Using series expansion evaluate the integral \(I=\int _{0}^{1}\ln \left ( \frac{1+x}{1-x}\right ) \frac{dx}{x}\)
Solution
We first need to find the Taylor series for \(\ln \left ( \frac{1+x}{1-x}\right ) \) expanded around \(x=0\). Since \begin{align} \ln \left ( \frac{1+x}{1-x}\right ) & =\ln \left ( \left ( 1+x\right ) \left ( \frac{1}{1-x}\right ) \right ) \nonumber \\ & =\ln \left ( 1+x\right ) +\ln \left ( \frac{1}{1-x}\right ) \nonumber \\ & =\ln \left ( 1+x\right ) -\ln \left ( 1-x\right ) \tag{1} \end{align}
Looking at \(\ln \left ( 1+x\right ) \), where now \(f\left ( x\right ) =\ln \left ( 1+x\right ) \), then we see that \(f^{\prime }\left ( x\right ) =\frac{1}{1+x},f^{\prime \prime }\left ( x\right ) =\frac{-1}{\left ( 1+x\right ) ^{2}},f^{\prime \prime \prime }\left ( x\right ) =\frac{2}{\left ( 1+x\right ) ^{3}},f^{\left ( 4\right ) }\left ( x\right ) =-\frac{2\cdot 3}{\left ( 1+x\right ) ^{4}},\cdots \), therefore\begin{align} \ln \left ( 1+x\right ) & =f\left ( 0\right ) +xf^{\prime }\left ( 0\right ) +\frac{x^{2}}{2}f^{\prime \prime }\left ( 0\right ) +\frac{x^{3}}{3!}f^{\prime \prime \prime }\left ( 0\right ) +\frac{x^{4}}{4!}f^{\left ( 4\right ) }\left ( 0\right ) +\cdots \nonumber \\ & =0+x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots \tag{2} \end{align}
Similarly for \(\ln \left ( 1-x\right ) \), where now \(f^{\prime }\left ( x\right ) =\frac{-1}{1-x},f^{\prime \prime }\left ( x\right ) =\frac{-1}{\left ( 1-x\right ) ^{2}},f^{\prime \prime \prime }\left ( x\right ) =\frac{-2}{\left ( 1-x\right ) ^{3}},f^{\left ( 4\right ) }\left ( x\right ) =-\frac{2\cdot 3}{\left ( 1-x\right ) ^{4}},\cdots \), therefore\begin{align} \ln \left ( 1-x\right ) & =f\left ( 0\right ) +xf^{\prime }\left ( 0\right ) +\frac{x^{2}}{2}f^{\prime \prime }\left ( 0\right ) +\frac{x^{3}}{3!}f^{\prime \prime \prime }\left ( 0\right ) +\frac{x^{4}}{4!}f^{\left ( 4\right ) }\left ( 0\right ) +\cdots \nonumber \\ & =0-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots \tag{3} \end{align}
Using (2,3) in (1) gives the series expansion for \(\ln \left ( \frac{1+x}{1-x}\right ) \) as\begin{align} \ln \left ( \frac{1+x}{1-x}\right ) & =\left ( x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots \right ) -\left ( -x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}+\cdots \right ) \nonumber \\ & =2x+\frac{2}{3}x^{3}+\frac{2}{5}x^{5}+\frac{2}{7}x^{7}+\cdots \tag{4} \end{align}
Using (4) in the integral given results in \begin{align*} I & =\int _{0}^{1}\left ( 2x+\frac{2}{3}x^{3}+\frac{2}{5}x^{5}+\frac{2}{7}x^{7}+\cdots \right ) \frac{dx}{x}\\ & =\int _{0}^{1}\left ( 2+\frac{2}{3}x^{2}+\frac{2}{5}x^{4}+\frac{2}{7}x^{6}+\cdots \right ) dx\\ & =\left [ 2x+\frac{2}{3}\frac{x^{3}}{3}+\frac{2}{5}\frac{x^{5}}{5}+\frac{2}{7}\frac{x^{7}}{7}+\cdots \right ] _{0}^{1} \end{align*}
Which simplifies to \begin{align} I & =2+\frac{2}{3}\frac{1}{3}+\frac{2}{5}\frac{1}{5}+\frac{2}{7}\frac{1}{7}+\cdots \nonumber \\ & =2+\frac{2}{3^{2}}+\frac{2}{5^{2}}+\frac{2}{7^{2}}+\frac{2}{9^{2}}+\cdots \nonumber \\ & =2\left ( 1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+\frac{1}{9^{2}}+\cdots \right ) \nonumber \\ & =2\sum _{n=0}^{\infty }\frac{1}{\left ( 2n+1\right ) ^{2}} \tag{5} \end{align}
