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2.7 HW 7

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2.7.1 Section 45, Problem 4

Figure 2.75:Problem statement

Solution

Solve for $y\left ( x,t\right )$ in

$$\begin{equation} y_{tt}=y_{xx}-2\beta y_{t}\qquad \left ( t>0,0<x<\pi \right ) \tag{1} \end{equation}$$ Boundary conditions $$\begin{align*} y\left ( 0,t\right ) & =0\\ y\left ( \pi ,t\right ) & =0 \end{align*}$$

Initial conditions

$$\begin{align*} y\left ( x,0\right ) & =f\left ( x\right ) \\ y_{t}\left ( x,0\right ) & =0 \end{align*}$$

Let $y=XT$. Substituting in (1) gives

$$T^{\prime \prime }X=X^{\prime \prime }T-2\beta T^{\prime }X$$ Dividing by $XT\neq 0$ $$\begin{align*} \frac{T^{\prime \prime }}{T} & =\frac{X^{\prime \prime }}{X}-2\beta \frac{T^{\prime }}{T}\\ \frac{T^{\prime \prime }}{T}+2\beta \frac{T^{\prime }}{T} & =\frac{X^{\prime \prime }}{X}=-\lambda \end{align*}$$

Where $\lambda$ is separation constant. Due to nature of boundary conditions being both homogeneous, then we know $\lambda >0$ is only possible case from earlier HW’s. The eigenvalue problem is

$$X^{\prime \prime }+\lambda X=0$$ Which we know has eigenvalues $\lambda =n^{2}$ for $n=1,2,\cdots$ with corresponding eigenfunctions $$\begin{equation} X_{n}=\sin \left ( nx\right ) \tag{1} \end{equation}$$ Now we solve the time ODE using these eigenvalues. $$\begin{align*} \frac{T^{\prime \prime }}{T}+2\beta \frac{T^{\prime }}{T} & =-n^{2}\\ T^{\prime \prime }+2\beta T^{\prime }+n^{2}T & =0 \end{align*}$$

This is standard second order ODE with positive damping $\beta$ and since $n^{2}$ is positive. The characteristic equation is

$$r^{2}+2\beta r+n^{2}=0$$ The roots are $$\begin{align*} r & =-\frac{b}{2a}\pm \frac{1}{2a}\sqrt{b^{2}-4ac}\\ & =-\frac{2\beta }{2}\pm \frac{1}{2}\sqrt{4\beta ^{2}-4n^{2}}\\ & =-\beta \pm \sqrt{\beta ^{2}-n^{2}}\\ & =-\beta \pm i\sqrt{n^{2}-\beta ^{2}} \end{align*}$$

Hence the solution is

$$\begin{align*} T_{n}\left ( t\right ) & =A_{n}e^{r_{1}t}+B_{n}e^{r_{2}t}\\ & =A_{n}e^{\left ( -\beta +i\sqrt{n^{2}-\beta ^{2}}\right ) t}+B_{n}e^{\left ( -\beta -i\sqrt{n^{2}-\beta ^{2}}\right ) t}\\ & =e^{-\beta t}\left ( A_{n}e^{i\sqrt{n^{2}-\beta ^{2}}t}+B_{n}e^{-i\sqrt{n^{2}-\beta ^{2}}t}\right ) \end{align*}$$

But the above can be rewritten using Euler relation as (the constants $A_{n},B_{n}$ will be different, but kept them the same names for simplicity)

