}{\pi }}\frac{1}{\alpha }(\frac{1}{\alpha }\sin \alpha -\cos {\alpha }^{})\sin \alpha xd\alpha \\ & =\frac{2}{\pi }{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}(\frac{1}{{\alpha }^2}\sin \alpha -\frac{1}{\alpha }\cos \alpha )\sin \alpha xd\alpha \\ \text{(A0)}& & =\frac{2}{\pi }{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}\left(\frac{\sin \alpha -\alpha \cos \alpha }{{\alpha }^2}\right)\sin \alpha xd\alpha \end{array}$$

Now we need to show that the above is the same as the inverse fourier transform found for problem 6. From back of the book, the IFT for problem 6 is given as

$$f(x)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin \alpha -\alpha \cos \alpha }{i\pi {\alpha }^2}{e}^{i\alpha x}d\alpha$$

Need to convert the above to $f_s(x)$. Since $e^{i\alpha x}=\cos \alpha x+i\sin \alpha x$

$$\begin{array}{rl}f(x)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\sin \alpha -\alpha \cos \alpha }{i\pi {\alpha }^2}(\cos \alpha x+i\sin \alpha x)d\alpha \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{(\sin \alpha -\alpha \cos \alpha )\cos \alpha x}{i\pi {\alpha }^2}+\frac{(\sin \alpha -\alpha \cos \alpha )i\sin \alpha x}{i\pi {\alpha }^2}d\alpha \\ \text{(1)}& & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{(\sin \alpha -\alpha \cos \alpha )\cos \alpha x}{i\pi {\alpha }^2}d\alpha +{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{(\sin \alpha -\alpha \cos \alpha )i\sin \alpha x}{i\pi {\alpha }^2}d\alpha \end{array}$$

Looking at the first integral,

$$\begin{array}{rl}& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{(\stackrel{\text{odd}}{\overbrace{\sin \alpha }}-\stackrel{\text{odd}}{\overbrace{\alpha }}\stackrel{\text{even}}{\overbrace{\cos \alpha }})\stackrel{\text{even}}{\overbrace{\cos \alpha x}}}{i\pi \stackrel{\text{even}}{\overbrace{{\alpha }^2}}}d\alpha \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{(\text{odd}-\text{odd}\times \text{even})\times \text{even}}{\text{even}}d\alpha \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{(\text{odd}-\text{odd})\times \text{even}}{\text{even}}d\alpha \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\text{odd }\times \text{even}}{\text{even}}d\alpha \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\text{odd}}{\text{even}}d\alpha \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\text{odd }d\alpha \end{array}$$

Hence the integral vanishes. Hence (1) becomes

$$\begin{array}{}\text{(2)}& f(x)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{(\sin \alpha -\alpha \cos \alpha )i\sin \alpha x}{i\pi {\alpha }^2}d\alpha \end{array}$$

Looking at the above

$$\begin{array}{rl}\frac{(\stackrel{\text{odd}}{\overbrace{\sin \alpha }}-\stackrel{\text{odd}}{\overbrace{\alpha }}\stackrel{\text{even}}{\overbrace{\cos \alpha }})i\stackrel{\text{odd}}{\overbrace{\sin \alpha x}}}{i\pi \stackrel{\text{even}}{\overbrace{{\alpha }^2}}}& =\frac{(\text{odd}-\text{odd}\times \text{even})\times \text{odd}}{\text{even}}\\ & =\frac{\text{odd}\times \text{odd}}{\text{even}}\\ & =\frac{\text{even}}{\text{even}}\\ & =\text{even}\end{array}$$

Since the integrand is even, then ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}=2{\int }_0^{\mathrm{\infty }}$ Hence (2) becomes

$$\begin{array}{rl}\text{(3)}& f(x)& =2{\int }_0^{\mathrm{\infty }}\frac{(\sin \alpha -\alpha \cos \alpha )i\sin \alpha x}{i\pi {\alpha }^2}d\alpha & =\frac{2}{\pi }{\int }_0^{\mathrm{\infty }}\frac{(\sin \alpha -\alpha \cos \alpha )}{{\alpha }^2}\sin \alpha xd\alpha \end{array}$$

comparing this to equation (A1) above, we see that

$$f_s(x)=f(x)$$

Which is what we are asked to show.

3.13.3 chapter 15, problem 4.20

Problem Find the fourier sin transform of the given $f\left(x\right)$ and write $f(x)$ as a fourier integral. Verify the answer is the same as the exponential fourier transform.

$$f(x)=\{\begin{array}{lll}\sin x& & \left|x\right|<\frac{\pi }{2}\\ & & \\ 0,& & \left|x\right|>\frac{\pi }{2}\end{array}$$

Solution

Let ${\mathcal{F}}_{s}f(x)$ be the Fourier sin transform of $f(x)$ defined as $g_s\left(\alpha \right)={\mathcal{F}}_{s}f(x)=\sqrt{\frac{2}{\pi }}{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}f(x)\sin \left(\alpha x\right)dx$. Hence, for the function above we get

$$g_s\left(\alpha \right)=\sqrt{\frac{2}{\pi }}{\displaystyle \underset{0}{\overset{1}{\int }}}\sin x\sin \left(\alpha x\right)dx$$

Notice, we integrate from zero, not from -1, since the sin transform is defined only for positive $x$. Since $\sin \beta \sin \gamma =\frac{1}{2}\cos (\beta -\gamma )-\frac{1}{2}\cos (\beta +\gamma )$ Then $\sin x\sin \left(\alpha x\right)=\frac{1}{2}\cos (x-\alpha x)-\frac{1}{2}\cos (x+\alpha x)$. Hence

$$\begin{array}{rl}{g}_s\left(\alpha \right)& =\sqrt{\frac{2}{\pi }}{\displaystyle \underset{0}{\overset{1}{\int }}}\frac{1}{2}\cos (x-\alpha x)-\frac{1}{2}\cos (x+\alpha x)dx\\ & =\frac{1}{2}\sqrt{\frac{2}{\pi }}\{{\left[\frac{\sin (x-\alpha x)}{1-\alpha }\right]}_0^1-{\left[\frac{\sin (x+\alpha x)}{1+\alpha }\right]}_0^1\}\\ & =\frac{1}{2}\sqrt{\frac{2}{\pi }}\{[\frac{\sin (1-\alpha )}{1-\alpha }-0]-{[\frac{\sin (1+\alpha )}{1+\alpha }-0]}_0^1\}\\ & =\frac{1}{2}\sqrt{\frac{2}{\pi }}(\frac{\sin (1-\alpha )}{1-\alpha }-\frac{\sin (1+\alpha )}{1+\alpha })\end{array}$$

Hence the Sin Fourier transform of $f(x)$ is

$$g_s\left(\alpha \right)=\frac{1}{2}\sqrt{\frac{2}{\pi }}(\frac{\sin (1-\alpha )}{1-\alpha }-\frac{\sin (1+\alpha )}{1+\alpha })$$

Therefore, to obtain $f(x)$ given its sin fourier transform $g_s\left(\alpha \right)$, we apply the inverse sin fourier transform

$$\begin{array}{rl}{f}_s(x)& =\sqrt{\frac{2}{\pi }}{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}{g}_s\left(\alpha \right)\sin \alpha xd\alpha \\ & =\sqrt{\frac{2}{\pi }}{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}\frac{1}{2}\sqrt{\frac{2}{\pi }}(\frac{\sin (1-\alpha )}{1-\alpha }-\frac{\sin (1+\alpha )}{1+\alpha })\sin \alpha xd\alpha \\ \text{(1)}& & =\frac{1}{\pi }{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}(\frac{\sin (1-\alpha )}{1-\alpha }-\frac{\sin (1+\alpha )}{1+\alpha })\sin \alpha xd\alpha \end{array}$$

Now we need to show that the above is the same as the exponential inverse fourier transform found for problem 12. The exponential IFT for problem 12 is

$$\begin{array}{}\text{(2)}& f(x)=\frac{1}{\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}\frac{\alpha i\cos \left(\alpha \frac{\pi }{2}\right)}{{\alpha }^2-1}{e}^{i\alpha x}d\alpha \end{array}$$

So Need to show that (1) and (2) are the same. Need to convert the above (1) to $f_s(x)$ in (2)$.$Since $e^{i\alpha x}=\cos \alpha x+i\sin \alpha x$, (2) can be written as

$$\begin{array}{rl}f(x)& =\frac{1}{\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\alpha i\cos \left(\alpha \frac{\pi }{2}\right)}{{\alpha }^2-1}(\cos \alpha x+i\sin \alpha x)d\alpha \\ \text{(3)}& & =\frac{1}{\pi }[{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\alpha i\cos \left(\alpha \frac{\pi }{2}\right)}{{\alpha }^2-1}\cos \alpha xd\alpha +{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\alpha i\cos \left(\alpha \frac{\pi }{2}\right)}{{\alpha }^2-1}i\sin \alpha xd\alpha ]\end{array}$$