The following are two methods to obtain closed form sum for \(\sum _{n=0}^{\infty }\frac{1}{\left ( 2n+1\right ) ^{2}}\). The first method is based on writing\begin{equation} \sum _{n=1}^{\infty }\frac{1}{n^{2}}=\sum _{n=1}^{\infty }\frac{1}{\left ( 2n\right ) ^{2}}+\sum _{n=0}^{\infty }\frac{1}{\left ( 2n+1\right ) ^{2}} \tag{6} \end{equation} Where the sum on the left is broken into odd and even terms on the right, as in\[ 1+\frac{1}{2^{2}}+\frac{1}{3^{2}}+\frac{1}{4^{2}}+\frac{1}{5^{2}}+\cdots =\left ( \frac{1}{2^{2}}+\frac{1}{4^{2}}+\cdots \right ) +\left ( \frac{1}{1^{2}}+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\cdots \right ) \] But, from lecture Sept. 12, 2018, we showed in class that \begin{equation} \sum _{n=1}^{\infty }\frac{1}{n^{2}}=\zeta \left ( 2\right ) =\frac{\pi ^{2}}{6} \tag{7} \end{equation} (This is called the Basel problem, and the above closed form sum was first given by Euler in 1734). Now using (7) into (6) results in\begin{align*} \sum _{n=0}^{\infty }\frac{1}{\left ( 2n+1\right ) ^{2}} & =\sum _{n=1}^{\infty }\frac{1}{n^{2}}-\sum _{n=1}^{\infty }\frac{1}{\left ( 2n\right ) ^{2}}\\ & =\sum _{n=1}^{\infty }\frac{1}{n^{2}}-\frac{1}{4}\sum _{n=1}^{\infty }\frac{1}{n^{2}}\\ & =\frac{3}{4}\left ( \sum _{n=1}^{\infty }\frac{1}{n^{2}}\right ) \\ & =\frac{3}{4}\left ( \frac{\pi ^{2}}{6}\right ) \\ & =\frac{\pi ^{2}}{8} \end{align*}
Another way to obtained closed form sum for \(\sum _{n=0}^{\infty }\frac{1}{\left ( 2n+1\right ) ^{2}}\) is to use Fourier series. Considering the Fourier series for the following periodic function\[ f\left ( x\right ) =\left \{ \begin{array} [c]{ccc}-x & & -\pi <x<0\\ 0 & & 0\leq x\leq \pi \end{array} \right . \] Using\[ f\left ( x\right ) =\frac{A_{0}}{2}+\sum _{n=1}^{\infty }A_{n}\cos \left ( nx\right ) +\sum _{n=1}^{\infty }B_{n}\sin \left ( nx\right ) \] Therefore \[ A_{0}=\frac{1}{\pi }\int _{-\pi }^{0}-xdx=\frac{-1}{\pi }\left ( \frac{x^{2}}{2}\right ) _{-\pi }^{0}=\frac{-1}{2\pi }\left ( x^{2}\right ) _{-\pi }^{0}=\frac{-1}{2\pi }\left ( -\pi ^{2}\right ) =\frac{1}{2}\pi \] And\begin{align*} A_{n} & =\frac{-1}{\pi }\int _{-\pi }^{0}x\cos \left ( nx\right ) dx=\frac{1+\left ( -1\right ) ^{n+1}}{n^{2}}\\ B_{n} & =\frac{-1}{\pi }\int _{-\pi }^{0}x\sin \left ( nx\right ) dx=\frac{\left ( -1\right ) ^{n+1}}{n}\pi \end{align*}
Hence the Fourier series for \(f\left ( x\right ) \) is\begin{align*} f\left ( x\right ) & =\frac{\pi }{4}-\frac{1}{\pi }\sum _{n=1}^{\infty }\frac{1+\left ( -1\right ) ^{n+1}}{n^{2}}\cos \left ( nx\right ) -\frac{1}{\pi }\sum _{n=0}^{\infty }\frac{\left ( -1\right ) ^{n+1}}{n}\pi \left ( \sin nx\right ) \\ & =\frac{\pi }{4}-\frac{1}{\pi }\sum _{n=1}^{\infty }\frac{1+\left ( -1\right ) ^{n+1}}{n^{2}}\cos \left ( nx\right ) -\sum _{n=0}^{\infty }\frac{\left ( -1\right ) ^{n+1}}{n}\sin \left ( nx\right ) \end{align*}
Evaluating the above at \(x=0\) then all the \(\sin \) terms vanish and we obtain\begin{align*} 0 & =\frac{\pi }{4}-\frac{1}{\pi }\sum _{n=1}^{\infty }\frac{1+\left ( -1\right ) ^{n+1}}{n^{2}}\\ & =\frac{\pi }{4}-\frac{2}{\pi }\left ( 1+\frac{1}{3^{2}}+\frac{1}{5^{2}}+\frac{1}{7^{2}}+\cdots \right ) \\ & =\frac{\pi }{4}-\frac{2}{\pi }\sum _{n=0}^{\infty }\frac{1}{\left ( 2n+1\right ) ^{2}} \end{align*}
Therefore\begin{align*} \frac{2}{\pi }\sum _{n=0}^{\infty }\frac{1}{\left ( 2n+1\right ) ^{2}} & =\frac{\pi }{4}\\ \sum _{n=0}^{\infty }\frac{1}{\left ( 2n+1\right ) ^{2}} & =\frac{\pi ^{2}}{8} \end{align*}
Now that we found closed form sum for \(\sum _{n=0}^{\infty }\frac{1}{\left ( 2n+1\right ) ^{2}}\), we can find the value of the integral. Since \(I=2\sum _{n=0}^{\infty }\frac{1}{\left ( 2n+1\right ) ^{2}}\), then\begin{align*} \int _{0}^{1}\ln \left ( \frac{1+x}{1-x}\right ) \frac{dx}{x} & =2\left ( \frac{\pi ^{2}}{8}\right ) \\ & =\frac{\pi ^{2}}{4} \end{align*}
Let \(I\left ( x\right ) =\int _{0}^{\infty }e^{xf\left ( t\right ) }dt\) with \(f\left ( t\right ) =t-\frac{e^{t}}{x}\), find a large \(x\) approximation for this integral.