$$T_{n}\left ( t\right ) =e^{-\beta t}\left ( A_{n}\cos \left ( \sqrt{n^{2}-\beta ^{2}}t\right ) +B_{n}\sin \left ( \sqrt{n^{2}-\beta ^{2}}t\right ) \right )$$ Let $\alpha _{n}=\sqrt{n^{2}-\beta ^{2}}$, then the above becomes $$\begin{equation} T_{n}\left ( t\right ) =e^{-\beta t}\left ( A_{n}\cos \left ( \alpha _{n}t\right ) +B_{n}\sin \left ( \alpha _{n}t\right ) \right ) \tag{2} \end{equation}$$ Since the PDE is linear and homogenous, then by superposition we obtain the final solution as $$\begin{align} y\left ( x,t\right ) & =\sum _{n=1}^{\infty }X_{n}T_{n}\nonumber \\ & =\sum _{n=1}^{\infty }e^{-\beta t}\left ( A_{n}\cos \left ( \alpha _{n}t\right ) +B_{n}\sin \left ( \alpha _{n}t\right ) \right ) \sin \left ( nx\right ) \tag{3} \end{align}$$

Now initial conditions are applied to determine $A_{n},B_{n}$. At $t=0$

$$f\left ( x\right ) =\sum _{n=1}^{\infty }A_{n}\sin \left ( nx\right )$$ Hence $A_{n}$ are the Fourier sine coefficient of the representation of $f\left ( x\right )$ which implies $$\begin{equation} A_{n}=\frac{2}{\pi }\int _{0}^{\pi }f\left ( x\right ) \sin \left ( nx\right ) dx\tag{4} \end{equation}$$ Taking time derivative of (3) gives $$y_{t}\left ( x,t\right ) =\sum _{n=1}^{\infty }\left [ -\beta e^{-\beta t}\left ( A_{n}\cos \left ( \alpha _{n}t\right ) +B_{n}\sin \left ( \alpha _{n}t\right ) \right ) +e^{-\beta t}\left ( -\alpha _{n}A_{n}\sin \left ( \alpha _{n}t\right ) +\alpha _{n}B_{n}\cos \left ( \alpha _{n}t\right ) \right ) \right ] \sin \left ( nx\right )$$ At $t=0$ the above becomes (since released from rest) $$0=\sum _{n=1}^{\infty }\left ( -\beta A_{n}+\alpha _{n}B_{n}\right ) \sin \left ( nx\right )$$ Therefore $$-\beta A_{n}+\alpha _{n}B_{n}=0$$ Hence $B_{n}=\frac{\beta A_{n}}{\alpha _{n}}$. Therefore (3) becomes $$\begin{align*} y\left ( x,t\right ) & =\sum _{n=1}^{\infty }e^{-\beta t}\left ( A_{n}\cos \left ( \alpha _{n}t\right ) +\frac{\beta A_{n}}{\alpha _{n}}\sin \left ( \alpha _{n}t\right ) \right ) \sin \left ( nx\right ) \\ & =e^{-\beta t}\sum _{n=1}^{\infty }A_{n}\left ( \cos \left ( \alpha _{n}t\right ) +\frac{\beta }{\alpha _{n}}\sin \left ( \alpha _{n}t\right ) \right ) \sin \left ( nx\right ) \end{align*}$$

Where $A_{n}=\frac{2}{\pi }\int _{0}^{\pi }f\left ( x\right ) \sin \left ( nx\right ) dx$ and $\alpha _{n}=\sqrt{n^{2}-\beta ^{2}}$. Which is the result required to show (Book used $B$ in place $A$, but it is the same thing, just different name for a constant).

2.7.2 Section 46, Problem 2

Figure 2.76:Problem statement

Solution

Part a

suppose $\omega \neq a$. Let

$$\begin{equation} y_{p}=A\cos \omega t+B\sin \omega t \tag{1} \end{equation}$$ Then $$\begin{align*} y_{p}^{\prime } & =-A\omega \sin \omega t+B\omega \cos \omega t\\ y_{p}^{\prime \prime } & =-A\omega ^{2}\cos \omega t-B\omega ^{2}\sin \omega t \end{align*}$$

