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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)

Rodrigo de Azevedo
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UGPhysics
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    What is your question? – Phira Sep 17 '11 at 08:31
  • How can I demonstrate the equality using the approach given. – UGPhysics Sep 17 '11 at 08:34
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    What have you tried? This is only slightly harder than the usual semicircular contour, you just have a small indent to dodge the singularity at the origin. The method is still identical to the contour integration problems you have done before. Also, is there a typo? You should be integrating $z^{-1} e^{iz}$. – Ragib Zaman Sep 17 '11 at 08:41
  • Far far?: Null. I need a full, correct answer to pore over to get an idea of the underlying concepts. (I'm using this as a pedagogical question. [Ref: R. Penrose, The Road to Reality, Ch. 7, p. 129, Problem [7.5]]). – UGPhysics Sep 17 '11 at 08:44
  • @UGP Have you seen any full treatment of a contour integration problem? – Phira Sep 17 '11 at 08:45
  • No. But I'm currently reading through Ch. 7 of Penrose's book: there's a modest but useful bit of info on the topic. Plus, I have several ugrad CA books- Complex Analysis, 3rd. [J. Bak, D. J. Newman] (Springer, 2010) amongst them. – UGPhysics Sep 17 '11 at 08:52
  • @UGPhysics In that case, I apologize for my previous comment. I has assumed you had some familiarity with simpler contour integration problems. We could show you a solution, but it would need some basic tools from complex analysis. Have you seen Cauchy's integral formula before? – Ragib Zaman Sep 17 '11 at 09:12
  • A related question. I'm quite sure this is a duplicate of something, but I can't find the other question... – J. M. ain't a mathematician Sep 17 '11 at 09:13
  • @Ragib: You needn't worry about my background. I will keep the response as a record for perusal as I learn the subject more. ... (However, cursorily, I've encountered the concept- albeit in a very sketchy manner - in Penrose's book.) – UGPhysics Sep 17 '11 at 09:21
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    Ah, here it is! – J. M. ain't a mathematician Sep 17 '11 at 09:41
  • @J.M.: Thanks! :) -cheers – UGPhysics Sep 17 '11 at 09:46
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    @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 09:48
  • Yup! :) ... I'm using it to appetize my tummy. :) ... Once I finish the chapter I plan on using the standard textbooks I have. – UGPhysics Sep 17 '11 at 09:51
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    @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
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    @Eric There are different solutions in answers there, I believe (and yes, maybe that question should be edited). – Grigory M Sep 17 '11 at 17:04
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    @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
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    I voted to reopen for the same reasons as Eric. – t.b. Sep 17 '11 at 17:17
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    Ditto Theo and Eric. – anon Sep 17 '11 at 22:36
  • Related: https://math.stackexchange.com/q/294490/339790 – Rodrigo de Azevedo Jun 06 '20 at 14:48

1 Answers1

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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,

Eric Naslund
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