I'm trying to evaluate the following integral:

$$\mathcal{J}=\int_{0}^{\infty}\frac{\sin x}{x\left ( 1+x^2 \right )^2}\,{\rm d}x$$

Well there are $3$ poles , one lying on the real line the other on the upper half plane and the other on the lower half plane. The residue at $z=i$ is $\displaystyle -\frac{3}{4e}$ . I'm integrating on a contour that looks like a semicircle on the upper half plane and has a branch about the origin .

Well I considered the function $\displaystyle f(z)=\frac{e^{iz}+1}{z\left ( z+1 \right )^2}$ which is clearly analytic expect for the poles. Hence if $\gamma$ denotes the contour , then:

$$\oint_{\gamma}f(z)\,{\rm d}z=\oint_{\gamma}\frac{e^{iz}+1}{z\left ( z^2+1 \right )^2}\,{\rm d}z =\oint_{\gamma}\frac{e^{iz}}{z\left ( z^2+1 \right )^2}\,{\rm d}z+\oint_{\gamma}\frac{{\rm d}z}{z\left ( z^2+1 \right )^2}=-2\pi i \frac{3}{4e}+2\pi i = 2\pi i \left ( 1-\frac{3}{4e} \right )=i \left (2\pi - \frac{3\pi}{2e} \right )$$

Hmm... applying the classical method I get that:

$$\int_{-\infty}^{\infty}f(x)\,{\rm d}x =2\pi - \frac{3\pi}{2e}$$

which is almost correct except for that $2$ in front of $\pi$. Where I have gone wrong?

P.S: I used the very obvious that the integrand is even.

  • $\begingroup$ Is the denominator $x(1+x)^2$ or $x(1+x^2)$? $\endgroup$ – Empy2 Jan 31 '15 at 16:03
  • $\begingroup$ Oups.. Typo.. It is $\displaystyle \frac{\sin x}{x\left ( 1+x^2 \right )^2}$. Mea culpa. $\endgroup$ – Tolaso Jan 31 '15 at 16:05

Depict carefully the path of integration: it is a semicircle in the upper half plane with a bulge at $z=0$ and a keyhole around $z=i$. This gives that you have to compute the residues of $f(z)=\frac{e^{iz}}{z(z^2+1)^2}$ at $z=0$ and $z=i$, but to consider only half the residue at $z=0$:

$$\mathcal{J}=\frac{1}{2}\text{Im}\int_{-\infty}^{+\infty}\frac{e^{iz}}{z(z^2+1)^2}\,dz = \frac{1}{2}\text{Im}\left(2\pi i\operatorname{Res}(f(z),z=i)+\pi i\operatorname{Res}(f(z),z=0)\right)$$ so: $$\mathcal{J} = \frac{1}{2}\text{Im}\left(2\pi i\cdot \frac{-3}{4e}+\pi i\right)=\frac{1}{2}\left(\pi-\frac{3\pi}{2e}\right)=\color{red}{\frac{\pi}{2}\left(1-\frac{3}{2e}\right)}.$$

  • $\begingroup$ @Tolaso: fixed. Luckily the proof is almost the same, just the residue in $z=i$ is different. $\endgroup$ – Jack D'Aurizio Jan 31 '15 at 16:08
  • $\begingroup$ Unfortunately, I'm a litte lost. Can you explain me how you computed the residue at $z=0$ and $z=1$ ? I cannot seem to handle the residues when they come with branch... If I could see the details.. that would be fine... Also why did you take the half of the residue at $z=0$? $\endgroup$ – Tolaso Jan 31 '15 at 16:10
  • $\begingroup$ Branch? What branch? $z=0$ and $z=i$ are a simple pole and a double pole for $\frac{e^{iz}}{z(z^2+1)^2}$. $\endgroup$ – Jack D'Aurizio Jan 31 '15 at 16:16
  • 2
    $\begingroup$ Can't you accept that W|A could be wrong and Jack right ? I got the same result as Jack's numerically. $\endgroup$ – Claude Leibovici Jan 31 '15 at 16:57
  • 1
    $\begingroup$ Actually I was wrong.. Jack was right all along.. I ran the calculations now...and i got my mistake. Thanks Jack.. ! Got it. $\endgroup$ – Tolaso Jan 31 '15 at 18:08

