I am meant to use the residue theorem to show that

$\int\limits_{-\infty}^\infty \frac{\cos t}{(t^2+1)^2}dt=\frac{\pi}{e}$.

So far I have deduced that I should take a contour over $\alpha$ the path from $-r$ to $r$ along with the semi-circle connecting $-r$ to $r$. Then I should take the limit as $r$ goes to infinity, show that the integral along the semicircle portion vanishes, and thus, by the residue theorem and the fact that the integrand has a simple pole at $i$, the improper integral is equal to $2\pi i$ times the residue of the integrand at $i$. Can someone tell me if I am approaching this correctly and possibly explain some of the details because I am having trouble.

  • $\begingroup$ Is it a simple pole? $\endgroup$ – Mhenni Benghorbal Feb 17 '16 at 8:04
  • 1
    $\begingroup$ Ah I forgot to consider this! If I am not mistaken it is a pole of order 2 so not a simple pole. $\endgroup$ – Tony S.F. Feb 17 '16 at 8:10

Try the complex function $\;f(z)=\frac{e^{iz}}{(x^2+1)^2}\;$ on the path

$$C_R=[-R,R]\cup\Gamma_R\;,\;\;\Gamma_R:=\{Re^{it}\in\Bbb C\;;\;0<t<\pi\}\;,\;\;R\in\Bbb R^+$$

It has a single double pole within the domain enclosed by this path, namely at $\;z=i\;$, and we get$${}$$

$$\text{res}(f)_{z=i}=\frac d{dz}\left((z-i)^2f(z)\right)_{z=i}=\left.\frac d{dz}\left(\frac{e^{iz}}{(z+i)^2}\right)\right|_{z=i}=\left.\frac{ie^{iz}(z+i)-2e^{iz}}{(z+i)^3}\right|_{z=i}=$$


and then$${}$$

$$-\frac{2\pi i\cdot i}{2e}=\frac\pi e=\oint_{C_r}f(z)\,dz=\int_{-R}^R\frac{e^{ix}}{(x^2+1)^2}dx+\int_{\Gamma_R}f(z)\,dz$$


$$\left|\int_{\Gamma_R}f(z)dz\right|\le\ell(\Gamma_R)\cdot\max_{z\in\Gamma_R}|f(z)|=\frac{\pi R^2}{\min\left|\left(R^2e^{2it}+1\right|\right)^2}\le\frac{\pi R^2}{R^4}\xrightarrow[R\to\infty]{}0$$


$$\frac\pi e=\lim_{R\to\infty}\oint_{C_R}f(z)\,dz=\int_{-\infty}^\infty\frac{\cos x+i\sin x}{(x^2+1)^2}dx$$

and comparing real parts you get your result.


You're mostly on the right track, but there are a few points:

  1. you can't use the residue theorem on $$f(z) = \frac{\cos z}{(z^2+1)^2}.$$ The integral over an added semi-circle will not tend to $0$ as $R \to \infty$ (regardless of the choice of half-plane). Instead, look at $$f(z) = \frac{e^{iz}}{(z^2+1)^2}$$ and take the real part of the resulting integral. With this choice, the integral over the semi-circle will tend to $0$ when $R\to\infty$ if you choose the correct half-plane. (What is needed for $|e^{iz}|$ to be small?)

  2. Your integrand has double poles at $z=\pm i$.

  • $\begingroup$ This is helpful thanks. I was already considering using $e^{iz}$ for the reasons you stated but I hadn't considered which half plane. I think I should use the upper half plane since then I will get $e^{i(ic)}=e^{-c}$ for some positive real $c$ which will go to 0 as $c\rightarrow\infty$ right? $\endgroup$ – Tony S.F. Feb 17 '16 at 16:41
  • $\begingroup$ @TonyS.F. Yes, that's right. $\endgroup$ – mrf Feb 17 '16 at 17:01

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.