I am looking to evaluate to evaluate the real integral $I=\int_{-\infty}^{\infty}\frac{x\sin x}{x^2+a^2}\,dx$ ($a>0$) using Cauchy's residue formula. My strategy is to use the residue theorem to calculate the value of a contour that encircles the a pole and has a side coincides with the real axis. My hope was that the other sides of the contour would be simple to evaluate, but I am struggling to find an appropriate contour.

Finding a Pole

Defining $f(z)=\frac{z\sin z}{z^2+a^2}$, we quickly see that $1/f$ vanishes at $z_{0}=ai$. Moreover one can check that $g(z)=\begin{cases} \frac{1}{f(z)} \text{ if } z\neq z_{0},\\ 0 \text{ if } z= z_{0}.\end{cases} $ is holomorphic, so $ai$ is indeed a pole of $f$.

Calculating the Residue

Writing $\frac{z\sin z}{z^2+a^2}=\frac{z\sin z}{(z+ai)(z-ai)}$ we see that $\text{res}_{z_{0}}f=$$\lim_{z\to ai}\frac{z\sin z}{(z+ai)}=\frac{\sin ai}{2}$.

Finding a "nice" Contour

By the residue theorem we know that $\int_{\gamma}f(z)\,dz=\pi i\sin ai $ where $\gamma$ is a closed contour that contains our pole (note $\gamma$ doesn't contain other poles). I was hoping to use the semi-circle of radius $R$ with its base along the real line so that $I=\pi i\sin ai -\lim_{R\to\infty}\int_{\gamma_{1}}f(z)\,dz$ where $\gamma_{1}$ is the contour along the arc of the semi-circle. I'm finding it hard to evaluate $\int_{\gamma_{1}}f(z)\,dz$, is there a better choice of contour?

  • $\begingroup$ Poles are at $z_{1/2}=\pm ai$. $\endgroup$
    – MrYouMath
    Oct 1, 2015 at 11:23
  • $\begingroup$ Am aware, but I only need a single pole for my approach. $\endgroup$ Oct 1, 2015 at 11:27

1 Answer 1



Consider $$f(z)=\frac{ze^{iz}}{1+z^2}$$ and integrate on $$\{[-r,r]\mid r>a\}\cup\{re^{i\theta}\mid \theta\in[0,\pi]\}.$$

  • $\begingroup$ Don't you need to take the imaginary part of your integral to get the OP integral? $\endgroup$
    – MrYouMath
    Oct 1, 2015 at 11:26
  • $\begingroup$ Yes, of course. $\endgroup$
    – Surb
    Oct 1, 2015 at 11:27
  • $\begingroup$ So using Cauchy integral formula I got $\frac{\pi i}{e}$ for the integral over the base, struggling with the arc though. $\endgroup$ Oct 1, 2015 at 11:54
  • $\begingroup$ Using the residue formula I have $\frac{1-2\pi i}{2e}$ for the integral over the arc $\endgroup$ Oct 1, 2015 at 12:02
  • $\begingroup$ I don't know, I didn't do the computation. If it's this, then take the limit when $r\to\infty $ and conclude. $\endgroup$
    – Surb
    Oct 1, 2015 at 12:05

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