Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Show that $\int^{\infty}_{0} x^{-1} \sin x dx = \frac\pi2$ by integrating $z^{-1}e^{iz}$ around a closed contour $\Gamma$ consisting of two portions of the real axis, from -$R$ to -$\epsilon$ and from $\epsilon$ to $R$ (with $R > \epsilon > 0$) and two connecting semi-circular arcs in the upper half-plane, of respective radii $\epsilon$ and $R$. Then let $\epsilon \rightarrow 0$ and $R \rightarrow \infty$.

[Ref: R. Penrose, The Road to Reality: a complete guide to the laws of the universe (Vintage, 2005): Chap. 7, Prob. [7.5] (p. 129)]

Note: Marked as "Not to be taken lightly", (i.e. very hard!)

Update: correction: $z^{-1}e^{iz}$ (Ref:

share|cite|improve this question
Ah, here it is! – J. M. Sep 17 '11 at 9:41
@UGP Penrose's book is a very interesting book, but it is not really a good resource to actually learn the basics of any topic, it is more like an appetizer or a menu with samples. – Phira Sep 17 '11 at 9:48
@Asaf: I don't think that this is a duplicate. That question asks specifically to solve it without contour integration, whereas this question is asking specifically for contour integration. – Eric Naslund Sep 17 '11 at 10:05
@Grigory: There is no (non-deleted) solution with complex analysis on that thread. This is not an exact duplicate. – Eric Naslund Sep 17 '11 at 17:09
I voted to reopen for the same reasons as Eric. – t.b. Sep 17 '11 at 17:17

What follows is a proof of how to use contour integration to get your identity.

First, we want to change $\sin x$ into $e^{ix}$. Notice that $$\int_{0}^\infty \frac{\sin x}{x}dx=\frac{1}{2i}\lim_{\epsilon\rightarrow 0}\lim_{R\rightarrow \infty}\left(\int_{-R}^\epsilon \frac{e^{iz}}{z}dz+\int_{\epsilon}^R \frac{e^{iz}}{z}dz\right).$$ The reason we put int the limit is because the integral $\int_{-\infty}^{\infty}\frac{e^{iz}}{z}dz$ does not converge. Now, consider the semi circle or radius $R$ in the upper half plane, and modify it by going around a semi circle of radius $\epsilon<R$ in the upper half plane to avoid the point $0$. Call this countour $\Gamma_{R,\epsilon}$. Also, let $C_\epsilon^+$ denote the semi circle of radius $\epsilon$ in the upper half plane. Then by using Jordans Lemma we can show that $$\lim_{\epsilon\rightarrow 0}\lim_{R\rightarrow \infty}\left(\int_{-R}^\epsilon \frac{e^{iz}}{z}dz+\int_{\epsilon}^R \frac{e^{iz}}{z}dz\right)=\lim_{R\rightarrow \infty}\lim_{\epsilon\rightarrow 0}\left(\int_{\Gamma_{R,\epsilon}} \frac{e^{iz}}{z}dz-\int_{C_\epsilon^+}\frac{e^{iz}}{z}dz\right).$$ Now, using the residue theorem and the fractional residue theorem we see that the right hand side above equals $\pi i$. Hence $$\int_{0}^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}.$$

Hope that helps,

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.