Looking at the first integral,

$$\begin{array}{rl}& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\stackrel{\text{odd}}{\overbrace{\alpha }}i\stackrel{\text{even}}{\overbrace{\cos \left(\alpha \frac{\pi }{2}\right)}}}{\stackrel{\text{even}}{\overbrace{{\alpha }^2}}-1}\stackrel{\text{even}}{\overbrace{\cos \alpha x}}d\alpha \\ & \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{(\text{odd}\times \text{even})\times \text{even}}{\text{even}}d\alpha \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\text{odd}\times \text{even}}{\text{even}}d\alpha \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\text{odd}}{\text{even}}d\alpha \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\text{odd}d\alpha \end{array}$$

Hence the integral vanishes. So (3) becomes

$$\begin{array}{rl}f(x)& =0+\frac{1}{\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\alpha i\cos \left(\alpha \frac{\pi }{2}\right)}{{\alpha }^2-1}i\sin \alpha xd\alpha \\ & =\frac{1}{\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\alpha i\cos \left(\alpha \frac{\pi }{2}\right)}{{\alpha }^2-1}i\sin \alpha xd\alpha \end{array}$$

Looking at the above integrand,

$$\begin{array}{rl}\frac{\stackrel{\text{odd}}{\overbrace{\alpha }}i\stackrel{\text{even}}{\overbrace{\cos \left(\alpha \frac{\pi }{2}\right)}}}{\stackrel{\text{even}}{\overbrace{{\alpha }^2}}-1}i\stackrel{\text{odd}}{\overbrace{\sin \alpha x}}& =\frac{(\text{odd}\times \text{even})\times \text{odd}}{\text{even}}\\ & =\frac{\text{odd}\times \text{odd}}{\text{even}}\\ & =\frac{\text{even}}{\text{even}}\\ & =\text{even}\end{array}$$

Since the integrand is even, then ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}=2{\int }_0^{\mathrm{\infty }}$. Hence (2) becomes

$$\begin{array}{rl}f(x)& =\frac{2}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\alpha i\cos \left(\alpha \frac{\pi }{2}\right)}{{\alpha }^2-1}i\sin \alpha xd\alpha \\ & =\frac{-2}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\alpha \cos \left(\alpha \frac{\pi }{2}\right)}{{\alpha }^2-1}\sin \alpha xd\alpha \\ & =\frac{2}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\alpha \cos \left(\alpha \frac{\pi }{2}\right)}{1-{\alpha }^2}\sin \alpha xd\alpha \end{array}$$

But $1-{\alpha }^2=(1+\alpha )(1-\alpha )$, therefore

$$f(x)=\frac{2}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\alpha \cos \left(\alpha \frac{\pi }{2}\right)}{(1+\alpha )(1-\alpha )}\sin \alpha xd\alpha$$

But $\cos \left(\alpha \frac{\pi }{2}\right)\sin \alpha x=\frac{1}{2}(-\sin (\alpha \frac{\pi }{2}-\alpha x)+\sin (\alpha \frac{\pi }{2}+\alpha x))$, hence the above becomes

$$\begin{array}{rl}f(x)& =\frac{2}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\alpha \frac{1}{2}(-\sin (\alpha \frac{\pi }{2}-\alpha x)+\sin (\alpha \frac{\pi }{2}+\alpha x))}{(1+\alpha )(1-\alpha )}d\alpha \\ f(x)& =\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\alpha (-\sin (\alpha \frac{\pi }{2}-\alpha x)+\sin (\alpha \frac{\pi }{2}+\alpha x))}{(1+\alpha )(1-\alpha )}d\alpha \\ f(x)& =\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\alpha \sin (\alpha \frac{\pi }{2}+\alpha x)}{(1+\alpha )(1-\alpha )}-\frac{\alpha \sin (\alpha \frac{\pi }{2}-\alpha x)}{(1+\alpha )(1-\alpha )}d\alpha \end{array}$$

3.13.4 chapter 15, problem 4.21

Problem Find the fourier transform of the given $f(x)$ $=e^{\frac{-x^2}{2{\sigma }^2}}$

Solution

Let $ϝ\left(\alpha \right)$ be the Fourier sin transform of $f(x)$ defined as
$$ϝ\left(\alpha \right)=\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}f(x)e^{-i\alpha x}dx$$

So, for the function above we get

$$\begin{array}{rl}ϝ\left(\alpha \right)& =\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{e}^{\frac{-x^2}{2{\sigma }^2}}{e}^{-i\alpha x}dx\\ & =\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{e}^{\frac{-x^2}{2{\sigma }^2}-i\alpha x}dx\\ & =\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{e}^{\frac{-x^2-i\alpha x\left(2{\sigma }^2\right)}{2{\sigma }^2}}dx\\ \text{(1)}& & =\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{e}^{\frac{-x^2-2i\alpha {\sigma }^{2}x}{2{\sigma }^2}}dx\end{array}$$

looking at the exponent $\frac{-x^2-2i\alpha {\sigma }^{2}x}{2{\sigma }^2}$. completing the square in $x$ gives

$$x^2+2i\alpha {\sigma }^{2}x={(x+Z)}^2-Y$$

Solving for $Z,Y$ gives

$$x^2+2i\alpha {\sigma }^{2}x=x^2+2xZ+Z^2-Y$$

Therefore $Z=i\alpha {\sigma }^2,Z^2-Y=0$, and $Y=-{\alpha }^2{\sigma }^4$. Hence Exponent can be written as

$$\begin{array}{rl}\frac{x^2+2i\alpha {\sigma }^{2}x}{-2{\sigma }^2}& =\frac{{(x+i\alpha {\sigma }^2)}^2-(-{\alpha }^2{\sigma }^4)}{-2{\sigma }^2}\\ & =\frac{{(x+i\alpha {\sigma }^2)}^2+{\alpha }^2{\sigma }^4}{-2{\sigma }^2}\\ & =\frac{{(x+i\alpha {\sigma }^2)}^2}{-2{\sigma }^2}-\frac{{\alpha }^2{\sigma }^4}{2{\sigma }^2}\\ & =\frac{{(x+i\alpha {\sigma }^2)}^2}{-2{\sigma }^2}-\frac{{\alpha }^2{\sigma }^2}{2}\end{array}$$

The integral (1) becomes

$$\begin{array}{rl}ϝ\left(\alpha \right)& =\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{e}^{\frac{{(x+i\alpha {\sigma }^2)}^2}{-2{\sigma }^2}-\frac{{\alpha }^2{\sigma }^2}{2}}dx\\ & =\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{e}^{\frac{{(x+i\alpha {\sigma }^2)}^2}{-2{\sigma }^2}}{e}^{-\frac{{\alpha }^2{\sigma }^2}{2}}dx\end{array}$$

Moving $e^{-\frac{{\alpha }^2{\sigma }^2}{2}}$ outside the integral because it does not depend on $x$ gives

$$ϝ\left(\alpha \right)=\frac{e^{-\frac{{\alpha }^2{\sigma }^2}{2}}}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{e}^{\frac{{(x+i\alpha {\sigma }^2)}^2}{-2{\sigma }^2}}dx$$

Let $y=x+i\alpha {\sigma }^2$, $dy=dx$ and the limits do not change. Hence we get

$$ϝ\left(\alpha \right)=\frac{e^{-\frac{{\alpha }^2{\sigma }^2}{2}}}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{e}^{\frac{y{}^2}{-2{\sigma }^2}}dy$$

Since the exponential function is raised to a square power, then we can write ${\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}{e}^{y^2}=2{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}{e}^{y^2}$ (since even function). Hence above integral becomes

$$ϝ\left(\alpha \right)=\frac{e^{-\frac{{\alpha }^2{\sigma }^2}{2}}}{\pi }{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}{e}^{\frac{y{}^2}{-2{\sigma }^2}}dy$$

Let $\zeta =\frac{y}{\sqrt{2}\sigma }$ $,$ then $y=\sqrt{2}\sigma \zeta$, and $y^2=2{\sigma }^2{\zeta }^2$. Hence $d\zeta =\frac{1}{\sqrt{2}\sigma }dy$ and the above integral becomes

$$\begin{array}{rl}ϝ\left(\alpha \right)& =\frac{e^{-\frac{{\alpha }^2{\sigma }^2}{2}}}{\pi }{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}{e}^{-{\zeta }^2}\sqrt{2}\sigma d\zeta \\ \text{(2)}& ϝ\left(\alpha \right)& =\sqrt{2}\sigma \frac{e^{-\frac{{\alpha }^2{\sigma }^2}{2}}}{\pi }{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}{e}^{-{\zeta }^2}d\zeta \end{array}$$

Now from equation 9.5 on page 468

$$\begin{array}{rl}\frac{2}{\sqrt{\pi }}{\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}{e}^{-{\zeta }^2}d\zeta & =1\\ {\displaystyle \underset{0}{\overset{\mathrm{\infty }}{\int }}}{e}^{-{\zeta }^2}d\zeta & =\frac{\sqrt{\pi }}{2}\end{array}$$

Hence (2) becomes

$$\begin{array}{rl}ϝ\left(\alpha \right)& =\sqrt{2}\sigma \frac{e^{-\frac{{\alpha }^2{\sigma }^2}{2}}}{\pi }\left(\frac{\sqrt{\pi }}{2}\right)\\ ϝ\left(\alpha \right)& =\frac{\sigma }{\sqrt{2\pi }}{e}^{-\frac{{\alpha }^2{\sigma }^2}{2}}\end{array}$$

Which is what we are asked to show.