Solution
\begin{align} I & =\int _{0}^{\infty }\exp \left ( xf\left ( t\right ) \right ) dt\nonumber \\ & =\int _{0}^{\infty }\exp \left ( x\left ( t-\frac{e^{t}}{x}\right ) \right ) dt\nonumber \\ & =\int _{0}^{\infty }\exp \left ( xt-e^{t}\right ) dt\nonumber \\ & =\int _{0}^{\infty }\exp \left ( F\left ( t\right ) \right ) dt \tag{1} \end{align}
Where \(F\left ( t\right ) =xt-e^{t}\). We need to find saddle point where \(F\left ( t\right ) \) is maximum. Hence\begin{align*} \frac{d}{dt}F\left ( t\right ) & =0\\ x-e^{t} & =0\\ e^{t} & =x\\ t_{0} & =\ln \left ( x\right ) \end{align*}
Where \(t_{0}\) is location of \(t\) where \(F\left ( t\right ) \) is maximum. We called this in class \(t_{peak}\). We now expand \(F\left ( t\right ) \) around \(t_{0}\) using Taylor series\begin{equation} F\left ( t\right ) =F\left ( t_{0}\right ) +F^{\prime }\left ( t_{0}\right ) \left ( t-t_{0}\right ) +\frac{1}{2}F^{\prime \prime }\left ( t_{0}\right ) \left ( t-t_{0}\right ) ^{2}+\cdots \tag{2} \end{equation} But \begin{align*} F\left ( t_{0}\right ) & =x\ln \left ( x\right ) -e^{\ln x}\\ & =x\ln x-x \end{align*}
And \(F^{\prime }\left ( t\right ) =x-e^{t}\), hence as expected \(F^{\prime }\left ( t_{0}\right ) =0\). And \(F^{\prime \prime }\left ( t\right ) =-e^{t}\), therefore \(F^{\prime \prime }\left ( t_{0}\right ) =-e^{\ln x}=-x\). We see also that \(F^{\prime \prime }\left ( t_{0}\right ) <0\), which means the saddle point was a maximum and not a minimum (since \(x\) is positive). Using these in (2) gives\begin{align*} F\left ( t\right ) & \approx \left ( x\ln x-x\right ) +\frac{1}{2}\left ( -x\right ) \left ( t-\ln x\right ) ^{2}\\ & =x\ln x-x-\frac{1}{2}x\left ( t-\ln x\right ) ^{2} \end{align*}
Substituting the above into (1) gives\begin{align} I & =\int _{0}^{\infty }\exp \left ( x\ln x-x-\frac{1}{2}x\left ( t-\ln x\right ) ^{2}\right ) dt\nonumber \\ & =\int _{0}^{\infty }\exp \left ( x\ln x\right ) \exp \left ( -x\right ) \exp \left ( -\frac{1}{2}x\left ( t-\ln x\right ) ^{2}\right ) dt\nonumber \\ & =\exp \left ( x\ln x\right ) \exp \left ( -x\right ) \int _{0}^{\infty }\exp \left ( -\frac{1}{2}x\left ( t-\ln x\right ) ^{2}\right ) dt\nonumber \\ & =x^{x}e^{-x}\int _{0}^{\infty }e^{-\frac{1}{2}x\left ( t-\ln x\right ) ^{2}}dt \tag{3} \end{align}
Now, since the peak value where \(F\left ( t\right ) \) occurs is on the positive real axis, because \(t_{0}=\ln \left ( x\right ) \), therefore \(x>1\) to have a maximum, and assuming a narrow peak, then all the contribution to the integral comes from \(x\) close to the peak location, so we can change \(\int _{0}^{\infty }e^{-\frac{1}{2}x\left ( t-\ln x\right ) ^{2}}dt\) to \(\int _{-\infty }^{\infty }e^{-\frac{1}{2}x\left ( t-\ln x\right ) ^{2}}dt\) without affecting the final result. Therefore (3) becomes\begin{equation} I=x^{x}e^{-x}\int _{-\infty }^{\infty }e^{-\frac{1}{2}x\left ( t-\ln x\right ) ^{2}}dt \tag{4} \end{equation} Now comparing \(\int _{-\infty }^{\infty }e^{-\frac{1}{2}x\left ( t-\ln x\right ) ^{2}}dt\) to the Gaussian integral \(\int _{-\infty }^{\infty }e^{-a\left ( t-b\right ) ^{2}}dt=\sqrt{\frac{\pi }{a}}\), shows that \(a=\frac{x}{2}\) for our case. Hence \[ \int _{-\infty }^{\infty }e^{-\frac{1}{2}x\left ( t-\ln x\right ) ^{2}}dt=\sqrt{\frac{2\pi }{x}}\] Therefore (4) becomes\[ I\approx x^{x}e^{-x}\sqrt{\frac{2\pi }{x}}\] For large \(x\).
