0
$\begingroup$

Mistake in calculating $\int_{-\infty}^{\infty}\frac{\sin(x)}{x(1+x^2)}$

I want to use Residue theorem. Consider the function $$f(z)=\frac{e^{iz}}{z(1+z^2)}$$ We integrate it over a semicircle $C_R$ with a small semicircle $C_{\epsilon}$ around $0$ to avoid the pole at $0$. I showed that the integral over $C_R$ goes to $0$ using Jordan's lemma. Now around $C_\epsilon$ we can use indentation lemma. The Residue at $0$ is $1$ thus the integral over $C_\epsilon$ as $\epsilon \to 0$ is $\pi$ (I think here is my mistake but I don't see how this is wrong). Thus we have $$\pi +\int_{-\infty}^{\infty}\frac{e^{iz}}{z(1+z^2)}=2\pi i(Residue)$$ The residue at $z=i$ is $-\frac{1}{2e}$ thus taking imaginary parts of both sides I get that the integral is $-\pi/e$ which is incorrect. I suspect that the integral over $C_{\epsilon}$ goes to $\pi *i$ but i cannot see why that would be the case.

$\endgroup$
2
  • 1
    $\begingroup$ The integral about a small arc of angle $\theta$ about $a\in\mathbb{C}$ as $\epsilon\to0$ is given by$$i\cdot\theta\cdot\text{res}(f,a)$$ $\endgroup$ Aug 14, 2020 at 15:22
  • $\begingroup$ @PeterForeman ugh, you are right! Thank you :P $\endgroup$
    – 2132123
    Aug 14, 2020 at 15:24

2 Answers 2

2
$\begingroup$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{-\infty}^{\infty}{\sin\pars{x} \over x\pars{1 + x^{2}}}\,\dd x & = \Im\int_{-\infty}^{\infty}{\expo{\ic x} - 1\over x\pars{x + \ic}\pars{x - \ic}}\,\dd x \\[5mm] & = \Im\bracks{2\pi\ic\, {\expo{\ic\pars{\ic}} - 1 \over \pars{\ic}\pars{\ic + \ic}}} = 2\pi\,{{\expo{-1} - 1 \over -2}} \\[5mm] & = \bbx{\large{\pi\pars{\expo{} - 1} \over \expo{}}} \approx 1.9859 \\ & \end{align} The integral path is "closed" with a semicircle in the upper complex plane.

$\endgroup$
2
  • $\begingroup$ That is a cool way of getting rid of the pole, as it does not influence the imaginary part! Thank you. $\endgroup$
    – 2132123
    Aug 14, 2020 at 19:22
  • $\begingroup$ @2132123 The main purpose is to avoid the integral divergence at $\displaystyle x = 0$. Thanks. $\endgroup$ Aug 14, 2020 at 19:31
1
$\begingroup$

Let's parameterize $C_{\epsilon}$ as $z(t) = \epsilon e^{i\pi - it}= -\epsilon e^{-it}$ for $0\leq t \leq \pi$. Note that $z(0) = -\epsilon$, $z(\pi) = \epsilon$, and $z(t)$ traverses the upper semicircle from left to right.

Since this means $dz = i\epsilon e^{-it}\; dt$, we then have

$$\int_{C_{\epsilon}}\frac{e^{iz}}{z(1+z^2)}\; dz = \int_0^{\pi}\; \frac{e^{-i\epsilon e^{-it}}}{-\epsilon e^{-it}(1+\epsilon^2 e^{-2it})} (i\epsilon e^{-it})\; dt$$ $$= -i\int_0^{\pi}\; \frac{e^{-i\epsilon e^{-it}}}{1+\epsilon^2 e^{-2it}} \; dt$$

The integrand converges to $1$ as $\epsilon \to 0$, so the integral converges to $$-it\Bigg|_{t=0}^{\pi} = -i\pi$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .