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Recently I met an integral which is $\int_0^\infty \left(\frac{\sin x}{x}\right)^3 \; dx$. I get the result is $3\pi/8$ by using Mathematica, but I cannot derive it independently. So I hope someone can help me. It is my first time to ask questions on math.stackexchange. Please tell me if I have done something wrong, thanks. At last, thanks all the people for your precious time.

Sasha
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dezhonger
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  • Have you tried Contour integrals? – Ali Jun 16 '13 at 04:55
  • You need to use complex analysis to do this integral. – Secret Math Jun 16 '13 at 04:58
  • @SecretMath Differentiation under the integral sign might also work. – Potato Jun 16 '13 at 04:59
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    Some integration by parts lets you reduce this problem to knowing $\int^{\infty}_{0} \dfrac{\sin x}{x} dx = \pi/2.$ This last integral can be done using contour integration or differentiation under the integral sign. – Ragib Zaman Jun 16 '13 at 05:04
  • Does Mathematica have an option where it shows details of its calculations? – Gerry Myerson Jun 16 '13 at 05:54
  • @Gerry Myerson I am a Mathematica beginner. I can only use it by inputing a math formula and wait result.I remember if has the option in http://www.wolframalpha.com/.Could you please how can I get details of its calculations in Mathematica software 9.0? thanks. – dezhonger Jun 16 '13 at 06:30
  • One way to compute $\int_0^{\infty}\left(\frac{\sin x}{x}\right)^m dx$ for $m \in \mathbb{Z}_{+}$ using contour integral can be found here. – achille hui Jun 16 '13 at 06:33
  • My question was not rhetorical. I don't know whether Mathematica has that option, I just thought it might be worth your while to find out. – Gerry Myerson Jun 16 '13 at 07:58

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Integrating by parts: $$ \int_0^\infty \frac{\sin^3(x)}{x^3} \mathrm{d}x = \int_0^\infty \sin^3(x) \mathrm{d} \left(-\frac{1}{2 x^2} \right) = \frac{3}{2} \int_0^\infty \cos(x) \frac{\sin^2(x)}{x^2} \mathrm{d}x = \ldots $$ Integrating by parts again: $$ \ldots = \frac{3}{2} \int_0^\infty \frac{2 \cos^2(x) \sin(x) - \sin^2(x)}{x} \mathrm{d}x = \ldots $$ Using $$ 2 \cos^2(x) \sin(x) - \sin^2(x) = \frac{1}{4} \left(3 \sin(3x)-\sin(x)\right) $$ we continue: $$ \ldots = \frac{3}{8} \left( \underbrace{\int_0^\infty \frac{3 \sin(3x)}{x} \mathrm{d}x }_{3 \frac{\pi}{2}}- \underbrace{\int_0^\infty \frac{\sin(x)}{x} \mathrm{d}x}_{\frac{\pi}{2}} \right) = \frac{3}{8} \pi $$ where we used the value of the Dirichlet integral $\int_0^\infty \sin(x) \frac{\mathrm{d}x}{x} = \frac{\pi}{2}$ and a change of variable $$ \int_0^\infty \frac{\sin(3x)}{x} \mathrm{d}x \stackrel{3x = y}{=} \int_0^\infty \frac{\sin(y)}{y} \mathrm{d}y = \frac{\pi}{2} $$

Sasha
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