Evaluate the following integrals with aid of residue theorem \(a\geq 0.\) (a) \(\int _{0}^{\infty }\frac{1}{x^{4}+1}dx\) (b) \(\int _{0}^{\infty }\frac{\cos \left ( ax\right ) }{x^{2}+1}dx\)
Since the integrand is even, then\[ I=\frac{1}{2}\int _{-\infty }^{\infty }\frac{1}{x^{4}+1}dx \] Now we consider the following contour
Therefore\[{\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=\left ( \lim _{R\rightarrow \infty }\int _{-R}^{0}f\left ( x\right ) dx+\lim _{\tilde{R}\rightarrow \infty }\int _{0}^{\tilde{R}}f\left ( x\right ) dx\right ) +\lim _{R\rightarrow \infty }\int _{C_{R}}f\left ( z\right ) dz \] Using Cauchy principal value the integral above can be written as\begin{align*}{\displaystyle \oint \limits _{C}} f\left ( z\right ) dz & =\lim _{R\rightarrow \infty }\int _{-R}^{R}f\left ( x\right ) dx+\lim _{R\rightarrow \infty }\int _{C_{R}}f\left ( z\right ) d\\ & =2\pi i\sum \operatorname{Residue} \end{align*}
Where \(\sum \operatorname{Residue}\) is sum of residues of \(\frac{1}{z^{4}+1}\) for poles that are inside the contour \(C\). Therefore the above becomes\begin{align} \lim _{R\rightarrow \infty }\int _{-R}^{R}f\left ( x\right ) dx & =2\pi i\sum \operatorname{Residue}-\lim _{R\rightarrow \infty }\int _{C_{R}}f\left ( z\right ) dz\nonumber \\ \int _{-\infty }^{\infty }\frac{1}{x^{4}+1}dx & =2\pi i\sum \operatorname{Residue}-\lim _{R\rightarrow \infty }\int _{C_{R}}\frac{1}{z^{4}+1}dz \tag{1} \end{align}
Now we will show that \(\lim _{R\rightarrow \infty }\int _{C_{R}}\frac{1}{z^{4}+1}dz=0\). Since \begin{align} \left \vert \int _{C_{R}}\frac{1}{z^{4}+1}dz\right \vert & \leq ML\nonumber \\ & =\left \vert f\left ( z\right ) \right \vert _{\max }\left ( \pi R\right ) \tag{2} \end{align}
But \[ f\left ( z\right ) =\frac{1}{\left ( z^{2}-i\right ) \left ( z^{2}+i\right ) }\] Hence, and since \(z=R\ e^{i\theta }\) then\[ \left \vert f\left ( z\right ) \right \vert _{\max }\leq \frac{1}{\left \vert z^{2}-i\right \vert _{\min }\left \vert z^{2}+i\right \vert _{\min }}\] But but inverse triangle inequality \(\left \vert z^{2}-i\right \vert \geq \left \vert z\right \vert ^{2}+1\) and \(\left \vert z^{2}+i\right \vert \geq \left \vert z\right \vert ^{2}-1\), and since \(\left \vert z\right \vert =R\) then the above becomes\begin{align*} \left \vert f\left ( z\right ) \right \vert _{\max } & \leq \frac{1}{\left ( R^{2}+1\right ) \left ( R^{2}-1\right ) }\\ & =\frac{1}{R^{4}-1} \end{align*}
Therefore (2) becomes\[ \left \vert \int _{C_{R}}\frac{1}{z^{4}+1}dz\right \vert \leq \frac{\pi R}{R^{4}-1}\] Then it is clear that as \(R\rightarrow \infty \) the above goes to zero since \(\lim _{R\rightarrow \infty }\frac{\pi R}{R^{4}-1}=\lim _{R\rightarrow \infty }\frac{\frac{\pi }{R^{3}}}{1-\frac{1}{R^{4}}}=\frac{0}{1}=0\). Then (1) now simplifies to\begin{equation} \int _{-\infty }^{\infty }\frac{1}{x^{4}+1}dx=2\pi i\sum \operatorname{Residue} \tag{2A} \end{equation} We just now need to find the residues of \(\frac{1}{z^{4}+1}\) located in upper half plane. The zeros of the denominator \(z^{4}+1=0\) are at \(z=-1^{\frac{1}{4}}=\left ( e^{i\pi }\right ) ^{\frac{1}{4}}\), then the first zero is at \(e^{i\frac{\pi }{4}}\), and the second zero at \(e^{i\left ( \frac{\pi }{4}+\frac{\pi }{2}\right ) }=e^{i\left ( \frac{3}{4}\pi \right ) }\) and the third zero at \(e^{i\left ( \frac{3}{4}\pi +\frac{\pi }{2}\right ) }=e^{i\left ( \frac{5}{4}\pi \right ) }\) and the fourth zero at \(e^{i\left ( \frac{5}{4}\pi +\frac{\pi }{2}\right ) }=e^{i\frac{7}{4}\pi }\). Hence poles are at
\begin{align*} z_{1} & =e^{i\frac{\pi }{4}}\\ z_{2} & =e^{i\frac{3}{4}\pi }\\ z_{3} & =e^{i\frac{5}{4}\pi }\\ z_{4} & =e^{i\frac{7}{4}\pi } \end{align*}
Out of these only the first two are in upper half plane \(z_{1}\) and \(z_{1}\). Hence\begin{align} \operatorname{Residue}\left ( z_{1}\right ) & =\lim _{z\rightarrow z_{1}}\left ( z-z_{1}\right ) f\left ( z\right ) \nonumber \\ & =\lim _{z\rightarrow z_{1}}\left ( z-z_{1}\right ) \frac{1}{z^{4}-1}\nonumber \end{align}
