I must solve $$\int_{-\infty}^{\infty}\frac{x \text{sin}x}{x^2+4} dx$$

I simplified this to $$\int_{-\infty}^{\infty}\frac{x \text{sin}x}{x^2+4} dx = \frac{1}{i}\int_{-\infty}^{\infty}\frac{x e^{ix}}{x^2+4} dx$$

Now consider only $$\int_{-\infty}^{\infty}\frac{x e^{ix}}{x^2+4} dx $$ I considered $C_R$ to be the simple closed contour traversing anticlockwise along the real axis from $-R$ to $R$ being $\Gamma_R$ and then along the half-circle $|z|=R$ in the upper half of the plane which is $\Delta_R$

This gives $$ \text{Res}(f,2i) = \int_{\Gamma_R} f(z) dz + \int_{\Delta_R} f(z) dz $$

The first integral along $\Gamma_R$ is $\int_{-\infty}^{\infty}\frac{x e^{ix}}{x^2+4} dx$ as $R$ tends to positive infinity. The residue of $f$ at $2i$ equals $\frac{\pi i}{e^2}$

So $$ \int_{-\infty}^{\infty}\frac{x e^{ix}}{x^2+4} dx = \frac{\pi i}{e^2} - \int_{\Delta_R} f(z) dz $$ where the second integral is taken as $R$ tends to positive infinity

So I compute the second integral by the ML Lemma where $L$ is $\pi R$ and $$|\frac{ze^{iz}}{z^2+4}| <= \frac{|ze^{iz}|}{R^2-4} = \frac{R|e^{iz}|}{R^2-4} = \frac{R}{R^2-4} = M$$

So $ML = \frac{\pi R^2}{R^2-4}$ which tends to $\pi$ as $R$ tends to positive infinity. This means the final answer is $\frac{\pi i}{e^2} - \pi$ but the solution says the answer is $\frac{\pi }{e^2}$ because the second integral is equal to $0$ as $R$ tends to positive infinity and only considering the imaginary part.

Can I have some help?

  • 2
    $\begingroup$ Take a look at Jordan's Lemma $\endgroup$ Apr 17 at 15:52
  • $\begingroup$ @Mr.GandalfSauron I did. So you are saying that the integral with respect to $\Delta_R$ is $0$ as $R$ tends to positive infinity by Jordan's Lemma? That gives me the required answer. $\endgroup$
    – adisnjo
    Apr 17 at 15:55
  • 1
    $\begingroup$ WIA ML for ? (WIA = What is acronym :) $\endgroup$
    – Jean Marie
    Apr 17 at 18:07
  • 1
    $\begingroup$ @JeanMarie ML Lemma $\endgroup$
    – adisnjo
    Apr 17 at 21:10
  • $\begingroup$ Your answer isn't helpful... Why don't you say (what I have finaly found) that L is the arc length of $\gamma$, M is an upper bound for $|f|$ on $\gamma$ ? $\endgroup$
    – Jean Marie
    Apr 18 at 8:10

2 Answers 2


What you've shown is that


not that the limit of the integral is $\pi$. You will need a better estimate to show that this integral does indeed tend to $0$ as $R\to\infty$. One such is given by the following version of Jordan's lemma (slightly modified from Wikipedia):

Lemma (Jordan's lemma). Let $f$ be a continuous complex-valued function in the closed upper half-plane, and for $R>0$ let $\Delta_R$ denote the semicircular contour $\{Re^{i\theta}:\theta\in[0,\pi]\}$. Then, for $a,R>0$, $$\left\lvert\int_{\Delta_R}f(z)e^{iaz}\,\mathrm{d}z\right\rvert\leq\frac{\pi}{a}\max_{z\in\Delta_R}\lvert f(z)\rvert.$$

For a proof of it you can see the linked Wikipedia page, but it really just amounts to parametrizing the curve and using the definition of contour integration. Using this we have that

$$\left\lvert\int_{\Delta_R}\frac{ze^{iz}}{z^2+1}\,\mathrm{d}z\right\rvert\leq\pi\max_{z\in\Delta_R}\left\lvert\frac{z}{z^2+1}\right\rvert\leq\pi\max_{z\in\Delta_R}\frac{\lvert z\rvert}{\lvert z\rvert^2-1}=\pi\frac{R}{R^2-1}\to0$$

as $R\to\infty$. This gives you the result.


You missed the fact that the $ML$ is just that; an inequality.

Anyway, your approach does now allow you to compute that integral, since you were not able to prove that $\displaystyle\lim_{R\to\infty}\oint_{\Delta_R}f(z)\,\mathrm dz$. But you can do as follows: take $R,R'>0$, and then consider the integral from $-R'$ to $R$, then the integral from $R$ to $R+(R+R')i$, then the integral from $R+(R+R')i$ to $-R'+(R+R')i$, and finally the integral from $-R'+(R+R')i$ (all these integrals are path integrals along straight lines). When $R,R'\to\infty$, the first integral approaches$$\int_{-\infty}^\infty\frac{ze^{iz}}{z^2+4}\,\mathrm dz;\label{a}\tag1$$the remaining three integrals approach $0$ (this is not obvious, but it is a standard Complex Analysis statement). Therefore,$$\eqref{a}=2\pi i\operatorname{res}_{z=2i}\left(\frac{ze^{iz}}{z^2+4}\,\mathrm dz\right)=2\pi i\times\frac1{2e^2}=\frac{\pi i}{e^2}.$$

  • $\begingroup$ Yes you are right, I realised that just after I posted the question. Thank you for your help. $\endgroup$
    – adisnjo
    Apr 17 at 15:58

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