2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $I'(1)$ ? $\endgroup$ – Lucian May 17 '14 at 0:42
  • $\begingroup$ It's an inspiring thought although I thought it will still be difficult to get appropriate estimate for $cos(bx)$ on the circle. $\endgroup$ – Daniel S. May 18 '14 at 7:25
4
$\begingroup$

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 $R \to \infty$.

$\endgroup$

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.