Substituting the above back into the given ODE gives

$$\begin{align} y_{p}^{\prime \prime }\left ( t\right ) +a^{2}y_{p}\left ( t\right ) & =b\sin \omega t\nonumber \\ \left ( -A\omega ^{2}\cos \omega t-B\omega ^{2}\sin \omega t\right ) +a^{2}\left ( A\cos \omega t+B\sin \omega t\right ) & =b\sin \omega t\nonumber \\ \cos \omega t\left ( -A\omega ^{2}+a^{2}A\right ) +\sin \omega t\left ( -B\omega ^{2}+a^{2}B\right ) & =b\sin \omega t \tag{2} \end{align}$$

By comparing coefficients, we see that

$$\begin{align*} -A\omega ^{2}+a^{2}A & =0\\ A\left ( a^{2}-\omega ^{2}\right ) & =0 \end{align*}$$

Since $\omega \neq a$ then this implies that $A=0$. And from (2), we see that

$$\begin{align*} -B\omega ^{2}+a^{2}B & =b\\ B & =\frac{b}{a^{2}-\omega ^{2}} \end{align*}$$

Therefore (1) becomes

$$\begin{equation} y_{p}=\frac{b}{a^{2}-\omega ^{2}}\sin \omega t \tag{3} \end{equation}$$ Now we need to find the complementary solution to $$y_{c}^{\prime \prime }+a^{2}y=0$$ Since $a^{2}>0$, then the solution is the standard one given by $$\begin{equation} y_{c}\left ( t\right ) =C_{1}\cos at+C_{2}\sin at \tag{4} \end{equation}$$ Adding (3,4) gives the general solution $$y\left ( t\right ) =C_{1}\cos at+C_{2}\sin at+\frac{b}{a^{2}-\omega ^{2}}\sin \omega t$$

Part (b)

Let

$$\begin{equation} y_{p}=At\cos \omega t+Bt\sin \omega t \tag{1} \end{equation}$$ Then $$\begin{align*} y_{p}^{\prime } & =A\cos \omega t-At\omega \sin \omega t+B\sin \omega t+Bt\omega \cos \omega t\\ y_{p}^{\prime \prime } & =-A\omega \sin \omega t-\left ( A\omega \sin \omega t+At\omega ^{2}\cos \omega t\right ) +B\omega \cos \omega t+\left ( B\omega \cos \omega t-Bt\omega ^{2}\sin \omega t\right ) \\ & =\left ( -At\omega ^{2}+2B\omega \right ) \cos \omega t+\left ( -2A\omega -Bt\omega ^{2}\right ) \sin \omega t \end{align*}$$

Substituting the above back into the given ODE gives

$$\begin{align} y_{p}^{\prime \prime }\left ( t\right ) +a^{2}y_{p}\left ( t\right ) & =b\sin \omega t\nonumber \\ \left ( \left ( -At\omega ^{2}+2B\omega \right ) \cos \omega t+\left ( -2A\omega -Bt\omega ^{2}\right ) \sin \omega t\right ) +a^{2}\left ( At\cos \omega t+Bt\sin \omega t\right ) & =b\sin \omega t\nonumber \\ \cos \omega t\left ( -At\omega ^{2}+2B\omega +a^{2}At\right ) +\sin \omega t\left ( -2A\omega -Bt\omega ^{2}+a^{2}Bt\right ) & =b\sin \omega t \tag{2} \end{align}$$

By comparing coefficients, we see that

$$\begin{align} -At\omega ^{2}+2B\omega +a^{2}At & =0\nonumber \\ At\left ( -\omega ^{2}+a^{2}\right ) +B\left ( 2\omega \right ) & =0 \tag{3} \end{align}$$

And from (2), we see also that

$$\begin{align} -2A\omega -Bt\omega ^{2}+a^{2}Bt & =b\nonumber \\ A\left ( -2\omega \right ) +Bt\left ( -\omega ^{2}+a^{2}\right ) & =b \tag{4} \end{align}$$

But since $\omega =a$, then (3) becomes

$$\begin{align*} B\left ( 2\omega \right ) & =0\\ B & =0 \end{align*}$$

And (4) becomes

$$\begin{align*} A\left ( -2\omega \right ) & =b\\ A & =\frac{-b}{2a} \end{align*}$$

