So we have $\int_{0}^{+\infty}\dfrac{x^2-a^2}{x^2+a^2}\cdot\dfrac{\sin x}{x}dx$. ($a>0$). I considered that we can just calculate the half of the imaginary part of $$\int_{-\infty}^{+\infty}\dfrac{x^2-a^2}{x^2+a^2}\cdot\dfrac{e^{ix}}{x}dx$$

I considered taking a contour of big upper half circle with radius $R$, and $[-R,-\epsilon]$,$[\epsilon,R]$, with a small lower half circle with radius $\epsilon$. I estimated that the integral of upper large circle is zero as $R$ tends to $\infty$. However, I can not seem to estimate the integral over the small circle since the rational function part is quite tricky.

Can anyone help with this? Any hint would be appreciated!

  • $\begingroup$ The complex integral you wrote doesn't converge, so I think this is not the right way to approach the problem. (Much of the time it would be, but not here.) $\endgroup$
    – Ian
    Commented Dec 2, 2015 at 21:30
  • $\begingroup$ The original problem evaluates from $0$ to infinity, is this the problem? I thought zero is a removable singularity so that I can extend it to the whole real line... I edited it, is it solvable now? $\endgroup$
    – jack
    Commented Dec 2, 2015 at 21:37
  • $\begingroup$ The singularity is removable in the original integral because of cancellations but not with the complex exponential. Now you have to come up with a contour that dodges the point $a i$... $\endgroup$
    – Ian
    Commented Dec 2, 2015 at 21:41
  • $\begingroup$ @Ian It's the right way, one considers it as a principal value integral (if one doesn't punch a hole in $\mathbb{R}$ and add the semicircle before replacing $\sin x$ with $e^{ix}$). $\endgroup$ Commented Dec 2, 2015 at 21:46

1 Answer 1


For functions $f$ with a simple pole at $z_0$, if we integrate along an arc $\gamma_{r,\theta}$ with radius $r$ and opening angle $\theta$, we have $$ \lim_{r\to 0} \int_{\gamma_{r,\theta}} f(z)\,dz = \frac{\theta}{2\pi} \operatorname{Res}(f,z_0). $$ See for example How to rigorously justify "picking up half a residue"? or Integrating $\frac {e^{iz}}{z}$ over a semicircle around $0$ of radius $\epsilon$ for some details.

In other words, you will pick up half the residue at $z=0$ (as well as the residue at $ia$ in your integral.


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