# Contour Integration Part

I'm trying to evaluate the following integral, and I'm getting stuck on one part. Here's the integral:

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

Basically, I'm converting this to the complex plane and performing a contour integration over the top half of the plane (semi-circle). Further, I'm looping around the singularity at z=0. Now, I'm fine with all of the integrals except for the integral involving the loop around the singularity. For this guy, I end up with the following. I have no idea how to evaluate this in the limit as r goes to 0. Any idea how to proceed?

$$\int_{\pi}^0 \frac{e^{ire^{i\theta}}}{r^2e^{i2\theta}+1} d\theta$$

• Since it's a simple pole, and you're going clockwise, it will evaluate to $-\pi i \text{Res}\left[ \frac{e^{iz}}{z(z^{2}+1)}, 0 \right]$. – Random Variable Mar 23 '14 at 4:42
• @Random Variable Wait, is that for the second integral? I don't think that will work, because the singularity is NOT enclosed. – Incognito Mar 23 '14 at 4:45
• Yes. It's sometimes referred to as the fractional residue theorem. Check out theorem 9 in the following paper. math.umn.edu/~edman/tex/CA_prelim.pdf – Random Variable Mar 23 '14 at 4:51

Write $$f(z)=\frac{e^{iz}}{z(z^2+1)}$$ and let $C$ be the clockwise semi-circular contour of radius $r$ about $0$. Now $f$ has a simple pole at $z=0$ with residue $1$, so $$f(z)-\frac{1}{z}$$ has a removable singularity at $z=0$. Therefore, for a suitable constant $c$, the function $$g(z)=\cases{f(z)-1/z&if z\ne0\cr c&if z=0\cr}$$ is continuous. Now $$\left|\int_C g(z)\,dz\right|\le (\pi r)\max_{|z|\le r}|g(z)|\to (\pi)(0)|c|=0$$ as $r\to0$, and $$\int_C\frac{1}{z}\,dz=\int_\pi^0 \frac{ie^{i\theta}}{e^{i\theta}}\,d\theta=-\pi i\ .$$ Therefore $$\int_C f(z)\,dz=\int_C \Bigl(g(z)+\frac{1}{z}\Bigr) dz\to -\pi i$$ as $r\to0$.
• how does $f(z)-\frac{1}{z}$ have a removable singularity? – Incognito Mar 23 '14 at 5:01
• Because $f(z)$ has a Laurent series $\frac{1}{z}+\cdots$, so if you subtract the $\frac{1}{z}$, the remaining series has no terms with negative powers of $z$. – David Mar 23 '14 at 5:06
$\ds{\Large\tt\mbox{ADDENDA}}$ ( Contour Integration ): \begin{align} \color{#00f}{\large% \pp\int_{-\infty}^{\infty}{\sin\pars{x} \over x\pars{x^{2} + 1}}\,\dd x} &= \lim_{\epsilon \to 0^{+}}\Im\bracks{% \int_{-\infty}^{-\epsilon}{\expo{\ic x} \over x\pars{x^{2} + 1}}\,\dd x + \int_{\epsilon}^{\infty}{\expo{\ic x} \over x\pars{x^{2} + 1}}\,\dd x} \\[3mm]&=\lim_{\epsilon \to 0^{+}}\Im\bracks{% 2\pi\ic\,{\expo{\ic\pars{\ic}} \over \ic\pars{\ic + \ic}}\ -\ \overbrace{\int_{\pi}^{0} {\exp\pars{\ic\epsilon\expo{\ic\theta}} \over \epsilon\expo{\ic\theta}\pars{\epsilon^{2}\expo{2\ic\theta} + 1}} \,\pars{\epsilon\expo{\ic\theta}\ic\,\dd\theta}} ^{\ds{\to -\ic\pi\ \mbox{when}\ \epsilon \to 0^{+}}}} \\[3mm]&=-\pi\expo{-1} + \pi =\color{#00f}{\large{\pars{-1 + \expo{}}\pi \over \expo{}}} \approx 1.9859 \end{align}
• @user108149 I wrote an alternative solution since I saw $\tt\mbox{@David}$ already did it by contour integration. – Felix Marin Mar 23 '14 at 20:24