Substituting these values we found for $A,B$, in (1) gives

$$y_{p}=\frac{-b}{2a}t\cos \omega t$$ But $\omega =a$, therefore $$\begin{equation} y_{p}=\frac{-b}{2a}t\cos at \tag{5} \end{equation}$$ The complementary solution do not change from part (a). Hence the general solution is $$y\left ( t\right ) =C_{1}\cos at+C_{2}\sin at-\frac{b}{2a}t\cos at$$ Which is the result required to show.

2.7.3 Section 46, Problem 3

Figure 2.77:Problem statement

Solution

The general solution from problem 2 is

$$y\left ( t\right ) =\left \{ \begin{array} [c]{cc}C_{1}\cos at+C_{2}\sin at+\frac{b}{a^{2}-\omega ^{2}}\sin \omega t & \omega \neq a\\ C_{1}\cos at+C_{2}\sin at-\frac{b}{2a}t\cos at & \omega =a \end{array} \right .$$ We need to find $C_{1},C_{2}$ when initial conditions are $y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =0$ for each of the above cases.

case $\omega \neq a$

$y\left ( 0\right ) =0$ gives

$$0=C_{1}$$ Hence solution now becomes $$y\left ( t\right ) =C_{2}\sin at+\frac{b}{a^{2}-\omega ^{2}}\sin \omega t$$ Taking time derivative gives $$y^{\prime }\left ( t\right ) =aC_{2}\cos at+\frac{\omega b}{a^{2}-\omega ^{2}}\cos \omega t$$ At $t=0$ the above gives $$\begin{align*} 0 & =aC_{2}+\frac{\omega b}{a^{2}-\omega ^{2}}\\ C_{2} & =\frac{1}{a}\frac{\omega b}{\omega ^{2}-a^{2}} \end{align*}$$

Using $C_{1},C_{2}$ found above, the solution becomes

$$\begin{align} y\left ( t\right ) & =\frac{1}{a}\frac{\omega b}{\omega ^{2}-a^{2}}\sin at+\frac{b}{a^{2}-\omega ^{2}}\sin \omega t\nonumber \\ & =\frac{b}{a^{2}-\omega ^{2}}\left ( \frac{\omega }{a}\sin at-\sin \omega t\right ) \tag{1} \end{align}$$

case $\omega =a$

$y\left ( 0\right ) =0$ gives

$$0=C_{1}$$ Hence solution now becomes $$y\left ( t\right ) =C_{2}\sin at-\frac{b}{2a}t\cos at$$ Taking time derivative gives $$y^{\prime }\left ( t\right ) =aC_{2}\cos at-\left ( \frac{b}{2a}\cos at-\frac{b}{2a}t^{2}\sin at\right )$$ At $t=0$ the above gives $$\begin{align*} 0 & =aC_{2}-\frac{b}{2a}\\ C_{2} & =\frac{1}{a}\frac{b}{2a} \end{align*}$$

Using $C_{1},C_{2}$ found above, the solution becomes

$$\begin{align} y\left ( t\right ) & =\frac{1}{a}\frac{b}{2a}\sin at-\frac{b}{2a}t\cos at\nonumber \\ & =\frac{b}{2a}\left ( \frac{1}{a}\sin at-t\cos at\right ) \tag{2} \end{align}$$

From (1,2) we see that

$$y\left ( t\right ) =\left \{ \begin{array} [c]{cc}\frac{b}{a^{2}-\omega ^{2}}\left ( \frac{\omega }{a}\sin at-\sin \omega t\right ) & \omega \neq a\\ \frac{b}{2a}\left ( \frac{1}{a}\sin at-t\cos at\right ) & \omega =a \end{array} \right .$$ Which is the result required to show.