Applying L’Hopitals\begin{align*} \operatorname{Residue}\left ( z_{1}\right ) & =\lim _{z\rightarrow z_{1}}\frac{1}{4z^{3}}\\ & =\frac{1}{4\left ( e^{i\frac{\pi }{4}}\right ) ^{3}}\\ & =\frac{1}{4e^{i\frac{3\pi }{4}}} \end{align*}
Similarly for the other residue\begin{align} \operatorname{Residue}\left ( z_{2}\right ) & =\lim _{z\rightarrow z_{2}}\left ( z-z_{2}\right ) f\left ( z\right ) \nonumber \\ & =\lim _{z\rightarrow z_{1}}\left ( z-z_{2}\right ) \frac{1}{z^{4}-1}\nonumber \end{align}
Applying L’Hopitals\begin{align*} \operatorname{Residue}\left ( z_{2}\right ) & =\lim _{z\rightarrow z_{2}}\frac{1}{4z^{3}}\\ & =\frac{1}{4\left ( e^{i\frac{3}{4}\pi }\right ) ^{3}}\\ & =\frac{1}{4e^{i\frac{9\pi }{4}}}\\ & =\frac{1}{4e^{i\frac{\pi }{4}}} \end{align*}
Hence (2A) becomes \begin{align*} \int _{-\infty }^{\infty }\frac{1}{x^{4}+1}dx & =2\pi i\left ( \frac{1}{4e^{i\frac{3\pi }{4}}}+\frac{1}{4e^{i\frac{\pi }{4}}}\right ) \\ & =2\pi i\left ( \frac{\sqrt{2}}{4i}\right ) \\ & =\frac{1}{2}\sqrt{2}\pi \end{align*}
But \(\int _{0}^{\infty }\frac{1}{x^{4}+1}dx=\frac{1}{2}\int _{-\infty }^{\infty }\frac{1}{x^{4}+1}dx\), therefore \begin{align*} \int _{0}^{\infty }\frac{1}{x^{4}+1}dx & =\frac{\sqrt{2}}{4}\pi \\ & =\frac{2}{4\sqrt{2}}\pi \\ & =\frac{\pi }{2\sqrt{2}} \end{align*}
Since the integrand is even, then \[ I=\frac{1}{2}\int _{-\infty }^{\infty }\frac{\cos \left ( ax\right ) }{x^{2}+1}dx \] We will evaluate \(\int _{-\infty }^{\infty }\frac{e^{iaz}}{x^{2}+1}dx\) and at the end take the real part of the answer. Considering the following contour
Then\[{\displaystyle \oint \limits _{C}} f\left ( z\right ) dz=\left ( \lim _{R\rightarrow \infty }\int _{-R}^{0}f\left ( x\right ) dx+\lim _{\tilde{R}\rightarrow \infty }\int _{0}^{\tilde{R}}f\left ( x\right ) dx\right ) +\lim _{R\rightarrow \infty }\int _{C_{R}}f\left ( z\right ) dz \] Using Cauchy principal value the integral above can be written as\begin{align*}{\displaystyle \oint \limits _{C}} f\left ( z\right ) dz & =\lim _{R\rightarrow \infty }\int _{-R}^{R}f\left ( x\right ) dx+\lim _{R\rightarrow \infty }\int _{C_{R}}f\left ( z\right ) d\\ & =2\pi i\sum \operatorname{Residue} \end{align*}
Where \(\sum \operatorname{Residue}\) is sum of residues of \(\frac{e^{iaz}}{x^{2}+1}\) for poles that are inside the contour \(C\). Therefore the above becomes\begin{align} \lim _{R\rightarrow \infty }\int _{-R}^{R}f\left ( x\right ) dx & =2\pi i\sum \operatorname{Residue}-\lim _{R\rightarrow \infty }\int _{C_{R}}f\left ( z\right ) dz\nonumber \\ \int _{-\infty }^{\infty }\frac{e^{iax}}{x^{2}+1}dx & =2\pi i\sum \operatorname{Residue}-\lim _{R\rightarrow \infty }\int _{C_{R}}\frac{e^{iaz}}{z^{2}+1}dz \tag{1} \end{align}
Now we will show that \(\lim _{R\rightarrow \infty }\int _{C_{R}}\frac{e^{iaz}}{z^{2}+1}dz=0\). Since \begin{align} \left \vert \int _{C_{R}}\frac{e^{iaz}}{z^{2}+1}dz\right \vert & \leq ML\nonumber \\ & =\left \vert f\left ( z\right ) \right \vert _{\max }\left ( \pi R\right ) \tag{2} \end{align}
But \begin{align*} f\left ( z\right ) & =\frac{e^{iaz}}{\left ( z-i\right ) \left ( z+i\right ) }\\ & =\frac{e^{ia\left ( x+iy\right ) }}{\left ( z-i\right ) \left ( z+i\right ) }\\ & =\frac{e^{iax-ay}}{\left ( z-i\right ) \left ( z+i\right ) }\\ & =\frac{e^{iax}e^{-ay}}{\left ( z-i\right ) \left ( z+i\right ) } \end{align*}
Hence\begin{align*} \left \vert f\left ( z\right ) \right \vert _{\max } & =\frac{\left \vert e^{iaz}\right \vert _{\max }\left \vert e^{-ay}\right \vert _{\max }}{\left \vert z-i\right \vert _{\min }\left \vert z+i\right \vert _{\min }}\\ & =\frac{\left \vert e^{-ay}\right \vert _{\max }}{\left ( R+1\right ) \left ( R-1\right ) }\\ & =\frac{\left \vert e^{-ay}\right \vert _{\max }}{R^{2}-1} \end{align*}
