calculate: $\int_{-\infty}^{\infty}\frac{\cos\frac{\pi}{2}x}{1-x^{2}}dx$ using complex analysis ; detect my mistake

Question

calculate: $\int_{-\infty}^{\infty}\frac{\cos\frac{\pi}{2}x}{1-x^{2}}dx$ using complex analysis. My try: $\int_{-\infty}^{\infty}\frac{\cos\frac{\pi}{2}x}{1-x^{2}}dx$

symetric therefore : $ \int_{-\infty}^{\infty}\frac{\cos\frac{\pi}{2}x}{1-x^{2}}dx=2\int_{0}^{\infty}\frac{\cos\frac{\pi}{2}x}{1-x^{2}}dx$

calculate instead: $2Re\int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz$

use pizza slice:$2Re\int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz=\int_{0}^{2\pi}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}d\theta+\int_{0}^{R}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}dR+\int_{0}^{R}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}dR$

take limits:

$2Re\int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz=Lim_{R\rightarrow\infty}\int_{0}^{2\pi}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}d\theta+Lim_{\theta\searrow0}\int_{0}^{R}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}dR+Lim_{\theta\nearrow0}\int_{0}^{R}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}dR$

$2Re\int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz=0+\int_{0}^{R}\frac{1}{1-e^{\pi\theta i}R^{2}}dR+\int_{0}^{R}\frac{1}{1-e^{\pi\theta i}R^{2}}dR$

According the residue theorem at$ \int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz=2\pi iRes_{z=-1}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}=0 $ therefore:$2Re\int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz=0$

hmm it might be part of the problem, might it be instead simply $1-z^2$? — hash man, Aug 12 '20 at 20:03
Indeed yes :) since ${e^{iz\pi}}$ would be ${(-1)^{z}}$, which ${\neq z^2}$ — Riemann'sPointyNose, Aug 12 '20 at 20:05
okay but later I converted it to $1-e^{\pi\theta i}R^{2}$. which seems to be fine. — hash man, Aug 12 '20 at 20:06
Without looking at the rest of your working just yet, at the end however you have not fixed the ${1-e^{iz\pi}}$ when calculating the residue — Riemann'sPointyNose, Aug 12 '20 at 20:13
hmmm I know the residue at this point is 0. don't I? @Riemann'sPointyNose — hash man, Aug 12 '20 at 20:29
You must start with $\displaystyle 2,\Re\int_{0}^{\infty}{\mathrm{e}^{\mathrm{i}\pi x/2} - \color{red}{\mathrm{i}} \over 1 - x^{2}},\mathrm{d}x$. — Felix Marin, Aug 12 '20 at 20:58

Felix Marin · Answer 1 · 2020-08-22T21:02:41.827

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\underline{\underline{Complex\ Integration}}:}$ \begin{align} &\bbox[10px,#ffd]{\int_{-\infty}^{\infty}{\cos\pars{\pi x/2} \over 1 - x^{2}}\,\dd x} = 2\int_{0}^{\infty}{\cos\pars{\pi x/2} \over 1 - x^{2}}\,\dd x = 2\,\Re\int_{0}^{\infty}{\expo{\pi x\ic/2} - \color{red}{\large\ic} \over 1 - x^{2}}\,\dd x \\[5mm] = &\ -\overbrace{\lim_{R \to \infty}\Re\int_{\large x\ \in\ R\expo{\pars{0,\pi/2}\,\ic}}{\expo{\pi x\ic/2} - \ic \over 1 - x^{2}}\,\dd x}^{\ds{=\ 0}}\ -\ 2\,\Re\int_{\infty}^{0}{\expo{\ic\pi\pars{\ic y}/2} - \ic \over 1 - \pars{\ic y}^{2}}\,\ic\,\dd y \\[5mm] = &\ 2\int_{0}^{\infty}{\dd y \over 1 + y^{2}} = 2\,{\pi \over 2} = \bbx{\large\pi} \\ & \end{align}

$\ds{\underline{\underline{Real\ Integration}}:}$ \begin{align} &\bbox[10px,#ffd]{\int_{-\infty}^{\infty}{\cos\pars{\pi x/2} \over 1 - x^{2}}\,\dd x} = {1 \over 2}\int_{-\infty}^{\infty}\bracks{% {\cos\pars{\pi x/2} \over 1 - x} + {\cos\pars{\pi x/2} \over 1 + x}}\,\dd x \\[5mm] = &\ -\int_{-\infty}^{\infty}{\cos\pars{\pi x/2} \over x - 1}\,\dd x = \int_{-\infty}^{\infty}{\sin\pars{\pi x/2} \over x}\,\dd x = \int_{-\infty}^{\infty}{\sin\pars{x} \over x}\,\dd x \\[5mm] = &\ \bbx{\large\pi} \\ & \end{align}

@hashman To avoid the divergence of the integral at $\displaystyle x = 1$. It doesn’t change the integral value because $\displaystyle\Re\left(\mathrm{i}\right) = 0$. — Felix Marin, Aug 13 '20 at 11:21
I try to get around this by taking two sloping rays which does not intersect the real line. why it is wrong? — hash man, Aug 13 '20 at 14:11

zhw. · Answer 2 · 2020-08-23T15:42:46.523

Let $f(z)=\dfrac{e^{i(\pi/2)z}}{1-z^2}.$

We want to find "$\int_{-\infty}^\infty f(x)\,dx$" and then take the real part. I have that in quotes as the integral is problematic unless we're careful about the singularities at $-1,1.$

A contour that will work contains the intervals $[-R,-1-r],$ $[-1+r,1-r],$ and $[1+r,R]$ (here $r,R>0$ and $r$ is much smaller than $R$). We also want the large semicircle described above. Around $-1$ we put the small semicircle of radius $r$ given by $-1-re^{it},0\le t \le \pi.$ Around $1$ we put the semicircle $1-re^{it},0\le t \le \pi.$ Hook these pieces up and orient the resulting closed contour positively. (It's good to draw a picture!)