3.13.5 chapter 15, problem 4.23

Problem Show that

${\int }_0^{\mathrm{\infty }}\frac{1-\cos \pi \alpha }{\alpha }\sin \alpha d\alpha =\frac{\pi }{2}$ and ${\int }_0^{\mathrm{\infty }}\frac{1-\cos \pi \alpha }{\alpha }\sin \pi \alpha d\alpha =\frac{\pi }{4}$

Solution

From problem 17, the Fourier sin transform for $f(x)$ shown in problem 3 is

$$g_s\left(\alpha \right)=\frac{\sqrt{2}(1-\cos \pi \alpha )}{\sqrt{\pi }\alpha }$$

From equation 4.14 page 651, $f(x)$ can be obtained from inverse sin transform is

$$\begin{array}{rl}{f}_s(x)& =\sqrt{\frac{2}{\pi }}{\int }_0^{\mathrm{\infty }}{g}_s\left(\alpha \right)\sin \alpha xd\alpha \\ & =\sqrt{\frac{2}{\pi }}{\int }_0^{\mathrm{\infty }}\left(\frac{\sqrt{2}(1-\cos \pi \alpha )}{\sqrt{\pi }\alpha }\right)\sin \alpha xd\alpha \\ \text{(1)}& & =\frac{2}{\pi }{\int }_0^{\mathrm{\infty }}\frac{(1-\cos \pi \alpha )}{\alpha }\sin \alpha xd\alpha \end{array}$$

Now, from the definition of $f(x)$, which is

$$f(x)=\{\begin{array}{lll}-1& & -\pi <x<0\\ 1& & 0<x<\pi \\ 0,& & \left|x\right|>\pi \end{array}$$

We see that for $x=1$, $f(x)=1$, hence substitute in (1) we get

$$\begin{array}{rl}1& =-\frac{2}{\pi }{\int }_0^{\mathrm{\infty }}\frac{(1-\cos \pi \alpha )}{\alpha }\sin \alpha d\alpha \\ \frac{\pi }{2}& ={\int }_0^{\mathrm{\infty }}\frac{(1-\cos \pi \alpha )}{\alpha }\sin \alpha d\alpha \end{array}$$

Which is the first result we required to show. For the second result, let $x=\pi$ hence $f\left(\pi \right)=$average value of $f(x)$ at $x=\pi$.  Which is given by $\frac{f\left({\pi }_{-}\right)+f\left({\pi }_{+}\right)}{2}=\frac{1+0}{2}=\frac{1}{2}$. Hence substitute in (1) we get

$$\begin{array}{rl}{f}_s(x=\pi )& =\frac{1}{2}=\frac{2}{\pi }{\int }_0^{\mathrm{\infty }}\frac{(1-\cos \pi \alpha )}{\alpha }\sin \alpha \pi d\alpha \\ \frac{\pi }{4}& ={\int }_0^{\mathrm{\infty }}\frac{(1-\cos \pi \alpha )}{\alpha }\sin \alpha \pi d\alpha \end{array}$$

Which is the second result we are asked to show.

3.13.6 chapter 15, problem 4.25

Problem Show that

(a) represent as an exponential fourier transform the function

$$f(x)=\{\begin{array}{lll}\sin x& & 0<x<\pi \\ & & \\ 0,& & \text{otherwise}\end{array}$$

(b) Show that the result can be written as

$$f(x)=\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\cos \alpha x+\cos \alpha (x-\pi )}{1-{\alpha }^2}d\alpha$$

Solution

The exponential Fourier transform is defined as

$$g\left(\alpha \right)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f\left(x\right)e^{-i\alpha x}dx$$

Applying the function $f(x)$ gives

$$g\left(\alpha \right)=\frac{1}{2\pi }{\int }_0^{\pi }\sin x{e}^{-i\alpha x}dx$$

But

$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$

Hence the transform can be written as

$$\begin{array}{rl}g\left(\alpha \right)& =\frac{1}{2\pi }{\int }_0^{\pi }\frac{e^{ix}-e^{-ix}}{2i}{e}^{-i\alpha x}dx\\ & =\frac{1}{4i\pi }{\int }_0^{\pi }{e}^{ix-i\alpha x}-e^{-ix-i\alpha x}dx\\ & =\frac{1}{4i\pi }{\int }_0^{\pi }{e}^{x(i-i\alpha )}-e^{x(-i-i\alpha )}dx\\ & =\frac{1}{4i\pi }({\left[\frac{e^{x(i-i\alpha )}}{i-i\alpha }\right]}_0^{\pi }-{\left[\frac{e^{x(-i-i\alpha )}}{-i-i\alpha }\right]}_0^{\pi })\\ & =\frac{1}{4i\pi }([\frac{e^{\pi (i-i\alpha )}}{i-i\alpha }-\frac{1}{i-i\alpha }]-[\frac{e^{\pi (-i-i\alpha )}}{-i-i\alpha }-\frac{1}{-i-i\alpha }])\\ & =\frac{1}{4i\pi }(\frac{1}{i-i\alpha }[e^{\pi (i-i\alpha )}-1]-\frac{1}{-i-i\alpha }[e^{\pi (-i-i\alpha )}-1])\\ & =\frac{1}{4i\pi }(\frac{1}{i-i\alpha }[e^{\pi (i-i\alpha )}-1]+\frac{1}{i+i\alpha }[e^{\pi (-i-i\alpha )}-1])\\ & =\frac{1}{i}\frac{1}{4i\pi }(\frac{1}{1-\alpha }[e^{\pi (i-i\alpha )}-1]+\frac{1}{1+\alpha }[e^{\pi (-i-i\alpha )}-1])\\ & =\frac{-1}{4\pi }(\frac{e^{\pi (i-i\alpha )}}{1-\alpha }-\frac{1}{1-\alpha }+\frac{e^{\pi (-i-i\alpha )}}{1+\alpha }-\frac{1}{1+\alpha })\\ & =\frac{-1}{4\pi }(\frac{(1+\alpha )e^{\pi (i-i\alpha )}+(1-\alpha )e^{\pi (-i-i\alpha )}}{(1-\alpha )(1+\alpha )}-\frac{(1+\alpha )+(1-\alpha )}{(1-\alpha )(1+\alpha )})\\ & =\frac{-1}{4\pi }(\frac{(1+\alpha )e^{\pi (i-i\alpha )}+(1-\alpha )e^{\pi (-i-i\alpha )}}{(1-{\alpha }^2)}-\frac{2}{(1-{\alpha }^2)})\\ & =\frac{-1}{(1-{\alpha }^2)4\pi }(e^{\pi (i-i\alpha )}+\alpha e^{\pi (i-i\alpha )}+e^{\pi (-i-i\alpha )}-\alpha e^{\pi (-i-i\alpha )}-2)\\ & =\frac{-1}{(1-{\alpha }^2)4\pi }(e^{\pi i}{e}^{-i\pi \alpha }+\alpha e^{\pi i}{e}^{-i\pi \alpha }+e^{-\pi i}{e}^{-i\pi \alpha }-\alpha e^{-\pi i}{e}^{-i\pi \alpha }-2)\end{array}$$

But $e^{\pi i}=-1$ and $e^{-\pi i}=-1$

$$\begin{array}{rl}g\left(\alpha \right)& =\frac{-1}{(1-{\alpha }^2)4\pi }(-e^{-i\pi \alpha }-\overbrace{\alpha e^{-i\pi \alpha }}-e^{-i\pi \alpha }+\overbrace{\alpha e^{-i\pi \alpha }}-2)\\ g\left(\alpha \right)& =\frac{-1}{(1-{\alpha }^2)4\pi }(-e^{-i\pi \alpha }-e^{-i\pi \alpha }-2)\\ g\left(\alpha \right)& =\frac{1}{(1-{\alpha }^2)4\pi }(e^{-i\pi \alpha }+e^{-i\pi \alpha }+2)\\ g\left(\alpha \right)& =\frac{2e^{-i\pi \alpha }+2}{(1-{\alpha }^2)4\pi }\end{array}$$

Hence the exponential fourier transform is

$$g\left(\alpha \right)=\frac{e^{-i\pi \alpha }+1}{(1-{\alpha }^2)2\pi }$$

Therefore $f(x)$ can be rewritten as

$$\begin{array}{rl}f(x)& ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}g\left(\alpha \right)e^{i\alpha x}d\alpha \\ & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{-i\pi \alpha }+1}{(1-{\alpha }^2)2\pi }{e}^{i\alpha x}d\alpha \\ \text{(1)}& & =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1+e^{-i\pi \alpha }}{(1-{\alpha }^2)}{e}^{i\alpha x}d\alpha \end{array}$$

Which is the answer required to show.

