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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?

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

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    $\begingroup$ Since it's a simple pole, and you're going clockwise, it will evaluate to $-\pi i \text{Res}\left[ \frac{e^{iz}}{z(z^{2}+1)}, 0 \right]$. $\endgroup$ – Random Variable Mar 23 '14 at 4:42
  • $\begingroup$ @Random Variable Wait, is that for the second integral? I don't think that will work, because the singularity is NOT enclosed. $\endgroup$ – Incognito Mar 23 '14 at 4:45
  • $\begingroup$ 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 $\endgroup$ – Random Variable Mar 23 '14 at 4:51
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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 $r\to0$.

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  • $\begingroup$ how does $f(z)-\frac{1}{z}$ have a removable singularity? $\endgroup$ – Incognito Mar 23 '14 at 5:01
  • $\begingroup$ 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 $z$. $\endgroup$ – David Mar 23 '14 at 5:06
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} \color{#00f}{\large% \int_{-\infty}^{\infty}{\sin\pars{x} \over x\pars{x^{2} + 1}}\,\dd x}&= \Im\pp\int_{-\infty}^{\infty}{\expo{\ic x} \over x\pars{x^{2} + 1}}\,\dd x = \Im\int_{-\infty}^{\infty}{\expo{\ic x} \over x^{2} + 1}\bracks{% {1 \over x + \ic 0^{+}} + \ic\pi\,\delta\pars{x}}\,\dd x \\[3mm]&= \Im\int_{-\infty}^{\infty}{\expo{\ic x} \over \pars{x + \ic 0^{+}}\pars{x^{2} + 1}} \,\dd x + \pi =\Im\bracks{2\pi\ic\,{\expo{\ic\pars{\ic}} \over \pars{\ic + \ic 0^{+}} \pars{\ic + \ic}}} + \pi \\[3mm]&=-\pi\expo{-1} + \pi =\color{#00f}{\large{\pars{-1 + \expo{}}\pi \over \expo{}}} \approx 1.9859 \end{align}

$\ds{\Large\tt\mbox{ADDENDA}}$ ( 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}

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  • $\begingroup$ need to do it with contour integration..... I figured it out though. $\endgroup$ – Incognito Mar 23 '14 at 18:16
  • $\begingroup$ @user108149 I wrote an alternative solution since I saw $\tt\mbox{@David}$ already did it by contour integration. $\endgroup$ – Felix Marin Mar 23 '14 at 20:24
  • $\begingroup$ @user108149 I already added the contour integration. Thanks. $\endgroup$ – Felix Marin Mar 23 '14 at 21:19

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