Since \(a>0\) and since in upper half \(y>0\) then \(\left \vert e^{-ay}\right \vert _{\max }=\left \vert e^{-aR}\right \vert _{\max }=1\). Jordan inequality was not needed here, since there is no extra \(x\) in the numerator of the integrand in this problem. The above now reduces to\[ \left \vert f\left ( z\right ) \right \vert _{\max }=\frac{1}{R^{2}-1}\] Equation (2) becomes\[ \left \vert \int _{C_{R}}\frac{e^{iaz}}{z^{2}+1}dz\right \vert \leq \frac{\pi R}{R^{2}-1}\] \(R\rightarrow \infty \) the above goes to zero since \(\lim _{R\rightarrow \infty }\frac{\pi R}{R^{2}-1}=\lim _{R\rightarrow \infty }\frac{\frac{\pi }{R^{2}}}{1-\frac{1}{R^{2}}}=\frac{0}{1}=0\). Equation (1) now simplifies to\[ \int _{-\infty }^{\infty }\frac{e^{iax}}{x^{4}+1}dx=2\pi i\sum \operatorname{Residue}\] We just now need to find the residues of \(\frac{1}{z^{2}+1}\) that are located in upper half plane. The zeros of the denominator \(z^{2}+1=0\) are at \(z=\pm i\), hence poles are at \begin{align*} z_{1} & =i\\ z_{2} & =-i \end{align*}
Only \(z_{1}\) is in upper half plane. Therefore\begin{align*} \operatorname{Residue}\left ( z_{1}\right ) & =\lim _{z\rightarrow z_{1}}\left ( z-z_{1}\right ) f\left ( z\right ) \\ & =\lim _{z\rightarrow z_{1}}\left ( z-z_{1}\right ) \frac{e^{iaz}}{\left ( z-z_{1}\right ) \left ( z-z_{2}\right ) }\\ & =\lim _{z\rightarrow z_{1}}\frac{e^{iaz}}{\left ( z-z_{2}\right ) }\\ & =\frac{e^{ia\left ( i\right ) }}{\left ( i+i\right ) }\\ & =\frac{e^{-a}}{2i} \end{align*}
Since \(\int _{-\infty }^{\infty }\frac{e^{ax}}{x^{4}+1}dx=2\pi i\sum \operatorname{Residue}\) then \begin{align*} \int _{-\infty }^{\infty }\frac{e^{iax}}{x^{4}+1}dx & =2\pi i\left ( \frac{e^{-a}}{2i}\right ) \\ & =\pi e^{-a} \end{align*}
Therefore\begin{align*} \int _{0}^{\infty }\frac{e^{iax}}{x^{4}+1}dx & =\frac{1}{2}\int _{-\infty }^{\infty }\frac{e^{ax}}{x^{4}+1}dx\\ & =\frac{\pi }{2}e^{-a} \end{align*}
But real part of the above is \[ \int _{0}^{\infty }\frac{\cos \left ( ax\right ) }{x^{4}+1}dx=\frac{\pi }{2}e^{-a}\]
Using residues evaluate(a) \(\int _{0}^{2\pi }\frac{1}{1+a\cos \theta }d\theta \) for \(\left \vert a\right \vert <1\) (b) \(\int _{0}^{\pi }\left ( \cos \left ( \theta \right ) \right ) ^{2n}d\theta \) for \(n\) integer.
Using contour which is anti-clockwise over the unit circle
Let \(z=e^{i\theta },\) hence \(dz=d\theta ie^{i\theta }=d\theta iz.\) Using \(\cos \theta =\frac{z+z^{-1}}{2}\) then the integral can be written in complex domain as\begin{align*}{\displaystyle \oint \limits _{C}} \frac{\frac{1}{iz}dz}{1+a\frac{z+z^{-1}}{2}} & =\frac{2}{i}{\displaystyle \oint \limits _{C}} \frac{\frac{1}{z}dz}{2+a\left ( z+\frac{1}{z}\right ) }\\ & =\frac{2}{i}{\displaystyle \oint \limits _{C}} \frac{dz}{2z+az^{2}+a}\\ & =\frac{2}{ai}{\displaystyle \oint \limits _{C}} \frac{dz}{z^{2}+\frac{2}{a}z+1}\\ & =\frac{2}{ai}{\displaystyle \oint \limits _{C}} \frac{dz}{\left ( z-z_{1}\right ) \left ( z-z_{2}\right ) } \end{align*}
Where \(z_{1},z_{2}\) are roots of \(z^{2}+\frac{2}{a}z+1=0\) which are found to be (using the quadratic formula) as\begin{align*} z_{1} & =\frac{-1-\sqrt{1-a^{2}}}{a}\\ z_{2} & =\frac{-1+\sqrt{1-a^{2}}}{a} \end{align*}
Since \(\left \vert a\right \vert <1\) then only \(z_{2}\) will be inside the unit disk for all \(a\) values. Therefore\begin{align} \frac{2}{ai}{\displaystyle \oint \limits _{C}} \frac{dz}{\left ( z-z_{1}\right ) \left ( z-z_{2}\right ) } & =\left ( \frac{2}{ai}\right ) 2\pi i\operatorname{Residue}\left ( z_{2}\right ) \nonumber \\ & =\frac{4}{a}\pi \operatorname{Residue}\left ( z_{2}\right ) \tag{1} \end{align}
Now we will find the \(\operatorname{Residue}\left ( z_{2}\right ) \) where in this case \(f\left ( z\right ) =\frac{1}{\left ( z-z_{1}\right ) \left ( z-z_{2}\right ) }\). Hence \begin{align*} \operatorname{Residue}\left ( z_{2}\right ) & =\lim _{z\rightarrow z_{2}}\left ( z-z_{2}\right ) f\left ( z\right ) \\ & =\lim _{z\rightarrow z_{2}}\left ( z-z_{2}\right ) \frac{1}{\left ( z-z_{1}\right ) \left ( z-z_{2}\right ) }\\ & =\lim _{z\rightarrow z_{2}}\frac{1}{\left ( z-z_{1}\right ) }\\ & =\frac{1}{\left ( \frac{-1+\sqrt{1-a^{2}}}{a}\right ) -\left ( \frac{-1-\sqrt{1-a^{2}}}{a}\right ) }\\ & =\frac{a}{2\sqrt{1-a^{2}}} \end{align*}
Using the above result in (1) gives \begin{align*} \int _{0}^{2\pi }\frac{1}{1+a\cos \theta }d\theta & =\left ( \frac{4}{a}\pi \right ) \frac{a}{2\sqrt{1-a^{2}}}\\ & =\frac{2\pi }{\sqrt{1-a^{2}}}\qquad a\neq 1 \end{align*}
Using Maple, verified that the above result is correct.