Part(b)

Now need to show that the above can be written as $$f(x)=\frac{1}{\pi }{\int }_0^{\mathrm{\infty }}\frac{\cos \alpha x+\cos \alpha (x-\pi )}{1-{\alpha }^2}d\alpha$$

From (1)

$$\begin{array}{rl}f(x)& =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1+e^{-i\pi \alpha }}{(1-{\alpha }^2)}{e}^{i\alpha x}d\alpha \\ & =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\alpha x}+e^{i\alpha x}{e}^{-i\pi \alpha }}{(1-{\alpha }^2)}d\alpha \\ & =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\alpha x}+e^{i\alpha x-i\pi \alpha }}{(1-{\alpha }^2)}d\alpha \\ & =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{e^{i\alpha x}+e^{i\alpha (x-\pi )}}{(1-{\alpha }^2)}d\alpha \\ & =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{2(\cos \alpha x+\cos \alpha (x-\pi ))}{(1-{\alpha }^2)}d\alpha \\ & =\frac{1}{\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{\cos \alpha x+\cos \alpha (x-\pi )}{(1-{\alpha }^2)}d\alpha \end{array}$$

Which is what is required to show.

3.13.7 chapter 15, problem 4.3

Problem Find the exponential fourier transform of the given $f\left(x\right)$ and write $f(x)$ as a fourier integral.$$f(x)=\{\begin{array}{lll}-1,& & -\pi <x<0\\ 1,& & 0<x<\pi \\ 0,& & \left|x\right|>\pi \end{array}$$ Solution

Let $ϝ\left(\alpha \right)$ be the Fourier transform of $f\left(x\right)$ defined as $ϝ\left(\alpha \right)=\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}f(x)e^{-i\alpha x}dx$, hence$$\begin{array}{rl}ϝ\left(\alpha \right)& =\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}f(x)e^{-i\alpha x}dx\\ 2\pi ϝ\left(\alpha \right)& ={\displaystyle \underset{-\pi }{\overset{0}{\int }}}-e^{-i\alpha x}dx+{\displaystyle \underset{0}{\overset{\pi }{\int }}}{e}^{-i\alpha x}dx\\ & =-{\left[\frac{e^{-i\alpha x}}{-i\alpha }\right]}_{-\pi }^0+{\left[\frac{e^{-i\alpha x}}{-i\alpha }\right]}_0^{\pi }\\ & =\frac{1}{i\alpha }{\left[e^{-i\alpha x}\right]}_{-\pi }^0-\frac{1}{i\alpha }{\left[e^{-i\alpha x}\right]}_0^{\pi }\\ & =\frac{1}{i\alpha }[e^0-e^{i\alpha \pi }]-\frac{1}{i\alpha }{[e^{-i\alpha \pi }-e^0]}_0^{\pi }\\ & =\frac{1}{i\alpha }[1-e^{i\alpha \pi }]-\frac{1}{i\alpha }{[e^{-i\alpha \pi }-1]}_0^{\pi }\\ & =\frac{1}{i\alpha }-\frac{e^{i\alpha \pi }}{i\alpha }-\frac{e^{-i\alpha \pi }}{i\alpha }+\frac{1}{i\alpha }\\ & =\frac{2}{i\alpha }-\frac{1}{i\alpha }(e^{i\alpha \pi }+e^{-i\alpha \pi })\end{array}$$

But $e^{i\alpha \pi }+e^{-i\alpha \pi }=2\cos \alpha \pi$. Hence$$\begin{array}{rl}2\pi ϝ\left(\alpha \right)& =\frac{2}{i\alpha }-\frac{1}{i\alpha }(2\cos \alpha \pi )\\ & =\frac{2}{i\alpha }(1-\cos \alpha \pi )\end{array}$$

Hence the Fourier transform of $f(x)$ is$$\begin{array}{rl}ϝ\left(\alpha \right)& =\frac{1}{2\pi }\left[\frac{2}{i\alpha }(1-\cos \alpha \pi )\right]\\ & =\frac{1}{\pi \alpha i}(1-\cos \alpha \pi )\end{array}$$

To obtain $f(x)$ given its fourier transform $ϝ\left(\alpha \right),$ then we apply the inverse fourier transform$$\begin{array}{rl}f(x)& ={\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}ϝ\left(\alpha \right)e^{i\alpha x}d\alpha \\ & ={\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}\frac{1}{\pi \alpha i}(1-\cos \alpha \pi )e^{i\alpha x}d\alpha \\ & =\frac{1}{\pi i}{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}\frac{1}{\alpha }(1-\cos \alpha \pi )e^{i\alpha x}d\alpha \end{array}$$

3.13.8 chapter 15, problem 4.5

Problem Find the exponential fourier transform of the given $f\left(x\right)$ and write $f(x)$ as a fourier integral.$$f(x)=\{\begin{array}{lll}1,& & 0<x<1\\ & & \\ 0,& & \text{otherwise}\end{array}$$ Solution

Let $ϝ(s)$ be the Fourier transform of $f\left(x\right)$ defined as $ϝ(s)=\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}f(x)e^{-i\alpha x}dx$$$\begin{array}{rl}ϝ(s)& =\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}f(x)e^{-isx}dx\\ 2\pi ϝ(s)& ={\displaystyle \underset{0}{\overset{1}{\int }}}{e}^{-isx}dx\\ & ={\left[\frac{e^{-isx}}{-is}\right]}_0^1\\ & =\frac{-1}{is}{\left[e^{-isx}\right]}_0^1\\ & =\frac{-1}{is}[e^{-is}-e^0]\\ & =\frac{-1}{is}[e^{-is}-1]\end{array}$$

Hence the Fourier transform of $f(x)$ is$$\begin{array}{rl}ϝ(s)& =\frac{1}{2\pi }\left[\frac{-1}{is}[e^{-is}-1]\right]\\ & =\frac{i}{2\pi s}(e^{-is}-1)\end{array}$$

To obtain $f(x)$ given its fourier transform $ϝ\left(s\right),$ then we apply the inverse fourier transform$$\begin{array}{rl}f(x)& ={\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}ϝ(s)e^{isx}ds\\ & ={\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}\frac{i}{2\pi s}(e^{-is}-1)e^{isx}ds\\ & =\frac{i}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}\frac{1}{s}(e^{-is}-1)e^{isx}ds\end{array}$$

3.13.9 chapter 15, problem 4.7

Problem Find the exponential fourier transform of the given $f\left(x\right)$ and write $f(x)$ as a fourier integral.

$$f(x)=\{\begin{array}{lll}\left|x\right|& & \left|x\right|<1\\ & & \\ 0,& & \left|x\right|>1\end{array}$$

Solution

Let $ϝ\left(\alpha \right)$ be the Fourier transform of $f\left(x\right)$ defined as $ϝ\left(\alpha \right)=\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}f(x)e^{-i\alpha x}dx$

$$\begin{array}{rl}ϝ\left(\alpha \right)& =\frac{1}{2\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}f(x)e^{-i\alpha x}dx\\ 2\pi ϝ\left(\alpha \right)& ={\displaystyle \underset{-1}{\overset{0}{\int }}}-x{e}^{-i\alpha x}dx+{\displaystyle \underset{0}{\overset{1}{\int }}}x{e}^{-i\alpha x}dx\end{array}$$

Integrating by parts, $u=x,v=\frac{e^{-i\alpha x}}{-i\alpha }$, hence $\int udv=$ $uv-\int duv$. The first integral is

$$\begin{array}{rl}{\displaystyle \underset{-1}{\overset{0}{\int }}}-x{e}^{-i\alpha x}dx& ={\left[(-x)\frac{e^{-i\alpha x}}{-i\alpha }\right]}_{-1}^0+{\displaystyle \underset{-1}{\overset{0}{\int }}}\frac{e^{-i\alpha x}}{-i\alpha }dx\\ & ={[(x)\frac{e^{-i\alpha x}}{i\alpha }]}_{-1}^0+{\displaystyle \underset{-1}{\overset{0}{\int }}}\frac{e^{-i\alpha x}}{-i\alpha }dx\\ & =\frac{1}{i\alpha }[0-(-1)\times e^{i\alpha }]-\frac{1}{i\alpha }{\displaystyle \underset{-1}{\overset{0}{\int }}}{e}^{-i\alpha x}dx\\ & =\frac{1}{i\alpha }\left[e^{i\alpha }\right]-\frac{1}{i\alpha }{\left[\frac{e^{-i\alpha x}}{-i\alpha }\right]}_{-1}^0\\ & =\frac{1}{i\alpha }\left[e^{i\alpha }\right]+\frac{1}{i^2{\alpha }^2}{\left[e^{-i\alpha x}\right]}_{-1}^0\\ & =\frac{1}{i\alpha }\left[e^{i\alpha }\right]-\frac{1}{{\alpha }^2}[1-e^{i\alpha }]\\ & =\frac{e^{i\alpha }}{i\alpha }-\frac{1}{{\alpha }^2}+\frac{e^{i\alpha }}{{\alpha }^2}\end{array}$$

