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I am trying to show

$$\int_{0}^{\infty}\frac{\sin^{3}(x)}{x^{3}}\mathrm dx = \frac{3\pi}{8}.$$

I believe the contour I should use is a semicircle in the upper half plane with a slight bump at the origin, so I miss the singularity.

Lastly I have the hint to consider $$\int_{0}^{\infty}\frac{e^{3iz}-3e^{iz}+2}{z^{3}}\mathrm dz$$

around the contour I mentioned. Thanks for any help or hints!


marked as duplicate by Jack D'Aurizio complex-analysis Nov 7 '17 at 14:22

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Your coutour will work perfectly, so I wonder why you're hesitating to proceed with your calculation. Anyway, here is a solution:

Since $\sin^3 z = (\sin 3z - 3\sin z)/4$, we have

$$\int_{0}^{\infty} \frac{\sin^3 x}{x^3} \, dx = \frac{1}{8} \Im \lim_{\epsilon \to 0} \int_{\mathbb{R} \backslash (-\epsilon, \epsilon)} \frac{3 e^{iz} - e^{3iz} -2}{z^3} \, dz,$$

where the term $-2$ is introduced in order to cancel out the pole of order 3, without affecting the value of the integral. Consider the counterclockwise-oriented upper semicircle $C$ of radius $R$, centered at the origin, with semicircular indent of radius $\epsilon$. Let $\Gamma_{R}^{+}$ and $\gamma_{\epsilon}^{-}$ denote semicircular arcs of $C$ of radius $R$ and $\epsilon$, respectively. If we put

$$f(z) = \frac{3 e^{iz} - e^{3iz} - 2}{z^3},$$

then we find that

  • On $\Gamma_R^+$, we have $|f(z)| \leq 6R^{-3}$ and thus $$\int_{\Gamma_{R}^{+}} f(z) \, dz \to 0 \quad \text{as } R \to \infty.$$

  • Notice that $$ f(z) = \frac{3}{z} + O(1) \quad \text{near } z = 0. $$ So by the direct computation, $$\int_{\gamma_{\epsilon}^{-}} f(z) \, dz = -\int_{0}^{\pi} f(\epsilon e^{i\theta}) i\epsilon e^{i\theta} \, d\theta = -\int_{0}^{\pi} (3i + O(\epsilon)) \, d\theta \to -3\pi i \quad \text{as } \epsilon \to 0.$$ (This is exactly the same as $-\pi i$ times the residue of $f$ at $z = 0$. The emergence of residue can be attributed to the fact that $f$ has only simple pole at $z = 0$.)

Since $f(z)$ has no pole on the region enclosed by $C$, we have $$\lim_{\epsilon \to 0} \int_{\mathbb{R} \backslash (-\epsilon, \epsilon)} \frac{3 e^{iz} - e^{3iz} - 2}{z^3} \, dz = 3\pi i.$$ This proves the desired identity.

  • $\begingroup$ The key step of this proof is the computation of the integral over $\gamma^-$. You do not explain this, although this is not a direct application of the (usual form of the) residue theorem. $\endgroup$ – Did Dec 4 '11 at 13:17
  • $\begingroup$ @pel, since you accepted this answer, I wonder how you justify the step I mentioned in my previous comment. $\endgroup$ – Did Dec 4 '11 at 13:19
  • $\begingroup$ @SangchulLee Sorry to ping on a six-year-old answer but I am having trouble with the part where you introduce the number $-2$. I recognize that this effectively cancels the pole of order three. Probably missing something simple but cannot justify how adding a non-zero number would not affect the value of the integral as you claim. Furthermore, for the integral of $\sin^5(x)/x^5$ we would have a higher-order pole, namely of order $5$. So this tecnique of won't work (would still be left with a non-simple pole of order $3$). $\endgroup$ – Linear Christmas Apr 26 '17 at 18:51

$$ {\rm J}\left(\alpha\right) \equiv. \int_{-\infty}^{\infty}{\sin^{3}\left(\alpha x\right) \over x^{3}}\,{\rm d}x\,, \qquad\qquad {\rm J}\left(0\right) = 0\,,\quad {\rm J}\left(1\right) = ? $$