Since integrand is even, then \(\int _{0}^{\pi }\left ( \cos \left ( \theta \right ) \right ) ^{2n}d\theta =\frac{1}{2}\int _{0}^{2\pi }\left ( \cos \left ( \theta \right ) \right ) ^{2n}d\theta \). Using same contour as in part (a), and letting \(z=e^{i\theta },\) hence \(dz=d\theta ie^{i\theta }=d\theta iz\) and using \(\cos \theta =\frac{z+z^{-1}}{2}\) then the integral can be written in complex domain as\begin{align*} \int _{0}^{2\pi }\left ( \cos \left ( \theta \right ) \right ) ^{2n}d\theta & ={\displaystyle \oint \limits _{C}} \left ( \frac{z+\frac{1}{z}}{2}\right ) ^{2n}\frac{dz}{iz}\\ & =\frac{1}{i}{\displaystyle \oint \limits _{C}} \frac{\left ( z+\frac{1}{z}\right ) ^{2n}}{2^{2n}}\frac{dz}{z}\\ & =\frac{1}{4^{n}i}{\displaystyle \oint \limits _{C}} \left ( z+\frac{1}{z}\right ) ^{2n}\frac{dz}{z}\\ & =\frac{1}{4^{n}i}{\displaystyle \oint \limits _{C}} \left ( \frac{z^{2}+1}{z}\right ) ^{2n}\frac{dz}{z}\\ & =\frac{1}{4^{n}i}{\displaystyle \oint \limits _{C}} \frac{\left ( z^{2}+1\right ) ^{2n}}{z^{2n}}\frac{dz}{z}\\ & =\frac{1}{4^{n}i}{\displaystyle \oint \limits _{C}} \frac{\left ( z^{2}+1\right ) ^{2n}}{z^{2n+1}}dz \end{align*}
Considering \(f\left ( z\right ) =\frac{\left ( z^{2}+1\right ) ^{2n}}{z^{2n+1}}\), this has a pole at \(z=0\) of order \(m=2n+1\). Therefore\begin{equation} \frac{1}{4^{n}i}{\displaystyle \oint \limits _{C}} \frac{\left ( z^{2}+1\right ) ^{2n}}{z^{2n+1}}dz=\left ( \frac{1}{4^{n}i}\right ) 2\pi i\operatorname{Residue}\left ( z=0\right ) \tag{1} \end{equation} So we now need to find residue of \(f\left ( z\right ) \) at \(z=0\) but for pole of order \(m=2n+1\). Using the formula for finding residue for pole of order \(m\) gives\[ \operatorname{Residue}\left ( z_{0}=0\right ) =\lim _{z\rightarrow z_{0}}\frac{d^{m-1}}{dz^{m-1}}\frac{\left ( z-z_{0}\right ) ^{m}f\left ( z\right ) }{\left ( m-1\right ) !}\] But \(m=2n+1\), and \(z_{0}=0\), hence the above becomes\begin{align*} \operatorname{Residue}\left ( 0\right ) & =\lim _{z\rightarrow 0}\frac{d^{2n}}{dz^{2n}}\frac{z^{2n+1}}{\left ( 2n\right ) !}\frac{\left ( z^{2}+1\right ) ^{2n}}{z^{2n+1}}\\ & =\frac{1}{\left ( 2n\right ) !}\lim _{z\rightarrow 0}\left ( \frac{d^{2n}}{dz^{2n}}\left ( z^{2}+1\right ) ^{2n}\right ) \end{align*}
Equation (1) becomes\[ \int _{0}^{2\pi }\left ( \cos \left ( \theta \right ) \right ) ^{2n}d\theta =\left ( \frac{1}{4^{n}}\right ) 2\pi \left ( \frac{1}{\left ( 2n\right ) !}\lim _{z\rightarrow 0}\left ( \frac{d^{2n}}{dz^{2n}}\left ( z^{2}+1\right ) ^{2n}\right ) \right ) \] Therefore\begin{align*} \int _{0}^{\pi }\left ( \cos \left ( \theta \right ) \right ) ^{2n}d\theta & =\frac{1}{2}\left ( \frac{1}{4^{n}}\right ) 2\pi \left ( \frac{1}{\left ( 2n\right ) !}\lim _{z\rightarrow 0}\left ( \frac{d^{2n}}{dz^{2n}}\left ( z^{2}+1\right ) ^{2n}\right ) \right ) \\ & =\frac{1}{4^{n}}\frac{\pi }{\left ( 2n\right ) !}\lim _{z\rightarrow 0}\left ( \frac{d^{2n}}{dz^{2n}}\left ( z^{2}+1\right ) ^{2n}\right ) \end{align*}
Will now try to obtained closed form solution. Trying for different \(n\) values in order to see the pattern. From few lectures ago, we learned also that \[ \Gamma \left ( n+\frac{1}{2}\right ) =\frac{1\cdot 3\cdot 5\cdot \cdots \cdot \left ( 2n-1\right ) }{2^{n}}\sqrt{\pi }\] Now will generate a table to see the pattern