And the second integral

$$\begin{array}{rl}{\displaystyle \underset{0}{\overset{1}{\int }}}x{e}^{-i\alpha x}dx& ={\left[x\frac{e^{-i\alpha x}}{-i\alpha }\right]}_0^1-{\displaystyle \underset{0}{\overset{1}{\int }}}\frac{e^{-i\alpha x}}{-i\alpha }dx\\ & =\frac{1}{-i\alpha }[1\times e^{-i\alpha }-0]+\frac{1}{i\alpha }{\displaystyle \underset{0}{\overset{1}{\int }}}{e}^{-i\alpha x}dx\\ & =\frac{1}{-i\alpha }\left[e^{-i\alpha }\right]+\frac{1}{i\alpha }{\left[\frac{e^{-i\alpha x}}{-i\alpha }\right]}_0^1\\ & =\frac{1}{-i\alpha }\left[e^{-i\alpha }\right]-\frac{1}{i^2{\alpha }^2}{\left[e^{-i\alpha x}\right]}_0^1\\ & =\frac{1}{-i\alpha }\left[e^{-i\alpha }\right]+\frac{1}{{\alpha }^2}[e^{-i\alpha }-1]\\ & =\frac{e^{-i\alpha }}{-i\alpha }+\frac{e^{-i\alpha }}{{\alpha }^2}-\frac{1}{{\alpha }^2}\end{array}$$

Hence

$$\begin{array}{rl}2\pi ϝ\left(\alpha \right)& =(\frac{e^{i\alpha }}{i\alpha }-\frac{1}{{\alpha }^2}+\frac{e^{i\alpha }}{{\alpha }^2})+(\frac{e^{-i\alpha }}{-i\alpha }+\frac{e^{-i\alpha }}{{\alpha }^2}-\frac{1}{{\alpha }^2})\\ 2\pi ϝ\left(\alpha \right)& =\frac{e^{i\alpha }}{i\alpha }-\frac{1}{{\alpha }^2}+\frac{e^{i\alpha }}{{\alpha }^2}-\frac{e^{-i\alpha }}{i\alpha }+\frac{e^{-i\alpha }}{{\alpha }^2}-\frac{1}{{\alpha }^2}\\ & =\frac{e^{i\alpha }}{i\alpha }+\frac{e^{i\alpha }}{{\alpha }^2}-\frac{e^{-i\alpha }}{i\alpha }+\frac{e^{-i\alpha }}{{\alpha }^2}-\frac{2}{{\alpha }^2}\\ & =\frac{1}{\alpha }(\frac{e^{i\alpha }}{i}-\frac{e^{-i\alpha }}{i})+\frac{1}{{\alpha }^2}(e^{i\alpha }+e^{-i\alpha })-\frac{2}{{\alpha }^2}\\ & =\frac{1}{\alpha }\left(\frac{e^{i\alpha }-e^{-i\alpha }}{i}\right)+\frac{1}{{\alpha }^2}(e^{i\alpha }+e^{-i\alpha })-\frac{2}{{\alpha }^2}\end{array}$$

But $e^{i\alpha }+e^{-i\alpha }=2\cos \alpha$ and $\frac{e^{i\alpha }-e^{-i\alpha }}{i}=2\sin \alpha$, Hence the above becomes

$$\begin{array}{rl}2\pi ϝ\left(\alpha \right)& =\frac{1}{\alpha }(2\sin \alpha )+\frac{1}{{\alpha }^2}(2\cos \alpha )-\frac{2}{{\alpha }^2}\\ & =\frac{2}{{\alpha }^2}[\alpha \sin \alpha +\cos \alpha -1]\\ ϝ\left(\alpha \right)& =\frac{1}{\pi {\alpha }^2}[\alpha \sin \alpha +\cos \alpha -1]\end{array}$$

Hence the Fourier transform of $f(x)$ is

$$ϝ\left(\alpha \right)=\frac{1}{\pi {\alpha }^2}[\alpha \sin \alpha +\cos \alpha -1]$$

To obtain $f(x)$ given its fourier transform $ϝ\left(\alpha \right),$ then we apply the inverse fourier transform

$$\begin{array}{rl}f(x)& ={\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}ϝ\left(\alpha \right)e^{i\alpha x}d\alpha \\ & ={\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}\frac{1}{\pi {\alpha }^2}[\alpha \sin \alpha +\cos \alpha -1]e^{i\alpha x}d\alpha \\ & =\frac{1}{\pi }{\displaystyle \underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}}\frac{1}{{\alpha }^2}[\alpha \sin \alpha +\cos \alpha -1]e^{i\alpha x}d\alpha \end{array}$$

3.13.10 chapter 15, problem 5.1

Problem Show that $g(t)⊛h\left(t\right)=h(t)⊛g(t)$

Solution

By definition,

$$\begin{array}{}\text{(1)}& g(t)⊛h(t)={\int }_0^{t}g(t-\tau )h\left(\tau \right)d\tau \end{array}$$

Let $u=t-\tau$, $du=-d\tau$, when $\tau =0$, $u=t$, when $\tau =t$, $u=0$, Hence The RHS becomes

$$\begin{array}{rl}{\int }_0^{t}g(t-\tau )h\left(\tau \right)d\tau & ={\int }_{u=t}^{u=0}g(u)h(t-u)(-du)\\ & =-{\int }_{u=t}^{u=0}g(u)h(t-u)du\\ & ={\int }_{u=0}^{u=t}g(u)h(t-u)du\end{array}$$

Since $u$ is a dummy variable of integration, call it anything we want, say $\tau$ so above integral becomes

$$\begin{array}{rl}{\int }_0^{t}g(t-\tau )h\left(\tau \right)d\tau & ={\int }_0^{t}g\left(\tau \right)h(t-\tau )d\tau \\ \text{(2)}& {\int }_0^{t}g(t-\tau )h\left(\tau \right)d\tau & ={\int }_0^{t}h(t-\tau )g\left(\tau \right)d\tau \end{array}$$

Hence from (2) $g(t)⊛h(t)=h\left(t\right)⊛g(t)$

3.13.11 chapter 15, problem 5.10

Problem

Use convolution integral to find the inverse transform of $\frac{1}{p{(p^2+a^2)}^2}$

Solution

$$\frac{1}{p{(p^2+a^2)}^2}=\frac{1}{p}\frac{1}{{(p^2+a^2)}^2}=GH$$

From Tables using L1 and L17 $g(t)=1$ and $h\left(t\right)=\frac{\sin at-at\cos at}{2a^3}$. Hence the inverse transform of $GH=$ $g(t)⊛h(t)$. Using L34

$$\begin{array}{}\text{(L34)}& g(t)⊛h(t)={\int }_0^{t}g(t-\tau )h\left(\tau \right)d\tau \end{array}$$

Hence

$$\begin{array}{rl}g(t)⊛h(t)& ={\int }_0^{t}1\times \frac{\sin a(t-\tau )-a(t-\tau )\cos a(t-\tau )}{2a^3}d\tau \\ & =\frac{1}{2a^3}{\int }_0^t\sin a(t-\tau )-a(t-\tau )\cos a(t-\tau )d\tau \\ \text{(1)}& & =\frac{1}{2a^3}[{\int }_0^t\sin a(t-\tau )d\tau -at{\int }_0^t\cos a(t-\tau )d\tau +a{\int }_0^t\tau \cos a(t-\tau )d\tau ]\end{array}$$

The last integral can be integrated by parts. $u=\tau$, $v=\frac{\sin a(t-\tau )}{-a}$

$$\begin{array}{rl}{\int }_0^t\tau \cos a(t-\tau )d\tau & ={\left[\tau \frac{\sin a(t-\tau )}{-a}\right]}_0^t-{\int }_0^t\frac{\sin a(t-\tau )}{-a}d\tau \\ & =\frac{-1}{a}{[\tau \sin a(t-\tau )]}_0^t+\frac{1}{a}{\int }_0^t\sin a(t-\tau )d\tau \\ & =\frac{-1}{a}{[\tau \sin a(t-\tau )]}_0^t+\frac{1}{a}{\left[\frac{-\cos a(t-\tau )}{-a}\right]}_0^t\\ & =\frac{-1}{a}{[\tau \sin a(t-\tau )]}_0^t+\frac{1}{a^2}{[\cos a(t-\tau )]}_0^t\\ & =\frac{-1}{a}[t\sin a(t-t)-0]+\frac{1}{a^2}[\cos a(t-t)-\cos a(t-0)]\\ & =\frac{-1}{a}\left[0\right]+\frac{1}{a^2}[\cos a\left(0\right)-\cos a(t)]\\ & =\frac{1}{a^2}[1-\cos at]\end{array}$$

Hence (1) becomes

$$\begin{array}{rl}g(t)⊛h(t)& =\frac{1}{2a^3}[{\int }_0^t\sin a(t-\tau )d\tau -at{\int }_0^t\cos a(t-\tau )d\tau +a\frac{1}{a^2}[1-\cos at]]\\ g(t)⊛h(t)& =\frac{1}{2a^3}[{\left[\frac{-\cos a(t-\tau )}{-a\tau }\right]}_0^t-at\left[\frac{\sin a(t-\tau )}{-a\tau }\right]+\frac{1}{a}[1-\cos at]]\\ & =\frac{1}{2a^3}[\frac{1}{a}{[\cos a(t-\tau )]}_0^t+t{[\sin a(t-\tau )]}_0^t+\frac{1}{a}[1-\cos at]]\\ & =\frac{1}{2a^3}[\frac{1}{a}[\cos a(t-t)-\cos a(t-0)]+t{[\sin a(t-t)-\sin a(t-0)]}_0^t+\frac{1}{a}[1-\cos at]]\\ & =\frac{1}{2a^3}[\frac{1}{a}[1-\cos at]+t[-\sin at]+\frac{1}{a}[1-\cos at]]\\ & =\frac{1}{2a^3}[\frac{1}{a}[1-\cos at]-t\sin at+\frac{1}{a}[1-\cos at]]\\ & =\frac{1}{2a^4}(2-2\cos at-at\sin at)\end{array}$$