2.7.4 Section 52, Problem 3

Figure 2.78:Problem statement

Solution

$$\begin{align*} f\left ( x\right ) & =\int _{0}^{\infty }\left ( A\left ( \alpha \right ) \cos \left ( \alpha x\right ) +B\left ( \alpha \right ) \sin \left ( \alpha x\right ) \right ) d\alpha \\ & =\int _{0}^{\infty }\left ( A\left ( \alpha \right ) \left ( \frac{e^{i\alpha x}+e^{-i\alpha x}}{2}\right ) -iB\left ( \alpha \right ) \left ( \frac{e^{i\alpha x}-e^{-i\alpha x}}{2}\right ) \right ) d\alpha \\ & =\int _{0}^{\infty }\left ( e^{i\alpha x}\left ( \frac{A\left ( \alpha \right ) -iB\left ( \alpha \right ) }{2}\right ) +e^{-i\alpha x}\left ( \frac{A\left ( \alpha \right ) +iB\left ( \alpha \right ) }{2}\right ) \right ) d\alpha \\ & =\int _{0}^{\infty }e^{i\alpha x}\frac{A\left ( \alpha \right ) -iB\left ( \alpha \right ) }{2}d\alpha +\int _{0}^{\infty }e^{-i\alpha x}\frac{A\left ( \alpha \right ) +iB\left ( \alpha \right ) }{2}d\alpha \\ & =\int _{0}^{\infty }e^{i\alpha x}\frac{A\left ( \alpha \right ) -iB\left ( \alpha \right ) }{2}d\alpha +\int _{0}^{\infty }e^{-i\alpha x}\frac{A\left ( \alpha \right ) +iB\left ( \alpha \right ) }{2}d\alpha \\ & =\int _{0}^{\infty }e^{i\alpha x}\frac{A\left ( \alpha \right ) -iB\left ( \alpha \right ) }{2}d\alpha +\int _{-\infty }^{0}e^{i\alpha x}\frac{A\left ( \alpha \right ) +iB\left ( \alpha \right ) }{2}d\alpha \\ & =\int _{-\infty }^{\infty }C\left ( \alpha \right ) e^{i\alpha x}d\alpha \end{align*}$$

Where

$$\begin{array} [c]{ccc}C\left ( \alpha \right ) =\frac{A\left ( \alpha \right ) -iB\left ( \alpha \right ) }{2} & , & C\left ( -\alpha \right ) =\frac{A\left ( \alpha \right ) +iB\left ( \alpha \right ) }{2}\end{array} \qquad \alpha >0$$ Expression (9) section (5) is $$\begin{align*} A\left ( \alpha \right ) & =\frac{1}{\pi }\int _{-\infty }^{\infty }f\left ( x\right ) \cos \left ( \alpha x\right ) dx\\ B\left ( \alpha \right ) & =\frac{1}{\pi }\int _{-\infty }^{\infty }f\left ( x\right ) \sin \left ( \alpha x\right ) dx \end{align*}$$

Substituting the above in $C\left ( \alpha \right ) =\frac{A\left ( \alpha \right ) -iB\left ( \alpha \right ) }{2}$ gives

$$\begin{align*} C\left ( \alpha \right ) & =\frac{1}{2}\left ( \frac{1}{\pi }\int _{-\infty }^{\infty }f\left ( x\right ) \cos \left ( \alpha x\right ) dx-i\frac{1}{\pi }\int _{-\infty }^{\infty }f\left ( x\right ) \sin \left ( \alpha x\right ) dx\right ) \\ & =\frac{1}{2\pi }\left ( \int _{-\infty }^{\infty }f\left ( x\right ) \cos \left ( \alpha x\right ) dx-\int _{-\infty }^{\infty }f\left ( x\right ) i\sin \left ( \alpha x\right ) dx\right ) \\ & =\frac{1}{2\pi }\int _{-\infty }^{\infty }f\left ( x\right ) \left ( \cos \left ( \alpha x\right ) -i\sin \left ( \alpha x\right ) \right ) dx \end{align*}$$

But using Euler relation $\cos \left ( \alpha x\right ) -i\sin \left ( \alpha x\right ) =e^{i\alpha x}$ then the above reduces to

$$C\left ( \alpha \right ) =\frac{1}{2\pi }\int _{-\infty }^{\infty }f\left ( x\right ) e^{i\alpha x}dx\qquad -\infty <\alpha <\infty$$ Which is what required to show.

