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 '15 at 11:23
  • $\begingroup$ Am aware, but I only need a single pole for my approach. $\endgroup$ Oct 1 '15 at 11:27


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 '15 at 11:26
  • $\begingroup$ Yes, of course. $\endgroup$
    – Surb
    Oct 1 '15 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 '15 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 '15 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 '15 at 12:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.