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

complex-analysisintegrationresidue-calculus

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.

Best Answer

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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.

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