So the inverse Laplace transform of $\frac{1}{p{(p^2+a^2)}^2}$ is $$\frac{1}{2a^4}(2-2\cos at-at\sin at)$$

3.13.12 chapter 15, problem 5.2

Problem Use L34 and L2 to find the inverse transform of $G\left(p\right)H(p)$ when $G(p)=\frac{1}{(p+a)}$ and $H(p)=\frac{1}{(p+b)}$ your result should be L7

Solution

$$\begin{array}{}\text{(L2)}& \mathcal{L}\left(e^{-at}\right)=\frac{1}{p+a}\end{array}$$

$$\begin{array}{}\text{(L34)}& g(t)⊛h(t)={\int }_0^{t}g(t-\tau )h\left(\tau \right)d\tau \end{array}$$

Using L2, $g(t)={\mathcal{L}}^{-1}\frac{1}{(p+a)}=e^{-at}$, and $h(t)={\mathcal{L}}^{-1}\frac{1}{(p+b)}=e^{-bt}$. Now Let $Y(p)=G(p)H(p)$, But

$$G(p)H(p)=\mathcal{L}\{g(t)⊛h(t)\}$$

Then

$$\begin{array}{rl}y(t)& =g(t)⊛h(t)={\int }_0^{t}g(t-\tau )h\left(\tau \right)d\tau \\ & ={\int }_0^t{e}^{-a(t-\tau )}{e}^{-b\tau }d\tau \\ & ={\int }_0^t{e}^{-at+a\tau }{e}^{-b\tau }d\tau \\ & ={\int }_0^t{e}^{-at}{e}^{a\tau }{e}^{-b\tau }d\tau \end{array}$$

$e^{-at}$ can be moved outside the integral

$$\begin{array}{rl}y(t)& =e^{-at}{\int }_0^t{e}^{a\tau }{e}^{-b\tau }d\tau \\ y(t)& =e^{-at}{\int }_0^t{e}^{a\tau -b\tau }d\tau \\ y(t)& =e^{-at}{\int }_0^t{e}^{\tau (a-b)}d\tau \\ y(t)& =e^{-at}{\left[\frac{e^{\tau (a-b)}}{a-b}\right]}_0^t\\ y(t)& =\frac{e^{-at}}{a-b}[e^{t(a-b)}-1]\\ y(t)& =\frac{e^{-at+t(a-b)}-e^{-at}}{a-b}\\ y(t)& =\frac{e^{-bt}-e^{-at}}{a-b}\\ y(t)& =\frac{e^{-at}-e^{-bt}}{b-a}\end{array}$$

Which is L7 as required to show.

3.13.13 chapter 15, problem 5.22

Problem

Verify Parseval’s theorem for $f(x)=e^{-\left|x\right|}$ and $g\left(\alpha \right)=$Fourier transform of $f(x)$

Solution

Parseval theorem says that total energy in a signal equal to the sum of the energies in the harmonics that make up the signal. i.e.

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left|f\left(x\right)\right|}^{2}dx$$

$$\begin{array}{rl}\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{|f(x)|}^{2}dx& =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left|e^{-\left|x\right|}\right|}^{2}dx\\ & =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-2\left|x\right|}dx\\ & =\frac{1}{2\pi }\{{\int }_{-\mathrm{\infty }}^0{e}^{2x}dx+{\int }_0^{\mathrm{\infty }}{e}^{-2x}dx\}\\ & =\frac{1}{2\pi }\{{\left[\frac{e^{2x}}{2}\right]}_{-\mathrm{\infty }}^0+{\left[\frac{e^{-2x}}{-2}\right]}_0^{\mathrm{\infty }}\}\\ & =\frac{1}{4\pi }\{{\left[e^{2x}\right]}_{-\mathrm{\infty }}^0-{\left[e^{-2x}\right]}_0^{\mathrm{\infty }}\}\\ & =\frac{1}{4\pi }\{[e^0-0]-[0-e^0]\}\\ & =\frac{1}{4\pi }\{1+1\}\\ \text{(1)}& & =\frac{1}{2\pi }\end{array}$$

Now we find the Fourier transform for $f(x)$

$$\begin{array}{rl}g\left(\alpha \right)& =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-\left|x\right|}{e}^{-ix\alpha }dx\\ & =\frac{1}{2\pi }[{\int }_{-\mathrm{\infty }}^0{e}^x{e}^{-ix\alpha }dx+{\int }_0^{\mathrm{\infty }}{e}^{-x}{e}^{-ix\alpha }dx]\\ & =\frac{1}{2\pi }[{\int }_{-\mathrm{\infty }}^0{e}^{x(1-i\alpha )}dx+{\int }_0^{\mathrm{\infty }}{e}^{x(-1-i\alpha )}dx]\\ & =\frac{1}{2\pi }({\left[\frac{e^{x(1-i\alpha )}}{1-i\alpha }\right]}_{-\mathrm{\infty }}^0+{\left[\frac{e^{x(-1-i\alpha )}}{-1-i\alpha }\right]}_0^{\mathrm{\infty }})\\ & =\frac{1}{2\pi }(\frac{1}{1-i\alpha }{\left[e^{x(1-i\alpha )}\right]}_{-\mathrm{\infty }}^0-\frac{1}{1+i\alpha }{\left[e^{x(-1-i\alpha )}\right]}_0^{\mathrm{\infty }})\\ & =\frac{1}{2\pi }(\frac{1}{1-i\alpha }{\left[e^{x(1-i\alpha )}\right]}_{-\mathrm{\infty }}^0-\frac{1}{1+i\alpha }{\left[e^{x(-1-i\alpha )}\right]}_0^{\mathrm{\infty }})\\ & =\frac{1}{2\pi }(\frac{1}{1-i\alpha }[1-e^{-\mathrm{\infty }(1-i\alpha )}]-\frac{1}{1+i\alpha }[e^{\mathrm{\infty }(-1-i\alpha )}-1])\\ & =\frac{1}{2\pi }(\frac{1}{1-i\alpha }\left[1\right]-\frac{1}{1+i\alpha }[-1])\\ & =\frac{1}{2\pi }(\frac{1}{1-i\alpha }+\frac{1}{1+i\alpha })\\ & =\frac{1}{2\pi }\left(\frac{1+i\alpha +1-i\alpha }{(1-i\alpha )(1+i\alpha )}\right)\\ & =\frac{1}{2\pi }\left(\frac{2}{1+{\alpha }^2}\right)\\ & =\frac{1}{\pi (1+{\alpha }^2)}\end{array}$$

So

$$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha & ={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left|\frac{1}{\pi (1+{\alpha }^2)}\right|}^{2}d\alpha \\ & =\frac{1}{{\pi }^2}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{{(1+{\alpha }^2)}^2}d\alpha \end{array}$$

But ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{{(1+{\alpha }^2)}^2}d\alpha =\frac{\pi }{2}$, Hence

$$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha & =\frac{1}{{\pi }^2}\left(\frac{\pi }{2}\right)\\ \text{(2)}& & =\frac{1}{2\pi }\end{array}$$

Comparing (1) and (2). They are the same. Hence$$$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left|g\left(\alpha \right)\right|}^{2}d\alpha =\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left|f\left(x\right)\right|}^{2}dx$$ was verified for this problem as required.

3.13.14 chapter 15, problem 5.4

Problem

Use convolution integral to find the inverse transform of $\frac{1}{(p+a){(p+b)}^2}$

Solution

$$\frac{1}{(p+a){(p+b)}^2}=\frac{1}{(p+a)}\frac{1}{{(p+b)}^2}$$

From but from L6

$$\begin{array}{}\text{(L2)}& \mathcal{L}\left(t^k{e}^{-at}\right)=\frac{k!}{{(p+a)}^{k+1}}\end{array}$$

Hence $\frac{1}{(p+a)}=\mathcal{L}\left(e^{-at}\right)$ and $\frac{1}{{(p+b)}^2}=\mathcal{L}\left(e^{-bt}\right)$. Hence the inverse transform of $\frac{1}{(p+a)}\frac{1}{{(p+b)}^2}=e^{-at}⊛t{e}^{-bt}$. Using L34

$$\begin{array}{}\text{(L34)}& g(t)⊛h(t)={\int }_0^{t}g(t-\tau )h\left(\tau \right)d\tau \end{array}$$

Hence

$$\begin{array}{rl}{e}^{-at}⊛t{e}^{-bt}& ={\int }_0^t{e}^{-a(t-\tau )}\tau e^{-b\tau }d\tau \\ & ={\int }_0^t{e}^{-at+a\tau }\tau e^{-b\tau }d\tau \\ & ={\int }_0^t{e}^{-at}{e}^{a\tau }\tau e^{-b\tau }d\tau \\ & =e^{-at}{\int }_0^t{e}^{a\tau }\tau e^{-b\tau }d\tau \\ & =e^{-at}{\int }_0^t\tau e^{\tau (a-b)}d\tau \end{array}$$

