# Show that $\int\nolimits^{\infty}_{0} x^{-1} \sin x dx = \frac\pi2$ [duplicate]

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: http://www.roadsolutions.ox.ac.uk/corrections.html)

## marked as duplicate by Guy Fsone, José Carlos Santos, Dando18, Nosrati, AquaNov 7 '17 at 17:44

• Ah, here it is! – J. M. is a poor mathematician 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

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 $$\int_{0}^\infty \frac{\sin x}{x}dx=\frac{\pi}{2}.$$