2.7.5 Section 53, Problem 4

Figure 2.79:Problem statement

Solution

Since $f\left ( x\right )$ is piecewise continuous and absolutely integrable (sine function), then

$$\frac{f\left ( x^{+}\right ) +f\left ( x^{-}\right ) }{2}=\frac{1}{\pi }\int _{0}^{\infty }\left ( \int _{-\infty }^{\infty }f\left ( s\right ) \cos \left ( \alpha \left ( s-x\right ) \right ) ds\right ) d\alpha$$ Substituting for $f\left ( s\right )$ inside the integral for the function given gives $$\frac{f\left ( x^{+}\right ) +f\left ( x^{-}\right ) }{2}=\frac{1}{\pi }\int _{0}^{\infty }\left ( \int _{0}^{\pi }\sin \left ( s\right ) \cos \left ( \alpha s-\alpha x\right ) ds\right ) d\alpha$$ Where we used $\int _{0}^{\pi }$ only, since the function is zero everywhere else. Using $2\sin A\cos B=\sin \left ( A+B\right ) +\sin \left ( A-B\right )$ then the above can be written as $$\begin{align} \frac{f\left ( x^{+}\right ) +f\left ( x^{-}\right ) }{2} & =\frac{1}{\pi }\int _{0}^{\infty }\left ( \frac{1}{2}\int _{0}^{\pi }\sin \left ( s+\alpha s-\alpha x\right ) +\sin \left ( s-\left ( \alpha s-\alpha x\right ) \right ) ds\right ) d\alpha \nonumber \\ & =\frac{1}{2\pi }\int _{0}^{\infty }\left ( \int _{0}^{\pi }\sin \left ( s+\alpha s-\alpha x\right ) +\sin \left ( s-\alpha s+\alpha x\right ) ds\right ) d\alpha \tag{1} \end{align}$$

But

$$\begin{align*} \int _{0}^{\pi }\sin \left ( s+\alpha s-\alpha x\right ) ds & =\left [ \frac{-\cos \left ( s+\alpha s-\alpha x\right ) }{1+\alpha }\right ] _{0}^{\pi }\\ & =\frac{-1}{1+\alpha }\left ( \cos \left ( \pi +\alpha \pi -\alpha x\right ) -\cos \left ( -\alpha x\right ) \right ) \\ & =\frac{-1}{1+\alpha }\left ( \cos \left ( \pi +\alpha \left ( \pi -x\right ) \right ) -\cos \left ( \alpha x\right ) \right ) \end{align*}$$

But $\cos \left ( \pi +\alpha \left ( \pi -x\right ) \right ) =-\cos \left ( \alpha \left ( \pi -x\right ) \right )$, and the above becomes

$$\begin{equation} \int _{0}^{\pi }\sin \left ( s+\alpha s-\alpha x\right ) ds=\frac{1}{1+\alpha }\left ( \cos \left ( \alpha \left ( \pi -x\right ) \right ) +\cos \left ( \alpha x\right ) \right ) \tag{2} \end{equation}$$ Similarly $$\begin{align} \int _{0}^{\pi }\sin \left ( s-\alpha s+\alpha x\right ) ds & =\left [ \frac{-\cos \left ( s-\alpha s+\alpha x\right ) }{1-\alpha }\right ] _{0}^{\pi }\nonumber \\ & =\frac{-1}{1-\alpha }\left ( \cos \left ( \pi -\alpha \pi +\alpha x\right ) -\cos \left ( \alpha x\right ) \right ) \nonumber \\ & =\frac{-1}{1-\alpha }\left ( \cos \left ( \pi -\alpha \left ( \pi +x\right ) \right ) -\cos \left ( \alpha x\right ) \right ) \nonumber \\ & =\frac{-1}{1-\alpha }\left ( -\cos \left ( -\alpha \left ( \pi +x\right ) \right ) -\cos \left ( \alpha x\right ) \right ) \nonumber \\ & =\frac{1}{1-\alpha }\left ( \cos \left ( \alpha \left ( \pi +x\right ) \right ) +\cos \left ( \alpha x\right ) \right ) \tag{3} \end{align}$$