Integrate by parts. $u=\tau$, $v=\frac{e^{\tau (a-b)}}{a-b}$

$$\begin{array}{rl}{e}^{-at}⊛t{e}^{-bt}& =e^{-at}\{{\left[\tau \frac{e^{\tau (a-b)}}{a-b}\right]}_0^t-{\int }_0^t\frac{e^{\tau (a-b)}}{a-b}d\tau \}\\ & =e^{-at}\{{\left[\tau \frac{e^{\tau (a-b)}}{a-b}\right]}_0^t-\frac{1}{a-b}{\left[\frac{e^{\tau (a-b)}}{a-b}\right]}_0^t\}\\ & =e^{-at}(\frac{1}{a-b}{\left[\tau e^{\tau (a-b)}\right]}_0^t-\frac{1}{{(a-b)}^2}{\left[e^{\tau (a-b)}\right]}_0^t)\\ & =e^{-at}(\frac{1}{a-b}\left[t{e}^{t(a-b)}\right]-\frac{1}{{(a-b)}^2}[e^{t(a-b)}-1])\\ & =e^{-at}(\frac{t{e}^{ta-tb}}{a-b}-\frac{e^{ta-tb}-1}{{(a-b)}^2})\\ & =\frac{t{e}^{-tb}}{a-b}-\frac{e^{-tb}-e^{-at}}{{(a-b)}^2}\\ & =\frac{(a-b)t{e}^{-tb}-e^{-tb}+e^{-at}}{{(a-b)}^2}\\ & =\frac{((a-b)t-1)e^{-tb}+e^{-at}}{{(a-b)}^2}\end{array}$$

So the inverse laplace transform of $\frac{1}{(p+a){(p+b)}^2}$ is $$\frac{((a-b)t-1)e^{-tb}+e^{-at}}{{(a-b)}^2}$$

3.13.15 chapter 15, problem 6.2

Problem

Find the inverse laplace transform using 6.6 of the function $\frac{1}{p^4-1}$

Solution

6.6 states that $f(t)$ = sum of all residues of $F\left(z\right)e^{zt}$ at all poles. Poles of $F(z)=\frac{1}{z^4-1}$are at $\pm 1,\pm i$. Hence

$$F(z)=\frac{e^{zt}}{(z-1)(z+1)(z-i)(z+i)}$$

Since each pole is of order 1, we use equation 6.1 page 599 which says

$$\text{Residue of }F(z)\text{at }z=z_0\text{is }\underset{z\to z_0}{lim}(z-z_0)F(z)$$

Hence sum of residue is

$$\begin{array}{rl}R& =\underset{z\to +1}{lim}\frac{e^{zt}}{(z+1)(z-i)(z+i)}+\underset{z\to -1}{lim}\frac{e^{zt}}{(z-1)(z-i)(z+i)}\\ & +\underset{z\to +i}{lim}\frac{e^{zt}}{(z-1)(z+1)(z+i)}+\underset{z\to -i}{lim}\frac{e^{zt}}{(z-1)(z+1)(z-i)}\\ & \\ & =\frac{e^t}{(1+1)(1-i)(1+i)}+\frac{e^{-t}}{(-1-1)(-1-i)(-1+i)}\\ & +\frac{e^{it}}{(i-1)(i+1)(i+i)}+\frac{e^{-it}}{(-i-1)(-i+1)(-i-i)}\\ & \\ & =\frac{e^t}{4}+\frac{e^{-t}}{-4}+\frac{e^{it}}{-4i}+\frac{e^{-it}}{4i}\\ & =\left(\frac{e^t-e^{-t}}{4}\right)-\frac{1}{2}\left(\frac{e^{it}-e^{-it}}{2i}\right)\\ & =\left(\frac{e^t-e^{-t}}{4}\right)-\frac{1}{2}(\sin t)\end{array}$$

3.13.16 chapter 15, problem 6.4

Problem

Find the inverse laplace transform using 6.6 of the function $\frac{p^3}{p^4-16}$

Solution

6.6 states that $f(t)$ = sum of all residues of $F\left(z\right)e^{zt}$ at all poles. Poles of $F(z)=\frac{z^3}{z^4-16}$are at $\pm 2,\pm 2i$, Hence

$$F(z)=\frac{z^3{e}^{zt}}{(z-2)(z+2)(z-2i)(z+2i)}$$

Since each pole is of order 1, we use equation 6.1 page 599 which says

$$\text{Residue of }F(z)\text{at }z=z_0\text{is }\underset{z\to z_0}{lim}(z-z_0)F(z)$$

Hence sum of residue is

$$\begin{array}{rl}R& =\underset{z\to +2}{lim}\frac{z^3{e}^{zt}}{(z+2)(z-2i)(z+2i)}+\underset{z\to -2}{lim}\frac{z^3{e}^{zt}}{(z-2)(z-2i)(z+2i)}\\ & +\underset{z\to +2i}{lim}\frac{z^3{e}^{zt}}{(z-2)(z+2)(z+2i)}+\underset{z\to -2i}{lim}\frac{z^3{e}^{zt}}{(z-2)(z+2)(z-2i)}\\ & \\ & \\ & =\frac{8e^{2t}}{(2+2)(2-2i)(2+2i)}+\frac{-8e^{-2t}}{(-2-2)(-2-2i)(-2+2i)}\\ & +\frac{{\left(2i\right)}^3{e}^{2it}}{(2i-2)(2i+2)(2i+2i)}+\frac{{(-2i)}^3{e}^{-2it}}{(-2i-2)(-2i+2)(-2i-2i)}\\ & \\ & \\ & =\frac{8e^{2t}}{(4)8}+\frac{-8e^{-2t}}{(-4)8}+\frac{(-8i)e^{2it}}{-8\left(4i\right)}+\frac{\left(8i\right)e^{-2it}}{-8(-4i)}\\ & =\frac{e^{2t}}{4}+\frac{e^{-2t}}{4}+\frac{e^{2it}}{4}+\frac{e^{-2it}}{4}\\ & =\frac{e^{2t}+e^{-2t}}{4}+\frac{1}{2}(\cos 2t)\end{array}$$

So inverse Laplace transform of $\frac{p^3}{p^4-16}$ is $$\frac{e^{2t}+e^{-2t}}{4}+\frac{1}{2}(\cos 2t)$$

3.13.17 chapter 15, problem 6.5

Problem

Find the inverse laplace transform using 6.6 of the function $\frac{3p^2}{p^3+8}$

Solution

6.6 states that $f(t)$ = sum of all residues of $F\left(z\right)e^{zt}$ at all poles. To find poles, look at $p^3+8=0,$ hence $p^3=-8$, $p=-8^{\frac{1}{3}}$. Let

$$8^{\frac{1}{3}}=r{e}^{i\theta }$$

Then the roots are

$$\begin{array}{rl}& =8^{{}^{\frac{1}{3}}}{e}^{\frac{i0}{3}},8^{{}^{\frac{1}{3}}}{e}^{\frac{i(0+2\pi )}{3}},8^{{}^{\frac{1}{3}}}{e}^{\frac{i(0+4\pi )}{3}}\\ & =8^{{}^{\frac{1}{3}}},8^{{}^{\frac{1}{3}}}(\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}),8^{{}^{\frac{1}{3}}}(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3})\\ & =2,2(\cos {120}^0+i\sin {120}^0),2(\cos {240}^0+i\sin {240}^0)\\ & =2,2(-\frac{1}{2}+i\frac{\sqrt{3}}{2}),2(-\frac{1}{2}+i\frac{-\sqrt{3}}{2})\\ & =2,-1+i\sqrt{3},-1-i\sqrt{3}\end{array}$$

Hence

$$p=-2,1-i\sqrt{3},1+i\sqrt{3}$$

And

$$\begin{array}{rl}F(z)& =\frac{3z^2}{z^3+8}{e}^{zt}\\ & =\frac{3z^2}{(z+2)(z-(1-i\sqrt{3}))(z-(1+i\sqrt{3}))}{e}^{zt}\end{array}$$

Since each pole is of order 1, we use equation 6.1 page 599 which says

$$\text{Residue of }F(z)\text{at }z=z_0\text{is }\underset{z\to z_0}{lim}(z-z_0)F(z)$$