$$ {\rm J}'\left(\alpha\right) = {3 \over 2}\int_{-\infty}^{\infty} {\sin\left(\alpha x\right)\sin\left(2\alpha x\right) \over x^{2}}\,{\rm d}x = {3 \over 4}\int_{-\infty}^{\infty} {\cos\left(\alpha x\right) - \cos\left(3\alpha x\right) \over x^{2}}\,{\rm d}x $$

$$ {\rm J}''\left(\alpha\right) = {3 \over 4}\int_{-\infty}^{\infty} {-\sin\left(\alpha x\right) + 3\sin\left(3\alpha x\right) \over x}\,{\rm d}x = {3 \over 4}\,2\pi\,{\rm sgn}\left(\alpha\right) = {3 \over 2}\,\pi\,{\rm sgn}\left(\alpha\right) $$

$$ {\rm J}'\left(\alpha\right) = {3 \over 2}\,\pi\left\vert\alpha\right\vert\,, \quad {\rm J}\left(1\right) = \int_{-\infty}^{\infty}{\sin^{3}\left(x\right) \over x^{3}}\,{\rm d}x = {3 \over 2}\,\pi\int_{0}^{1}\left\vert\alpha\right\vert\,{\rm d}\alpha = {3 \over 4}\,\pi $$

$$ \int_{0}^{\infty}{\sin^{3}\left(x\right) \over x^{3}}\,{\rm d}x = {1 \over 2}\,\int_{-\infty}^{\infty}{\sin^{3}\left(x\right) \over x^{3}}\,{\rm d}x = {3 \over 8}\,\pi $$


$\int \frac{\sin^3x}{x^3}\,\mathrm dx$

$=\frac{1}{4}\int\frac{3\sin x-\sin3x}{x^3}\,\mathrm dx$

$=\frac{1}{4}\left[(3\sin x-\sin3x)(\frac{-1}{2x^2})+\frac12\int \frac{3\cos x-3\cos3x}{x^2}\right]\,\mathrm dx$

$=\frac14\left[(3\sin x-\sin3x)(\frac{-2}{x^2})+\frac32\left[(\cos x-\cos3x)(\frac{-1}{x})+\int \frac{-\sin x+3\sin3x}{x}\right]\right]\,\mathrm dx$

$=\frac14\left[(3\sin x-\sin3x)(\frac{-2}{x^2})+\frac32(\cos x-\cos3x)(\frac{-1}{x})+\frac32\int \frac{-\sin x+3\sin3x}{x}\right]\,\mathrm dx$

For limits $x=0$ and $x \to \infty$ the first two terms disappear[each vanishes for both the limits].

The third term, on considering the integral $\int_0^\infty\frac{\sin x}{x}\, \mathrm dx=\pi/2$ evaluates to:

$\frac{3}{8}[-\pi/2+3\pi/2] $


  • $\begingroup$ Anamitra, if I may: the trigonometric functions in $\LaTeX$ look much better when you use \sin and \cos instead of sin and cos. :) $\endgroup$ – J. M. is a poor mathematician Dec 4 '11 at 12:28
  • $\begingroup$ I tried to make it use align, but for some reason the live preview in the editor showed it as completely broken, so I ended up just making it use \cos and \sin. $\endgroup$ – kahen Dec 4 '11 at 14:24
  • $\begingroup$ On Evaluation of $\int_0^{\infty}\frac{\sin x}{x}$:Consider the Fourier Transform of f(x)=1 for|x|<=1 and zero otherwise.$F[f(x)]=2\frac{\sin s}{s}$ for $s\ne 0$ and 2 if s=0. By inversion we get,$\frac{1}{2\pi}\int_{-\infty}^{\infty}\frac{2\sin s}{s}e^{-isx} ds=1$ for |x|<=1. For x=0 we have,$\frac{1}{2\pi}\int_{-\infty}^{\infty} \frac{2 \sin s}{s}ds=1$. The result follows.You may obtain the reult by ordinary integration also. $\endgroup$ – Anamitra Palit Dec 4 '11 at 15:42

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