| \(n\) | \(\frac{1}{4^{n}}\frac{\pi }{\left ( 2n\right ) !}\lim _{z\rightarrow 0}\left ( \frac{d^{2n}}{dz^{2n}}\left ( z^{2}+1\right ) ^{2n}\right ) \) | result of integral | \(\Gamma \left ( n+\frac{1}{2}\right ) \) |
| \(1\) | \(\frac{1}{4}\frac{\pi }{2!}\lim _{z\rightarrow 0}\frac{d^{2}}{dz^{2}}\left ( z^{2}+1\right ) ^{2}\) | \(\frac{\pi }{2}\) | \(\Gamma \left ( 1+\frac{1}{2}\right ) =\frac{\sqrt{\pi }}{2}\) |
| \(2\) | \(\frac{1}{4^{2}}\frac{\pi }{4!}\lim _{z\rightarrow 0}\frac{d^{4}}{dz^{4}}\left ( z^{2}+1\right ) ^{4}\) | \(\frac{3\pi }{8}\) | \(\Gamma \left ( 2+\frac{1}{2}\right ) =\frac{3\sqrt{\pi }}{4}\) |
| \(3\) | \(\frac{1}{4^{3}}\frac{\pi }{6!}\lim _{z\rightarrow 0}\frac{d^{6}}{dz^{6}}\left ( z^{2}+1\right ) ^{6}\) | \(\frac{5\pi }{16}\) | \(\Gamma \left ( 3+\frac{1}{2}\right ) =\frac{15\sqrt{\pi }}{8}\) |
| \(4\) | \(\frac{1}{4^{4}}\frac{\pi }{8!}\lim _{z\rightarrow 0}\frac{d^{8}}{dz^{8}}\left ( z^{2}+1\right ) ^{8}\) | \(\frac{35\pi }{128}\) | \(\Gamma \left ( 4+\frac{1}{2}\right ) =\frac{105\sqrt{\pi }}{16}\) |
| \(5\) | \(\frac{1}{4^{5}}\frac{\pi }{10!}\lim _{z\rightarrow 0}\frac{d^{10}}{dz^{10}}\left ( z^{2}+1\right ) ^{10}\) | \(\frac{63\pi }{256}\) | \(\Gamma \left ( 5+\frac{1}{2}\right ) =\frac{945\sqrt{\pi }}{32}\) |
| \(\vdots \) | \(\vdots \) | \(\vdots \) | \(\vdots \) |
Based on the above, we see that \(I=\frac{\sqrt{\pi }\Gamma \left ( n+\frac{1}{2}\right ) }{n!}\), which is verified as follows
| \(n\) | result of integral | \(\Gamma \left ( n+\frac{1}{2}\right ) \) | \(\frac{\sqrt{\pi }\Gamma \left ( n+\frac{1}{2}\right ) }{n!}\) |
| \(1\) | \(\frac{\pi }{2}\) | \(\Gamma \left ( 1+\frac{1}{2}\right ) =\frac{\sqrt{\pi }}{2}\) | \(\frac{\sqrt{\pi }\left ( \frac{\sqrt{\pi }}{2}\right ) }{1}=\frac{1}{2}\pi \) |
| \(2\) | \(\frac{3\pi }{8}\) | \(\Gamma \left ( 2+\frac{1}{2}\right ) =\frac{3\sqrt{\pi }}{4}\) | \(\frac{\sqrt{\pi }\left ( \frac{3\sqrt{\pi }}{4}\right ) }{2!}=\frac{3}{8}\pi \) |
| \(3\) | \(\frac{5\pi }{16}\) | \(\Gamma \left ( 3+\frac{1}{2}\right ) =\frac{15\sqrt{\pi }}{8}\) | \(\frac{\sqrt{\pi }\left ( \frac{15\sqrt{\pi }}{8}\right ) }{3!}=\frac{15\pi }{\left ( 6\right ) \left ( 8\right ) }=\frac{15\pi }{48}=\frac{3}{16}\pi \) |
| \(4\) | \(\frac{35\pi }{128}\) | \(\Gamma \left ( 4+\frac{1}{2}\right ) =\frac{105\sqrt{\pi }}{16}\) | \(\frac{\sqrt{\pi }\left ( \frac{105\sqrt{\pi }}{16}\right ) }{4!}=\frac{\sqrt{\pi }\left ( 105\sqrt{\pi }\right ) }{\left ( 24\right ) \left ( 16\right ) }=\frac{105\pi }{384}=\frac{35}{128}\pi \) |
| \(5\) | \(\frac{63\pi }{256}\) | \(\Gamma \left ( 5+\frac{1}{2}\right ) =\frac{945\sqrt{\pi }}{32}\) | \(\frac{\sqrt{\pi }\left ( \frac{945\sqrt{\pi }}{32}\right ) }{5!}=\frac{945\pi }{\left ( 120\right ) \left ( 32\right ) }=\frac{945\pi }{3840}\) \(=\frac{63}{256}\pi \) |
| \(\vdots \) | \(\vdots \) | \(\vdots \) | \(\vdots \) |
Therefore\[ \int _{0}^{\pi }\left ( \cos \left ( \theta \right ) \right ) ^{2n}d\theta =\frac{\sqrt{\pi }\Gamma \left ( n+\frac{1}{2}\right ) }{n!}\] Tried to do pole/zero cancellation on the integrand of \({\displaystyle \oint \limits _{C}} \frac{\left ( z^{2}+1\right ) ^{2n}}{z^{2n+1}}dz\) in order to find a simpler method than the above but was not able to. The above result was verified using the computer