Call this contour $\gamma=\gamma_{r,R}.$ Note that $\gamma$ does not contain either of $-1,1$ in its interior. Thus by Cauchy's theorem, $\int_\gamma f(z)\,dz =0.$

There are three intervals in this contour; let's denote the integral of $f$ over the union of all of them by $I(r,R).$ Note that $I(r,R)$ is real.

The first small semicircle:

$$\int_{0}^{\pi} f(-1-re^{it})(-ire^{it})\,dt=-\int_{0}^{\pi}\frac{\exp[i(\pi/2)(-1-re^{it})]ire^{it}}{1-(-1-re^{it})^2}\,dt$$ $$ = -\int_{0}^{\pi}\frac{i\exp[i(\pi/2)(-1-re^{it})]}{-2+re^{it}}\,dt.$$

As $r\to 0^+,$ the last integrand converges nicely to $\dfrac{i\exp[-i(\pi/2]}{-2} = 1/2.$ Thus the integral converges to $-\pi\cdot(1/2)=-\pi/2.$

The big semicircle:

$$\int_{0}^{\pi} f(Re^{it})iRe^{it}\,dt= \int_{0}^{\pi} \frac{\exp[i(\pi/2)Re^{it}]iRe^{it}}{1-R^2e^{2it}}\,dt.$$

This one's easy to estimate: Slap absolute values on everything to see the integrand is bounded above by $R/(R^2-1).$ (The fact that $\sin t\ge 0$ in $[0,\pi]$ comes in here.) As $R\to \infty,$ the integral $\to 0.$

The second small semicircle: Just like the first, giving a limit of $-\pi/2.$

So we have

$$I(r,R) + \text{ integrals over semicirles } = 0.$$

Our works shows that if $R\to \infty$ and $r\to 0$ (let $r=1/R$ if you like) we get

$$\int_{-\infty}^\infty \frac{\cos(\pi/2)x}{1-x^2 } = -(-\pi/2-\pi/2) =\pi.$$

Added later: Comment on errors you may have made. The problems start with "calculate instead"

$$2Re\int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz.$$

I'm not sure why you changed $x$ to $z;$ we're still on the real axis at this point. But that's a minor thing. The big problem is the denominator. As others pointed out, it should be $1-z^2.$ It's important to get this right.

Onward to the pizza slice:

$$2Re\int_{0}^{\infty}\frac{e^{\frac{\pi}{2}zi}}{1-e^{\pi zi}}dz=\int_{0}^{2\pi}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}d\theta+\int_{0}^{R}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}dR+\int_{0}^{R}\frac{e^{\frac{\pi}{2}\theta i}}{1-e^{\pi\theta i}R^{2}}dR.$$

The minor things: You have the same integral added to itself at the end? Also, $dR$ is strange, as $R$ is a limit of integration. And we have an integral over $[0,\infty)$ equal to a sum of integrals over finite intervals?

I'll stop here for now. Can you explain the strategy? What is the pizza slice contour? We can converse on this if you like.

$e^{\pi z/2}$ should be $e^{\pi i z /2}$. $\int_{-\infty}^\infty f(x) dx$ doesn't converge conditionally either, it diverges because of the poles at $\pm 1$. We can show that the original integral is equal to $\operatorname {v. ! p.} \int_{-\infty}^\infty f(x) dx$ though. Or to $$\int_{-\infty - i0}^{\infty - i0} \frac {e^{\pi i x/2}} {2 (1 - x^2)} dx.$$ — Maxim, Aug 22 '20 at 15:30

score 1 · Accepted Answer · answered Aug 25 '20 at 08:17

You were really close. only one issue: let's say that the function within the pizza is $f_n$ and the limit is $f$. You assume that there $f_{n}\begin{array}{c} loc\\ \nRightarrow \end{array}f$ (locally uniformly convergence). which isn't correct. so is the solution completely wrong? no. if we split a circle from this area, with radius as small as we want: $\lim_{\delta\rightarrow0}\mathfrak{R\textrm{ }\int_{|\textrm{z-1|=\ensuremath{\delta}}}}\frac{e^{\frac{\pi}{2}z}dz}{z^{2}-1}=\lim_{\delta\rightarrow0}\mathfrak{R\textrm{ }\int_{0}^{2\pi}}\frac{e^{\frac{\pi}{2}e^{\theta i}\delta i+1}dz}{e^{\theta i}+2}d\theta=\mathfrak{R\textrm{ }}\int_{0}^{2\pi}\frac{1}{2}=\pi$ which leads to: $\int_{-\infty}^{\infty}\frac{\cos\frac{\pi}{2}x}{x^{2}-1}=\pi$

calculate: $\int_{-\infty}^{\infty}\frac{\cos\frac{\pi}{2}x}{1-x^{2}}dx$ using complex analysis ; detect my mistake

3 Answers3