Substituting (2,3) back in (1) gives

$$\begin{align*} \frac{f\left ( x^{+}\right ) +f\left ( x^{-}\right ) }{2} & =\frac{1}{2\pi }\int _{0}^{\infty }\left ( \frac{1}{1+\alpha }\left ( \cos \left ( \alpha \left ( \pi -x\right ) \right ) +\cos \left ( \alpha x\right ) \right ) +\frac{1}{1-\alpha }\left ( \cos \left ( \alpha \left ( \pi +x\right ) \right ) +\cos \left ( \alpha x\right ) \right ) \right ) d\alpha \\ & =\frac{1}{2\pi }\int _{0}^{\infty }\left ( \cos \left ( \alpha \left ( \pi -x\right ) \right ) \left ( \frac{1}{1+\alpha }+\frac{1}{1-\alpha }\right ) +\cos \left ( \alpha x\right ) \left ( \frac{1}{1+\alpha }+\frac{1}{1-\alpha }\right ) \right ) d\alpha \\ & =\frac{1}{2\pi }\int _{0}^{\infty }\left ( \cos \left ( \alpha \left ( \pi -x\right ) \right ) \left ( \frac{2}{1-\alpha ^{2}}\right ) +\cos \left ( \alpha x\right ) \left ( \frac{2}{1-\alpha ^{2}}\right ) \right ) d\alpha \\ & =\frac{1}{\pi }\int _{0}^{\infty }\frac{\cos \left ( \alpha \left ( \pi -x\right ) \right ) +\cos \left ( \alpha x\right ) }{1-\alpha ^{2}}d\alpha \end{align*}$$

But $f\left ( x\right )$ is continuous then $\frac{f\left ( x^{+}\right ) +f\left ( x^{-}\right ) }{2}=f\left ( x\right )$ and the above becomes

$$f\left ( x\right ) =\frac{1}{\pi }\int _{0}^{\infty }\frac{\cos \left ( \alpha \left ( \pi -x\right ) \right ) +\cos \left ( \alpha x\right ) }{1-\alpha ^{2}}d\alpha$$ When $x=\frac{\pi }{2}$ the above gives $$f\left ( \frac{\pi }{2}\right ) =\frac{1}{\pi }\int _{0}^{\infty }\frac{\cos \left ( \alpha \left ( \pi -\frac{\pi }{2}\right ) \right ) +\cos \left ( \alpha \frac{\pi }{2}\right ) }{1-\alpha ^{2}}d\alpha$$ But $f\left ( \frac{\pi }{2}\right ) =\sin \left ( \frac{\pi }{2}\right ) =1$, hence $$\begin{align*} 1 & =\frac{1}{\pi }\int _{0}^{\infty }\frac{\cos \left ( \alpha \frac{\pi }{2}\right ) +\cos \left ( \alpha \frac{\pi }{2}\right ) }{1-\alpha ^{2}}d\alpha \\ & =\frac{1}{\pi }\int _{0}^{\infty }\frac{2\cos \left ( \alpha \frac{\pi }{2}\right ) }{1-\alpha ^{2}}d\alpha \end{align*}$$

Therefore

$$\frac{\pi }{2}=\int _{0}^{\infty }\frac{\cos \left ( \alpha \frac{\pi }{2}\right ) }{1-\alpha ^{2}}d\alpha$$