Another way to evaluate this integral is to use Parseval's theorem, which states that for functions $f$ and $g$ with respective Fourier transforms $F$ and $G$, we have

$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k) $$

Here $f(x) = \sin{x}/x$ and $g(x) = (1+x^2)^{-2}$. Thus, $F(k) = \pi$ when $k \in [-1.1]$ and $0$ otherwise and $G(k) = (\pi/2) (|k|+1) e^{-|k|}$. The integral is then

$$\frac{\pi}{8} \int_{-1}^1 dk \, (|k|+1) e^{-|k|} = \frac{\pi}{4} \int_0^1 dk (k+1) e^{-k} $$

$$\int_0^1 dk \, e^{-k} = 1-e^{-1}$$ $$\int_0^1 dk \,k \, e^{-k} = -e^{-1}+ \int_0^1 dk \, e^{-k} = 1-2 e^{-1}$$

Therefore, the integral is equal to

$$\frac{\pi}{4} \left (2-\frac{3}{e} \right ) = \frac{\pi}{2} \left (1-\frac{3}{2 e} \right )$$

  • $\begingroup$ Very nice.. I realy like the Fourier approach. $\endgroup$ – Tolaso Jan 31 '15 at 18:09

Another approach: Parameterize the integral as $$I(a)=\int_{0}^{\infty}\frac{\sin ax}{x\left ( 1+x^2 \right )^2}\,{\rm d}x$$ Take the Laplace transform, find your partial fractions, and take the inverse transform: $$\begin{align*}\mathcal{L}_s\{I(a)\}&=\int_0^\infty\int_{0}^{\infty}\frac{\sin ax}{x\left ( 1+x^2 \right )^2}e^{-as}\,{\rm d}a\,{\rm d}x\\ &=\int_0^\infty\frac{\mathcal{L}_s\{\sin ax\}}{x\left ( 1+x^2 \right )^2}\,{\rm d}x\\ &=\int_0^\infty\frac{x}{x\left ( 1+x^2 \right )^2\left(s^2+x^2\right)}\,{\rm d}x\\ &=\int_0^\infty\frac{{\rm d}x}{\left ( 1+x^2 \right )^2\left(s^2+x^2\right)}\\ &=-\frac{1}{(s^2-1)^2}\int_0^\infty\left(\frac{1}{1+x^2}-\frac{s^2-1}{(1+x^2)^2}-\frac{1}{s^2+x^2}\right)\,{\rm d}x\\ &=-\frac{1}{(s^2-1)^2}\left(\frac{\pi}{2}-\frac{\pi(s^2-1)}{4}-\frac{\pi}{2s}\right)\\ &=\frac{\pi}{4}\left(\frac{2}{s}-\frac{2}{s+1}-\frac{1}{(s+1)^2}\right)\\ I(a)&=\frac{\pi}{4}{\mathcal{L}^{-1}}_a\left\{\frac{2}{s}-\frac{2}{s+1}-\frac{1}{(s+1)^2}\right\}\\ &=\frac{\pi}{4}(2+-2e^{-a}-ae^{-a})\end{align*}$$ Finally, since $\mathcal{J}=I(1)$, you have $$\mathcal{J}=\frac{\pi}{2}-\frac{3\pi}{4e}$$

  • $\begingroup$ Very nich approach using Laplace Transformation. $\endgroup$ – Tolaso Feb 1 '15 at 18:13

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}{\sin\pars{x} \over x\pars{1 + x^{2}}^{2}}\,\dd x} = {1 \over 2}\,\Im\int_{-\infty}^{\infty}{\expo{\ic x} - 1 \over x\pars{x - \ic}^{2}\pars{x + \ic}^{2}}\,\dd x \\[5mm] = &\ {1 \over 2}\,\Im\braces{2\pi\ic\,\lim_{x\ \to\ \ic}\,\,\, \totald{}{x}\bracks{\expo{\ic x} - 1 \over x\pars{x + \ic}^{2}}} \\[5mm] = &\ \bbx{{\pi \over 2} - {3\pi \over 4\expo{}}} \approx 0.7040 \\ & \end{align}


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