# Evaluate an integral with Residue theorem

Evaluate the following integral with Residue Theorem $$\int_0^{\infty}\frac{x\sin x}{x^2+a^2}$$ where $a$ is real number.

Basic instinct tells us to consider the hemicircle contour $\gamma=[-R,R]+Re^{it},t\in[0,\pi]$,which contain a simple pole $ai(a>0)$ inside,and if let $R$ goes to $\infty$ the integral on $Re^{it}$ should goes to zero,then we can find the integral on the real axis.But the problem here is that on the circle $|z\sin z|$ basically behaves like $|z|^2$ which is the same as the numerator.

If we substitute $\sin z$ with $e^{iz}$,then the imaginary part of the integral on real axis will equal to the required one.But still there's no satisfying estimate on the circle for $e^{iz}$.I was stuck here and seems that consider a rectangular contour will not avoid such a problem.

Any kind of help will be great.

• Would it be easier for you to show that $I(b)=\displaystyle\int_{-\infty}^\infty\frac{\cos\big(bx\big)}{x^2+a^2}dx = \frac\pi ae^{-ab}$ , and then compute $I'(1)$ ? – Lucian May 17 '14 at 0:42
• It's an inspiring thought although I thought it will still be difficult to get appropriate estimate for $cos(bx)$ on the circle. – Daniel S. May 18 '14 at 7:25

Semicircle does just fine for this problem. On the semicircle, for large $R$, the integral has a magnitude bounded by
$$R^2 \int_0^{\pi} d\theta \, \frac{e^{-R \sin{\theta}}}{R^2-a^2} \le \frac{2 R^2}{R^2-a^2} \int_0^{\pi/2} d\theta \, e^{-2 R \theta/\pi} \le \frac{\pi}{R}$$
which obviously vanishes as $R \to \infty$.