Hence sum of residue is

$$\begin{array}{rl}R& =\underset{z\to -2}{lim}\frac{3z^2}{(z-(1-i\sqrt{3}))(z-(1+i\sqrt{3}))}{e}^{zt}+\underset{z\to (1-i\sqrt{3})}{lim}\frac{3z^2}{(z+2)(z-(1+i\sqrt{3}))}{e}^{zt}\\ & +\underset{z\to (1+i\sqrt{3})}{lim}\frac{3z^2}{(z+2)(z-(1-i\sqrt{3}))}{e}^{zt}\\ & =\frac{3{(-2)}^2}{(-2-(1-i\sqrt{3}))(-2-(1+i\sqrt{3}))}{e}^{-2t}+\frac{3{(1-i\sqrt{3})}^2}{((1-i\sqrt{3})+2)((1-i\sqrt{3})-(1+i\sqrt{3}))}{e}^{(1-i\sqrt{3})t}\\ & +\frac{3{(1+i\sqrt{3})}^2}{((1+i\sqrt{3})+2)((1+i\sqrt{3})-(1-i\sqrt{3}))}{e}^{(1+i\sqrt{3})t}\\ & =\frac{12}{(-3+i\sqrt{3})(-3-i\sqrt{3})}{e}^{-2t}+\frac{3(-2-2i\sqrt{3})}{(3-i\sqrt{3})(-2i\sqrt{3})}{e}^{(1-i\sqrt{3})t}+\frac{3(-2+2i\sqrt{3})}{(3+i\sqrt{3})\left(2i\sqrt{3}\right)}{e}^{(1+i\sqrt{3})t}\\ & =\frac{12}{12}{e}^{-2t}+\frac{-6-6i\sqrt{3}}{-6i\sqrt{3}-6}{e}^{(1-i\sqrt{3})t}+\frac{-6+6i\sqrt{3}}{6i\sqrt{3}-6}{e}^{(1+i\sqrt{3})t}\\ & =e^{-2t}+e^{(1-i\sqrt{3})t}+e^{(1+i\sqrt{3})t}\\ & =e^{-2t}+e^t{e}^{-i\sqrt{3}t}+e^t{e}^{i\sqrt{3}t}\\ & =e^{-2t}+e^t(e^{i\sqrt{3}t}+e^{-i\sqrt{3}t})\\ & =e^{-2t}+e^t(2\cos \sqrt{3}t)\end{array}$$

So inverse Laplace transform of $\frac{3p^2}{p^3+8}$ is $$e^{-2t}+e^t(2\cos \sqrt{3}t)$$

3.13.18 chapter 15, problem 6.9

Problem

Find the inverse laplace transform using 6.6 of the function $\frac{p}{p^4-1}$

Solution

6.6 states that $f(t)$ = sum of all residues of $F\left(z\right)e^{zt}$ at all poles. To find poles, look at $p^4-1=0,$ hence $p^4=1$, $p=1^{\frac{1}{4}}$. Let

$$1^{\frac{1}{4}}=r{e}^{i\theta }$$

Then roots are

$$\begin{array}{rl}& =1^{{}^{\frac{1}{4}}}{e}^{\frac{i0}{4}},1^{{}^{\frac{1}{4}}}{e}^{\frac{i(0+2\pi )}{4}},1^{{}^{\frac{1}{4}}}{e}^{\frac{i(0+4\pi )}{4}}{,}^{{}^{\frac{1}{4}}}{e}^{\frac{i(0+6\pi )}{4}}\\ & =1,1(\cos \frac{2\pi }{4}+i\sin \frac{2\pi }{4}),1(\cos \frac{4\pi }{4}+i\sin \frac{4\pi }{4}),1(\cos \frac{6\pi }{4}+i\sin \frac{6\pi }{4})\\ & =1,(0+i),(-1+i0),(0-i1)\\ & =1,i,-1,-i\end{array}$$

Therefore

$$p=1,i,-1,-i$$

Hence

$$\begin{array}{rl}F(z)& =\frac{z}{z^4-1}{e}^{zt}\\ & =\frac{z}{(z-1)(z-i)(z+1)(z+i)}{e}^{zt}\end{array}$$

Since each pole is of order 1, we use equation 6.1 page 599 which says

$$\text{Residue of }F(z)\text{at }z=z_0\text{is }\underset{z\to z_0}{lim}(z-z_0)F(z)$$

Hence sum of residue is

$$\begin{array}{rl}R& =\underset{z\to 1}{lim}\frac{z}{(z-i)(z+1)(z+i)}{e}^{zt}+\underset{z\to i}{lim}\frac{z}{(z-1)(z+1)(z+i)}{e}^{zt}\\ & +\underset{z\to -1}{lim}\frac{z}{(z-1)(z-i)(z+i)}{e}^{zt}+\underset{z\to -i}{lim}\frac{z}{(z-1)(z-i)(z+1)}{e}^{zt}\\ & \\ & =\frac{1}{(1-i)(1+1)(1+i)}{e}^t+\frac{i}{(i-1)(i+1)(i+i)}{e}^{it}\\ & +\frac{-1}{(-1-1)(-1-i)(-1+i)}{e}^{-t}+\frac{-i}{(-i-1)(-i-i)(-i+1)}{e}^{-it}\\ & \\ & =\frac{1}{4}{e}^t+\frac{i}{-4i}{e}^{it}+\frac{1}{4}{e}^{-t}+\frac{-i}{4i}{e}^{-it}\\ & =\frac{1}{4}(e^t+e^{-t})-\frac{1}{2}(\cos t)\end{array}$$

So inverse Laplace transform of $\frac{p}{p^4-1}$ is $$\frac{1}{4}(e^t+e^{-t})-\frac{1}{2}(\cos t)$$

3.13.19 chapter 15, problem 7.11

Problem

Using the $\delta$ function method, Find the response of the following system to a unit impulse. $\frac{d^{4}y}{d{y}^4}-y=\delta (t-t_0)$

Solution

Taking the laplace transform of each side gives (assuming initial conditions for the system are at rest)

$$\begin{array}{rl}Y{p}^4-Y& =e^{-p{t}_0}\\ Y& =\frac{e^{-p{t}_0}}{p^4-1}\\ Y& =\frac{e^{-p{t}_0}}{(p^2-1)(p^2+1)}\end{array}$$

Finding the inverse laplace of $\frac{1}{(p-1)(p+1)(p^2+1)}=\frac{1}{(p-1)(p+1)}\frac{1}{(p^2+1)}=GH$. Then $g(t)=\frac{e^t-e^{-t}}{2}$using L7 and$,$ $h(t)=\sin t$ using L3. Hence the inverse transform is $$

$$\begin{array}{rl}g(t)⊛h(t)& ={\int }_0^t\frac{e^{\tau }-e^{-\tau }}{2}\sin (t-\tau )d\tau \\ & =\frac{1}{2}(\sinh t-\sin t)\end{array}$$

Using L28 with the result above we get

$$y(t)=\{\begin{array}{lll}\frac{1}{2}(\sinh (t-t_0)-\sin (t-t_0))& & t>t_0\\ 0& & t<t_0\end{array}$$

Or by expressing $\sinh$ using $\exp$, the above becomes

$$y(t)=\{\begin{array}{lll}\frac{1}{4}(e^{t-t_0}-e^{-t+t_0}-2\sin (t-t_0))& & t>t_0\\ 0& & t<t_0\end{array}$$

3.13.20 chapter 15, problem 7.7

Problem

Using the $\delta$ function method, Find the response of the following system to a unit impulse. $y^{\prime \prime }+2y^{\prime }+y=\delta (t-t_0)$

Solution

Take the laplace transform of each side we get (assume initial conditions for a system at rest)

$$\begin{array}{rl}Y{p}^2+2Yp+Y& =e^{-p{t}_0}\\ Y& =\frac{e^{-p{t}_0}}{p^2+2p+1}\\ Y& =\frac{e^{-p{t}_0}}{{(p+1)}^2}\end{array}$$

Using L28 and L6 (for $k=1$)

$$y(t)=\{\begin{array}{lll}(t-t_0)e^{-(t-t_0)}& & t>t_0\\ 0& & t<t_0\end{array}$$

3.13.21 chapter 15, problem 7.9

Problem

Using the $\delta$ function method, Find the response of the following system to a unit impulse. $y^{\prime \prime }+2y^{\prime }+10y=\delta (t-t_0)$

Solution

Taking the laplace transform of each side we get (assume initial conditions for a system at rest)$$\begin{array}{rl}Y{p}^2+2Yp+10Y& =e^{-p{t}_0}\\ Y& =\frac{e^{-p{t}_0}}{p^2+2p+10}\\ Y& =\frac{e^{-p{t}_0}}{(p-a)(p-b)}\end{array}$$

Where $a=-1+3i,b=-1-3i$ the roots of $p^2+2p+10$. Using L28 and L7$$y(t)=\{\begin{array}{lll}\frac{e^{a(t-t_0)}-e^{b(t-t_0)}}{(-b)-(-a)}& & t>t_0\\ 0& & t<t_0\end{array}$$ Replacing values for $a,b$ gives$$\begin{array}{rl}y(t)& =\frac{e^{a(t-t_0)}-e^{b(t-t_0)}}{a-b}\\ & =\frac{e^{(-1+3i)(t-t_0)}-e^{(-1-3i)(t-t_0)}}{(-1+3i)-(-1-3i)}\\ & =\frac{e^{(-1+3i)(t-t_0)}-e^{(-1-3i)(t-t_0)}}{6i}\\ & =\frac{e^{-t+t_0+3it-3i{t}_0}-e^{-t+t_0-3it+3i{t}_0}}{6i}\\ & =e^{-t+t_0}\frac{e^{3i(t-t_0)}-e^{-3i(t-t_0)}}{6i}\\ & =e^{-t+t_0}\left(\frac{1}{3}\right)\left(\frac{e^{3i(t-t_0)}-e^{-3i(t-t_0)}}{2i}\right)\\ & =\frac{e^{-t+t_0}}{3}\sin 3(t-t_0)\end{array}$$

Therefore$$y(t)=\{\begin{array}{lll}\frac{e^{-t+t_0}}{3}\sin 3(t-t_0)& & t>t_0\\ 0& & t<t_0\end{array}$$