show that

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

using different ways

thanks for all

  • $\begingroup$ A different way to what? What have you tried? $\endgroup$ – Dan Rust Jul 27 '13 at 1:16
  • $\begingroup$ @DanielRust I think by complex integration we can solve and I tried with Divide into two parts but I write using different way just to improve my skils $\endgroup$ – mnsh Jul 27 '13 at 1:19
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    $\begingroup$ This question asks a more general question, $$\int_0^\infty\left(\frac{\sin(x)}{x}\right)^n\,\mathrm{d}x$$ $\endgroup$ – robjohn Jul 27 '13 at 4:15

Let $$f(y) = \int_{0}^{\infty} \frac{\sin^3{yx}}{x^3} \mathrm{d}x$$ Then, $$f'(y) = 3\int_{0}^{\infty} \frac{\sin^2{yx}\cos{yx}}{x^2} \mathrm{d}x = \frac{3}{4}\int_{0}^{\infty} \frac{\cos{yx} - \cos{3yx}}{x^2} \mathrm{d}x$$ $$f''(y) = \frac{3}{4}\int_{0}^{\infty} \frac{-\sin{yx} + 3\sin{3yx}}{x} \mathrm{d}x$$ Therefore, $$f''(y) = \frac{9}{4} \int_{0}^{\infty} \frac{\sin{3yx}}{x} \mathrm{d}x - \frac{3}{4} \int_{0}^{\infty} \frac{\sin{yx}}{x} \mathrm{d}x$$

Now, it is quite easy to prove that $$\int_{0}^{\infty} \frac{\sin{ax}}{x} \mathrm{d}x = \frac{\pi}{2}\mathop{\mathrm{signum}}{a}$$

Therefore, $$f''(y) = \frac{9\pi}{8} \mathop{\mathrm{signum}}{y} - \frac{3\pi}{8} \mathop{\mathrm{signum}}{y} = \frac{3\pi}{4}\mathop{\mathrm{signum}}{y}$$ Then, $$f'(y) = \frac{3\pi}{4} |y| + C$$ Note that, $f'(0) = 0$, therefore, $C = 0$. $$f(y) = \frac{3\pi}{8} y^2 \mathop{\mathrm{signum}}{y} + D$$ Again, $f(0) = 0$, therefore, $D = 0$.

Hence, $$f(1) = \int_{0}^{\infty} \frac{\sin^3{x}}{x^3} = \frac{3\pi}{8}$$

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    $\begingroup$ +1 for using the parametrization trick. $\endgroup$ – Shuhao Cao Jul 27 '13 at 4:40

Use Parseval's theorem:

$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G^*(k)$$

where $f$, $g$ and $F$, $G$ are respective Fourier transform pairs, e.g.,

$$F(k) = \int_{-\infty}^{\infty} dx \, f(x) \, e^{i k x}$$

etc. If $f(x) = \sin{x}/x$, then

$$F(k) = \begin{cases} \pi & |k| \le 1\\0 & |k| \gt 1 \end{cases}$$

Further, if $g(x) = \sin^2{x}/x^2$, then

$$G(k) = \begin{cases}\pi \left (1-\frac{|k|}{2} \right ) & |k| \le 2 \\ 0& |k| \gt 2\end{cases}$$


$$\int_{-\infty}^{\infty} dx \,\frac{\sin^3{x}}{x^3} = \frac{1}{2 \pi} \int_{-1}^1 dk \, \pi^2 \left (1-\frac{|k|}{2} \right ) = \pi - \frac{\pi}{2} \int_0^1 dk \,k = \pi-\frac{\pi}{4}$$


$$\int_{0}^{\infty} dx \,\frac{\sin^3{x}}{x^3} = \frac{3 \pi}{8}$$


You can also use contour integration techniques. For the integral

$$\int_0^{\infty} dt \frac{\sin^3{ \pi t}}{(\pi t)^3} \cos{u t}$$

I have derived a complete solution to the problem of its evaluation here using both contour integral techniques as well as the convolution theorem. You will see that the results agree for $u=0$ by a simple rescaling of the integral.

  • $\begingroup$ thanks but sorry I don't know alot about this theorem $\endgroup$ – mnsh Jul 27 '13 at 1:29
  • $\begingroup$ Fundamental for Fourier transforms; allows evaluation of a wide range of integrals that are otherwise difficult. See en.wikipedia.org/wiki/… $\endgroup$ – Ron Gordon Jul 27 '13 at 1:32

In this answer, the more general integral $$ \int_0^\infty\left(\frac{\sin(x)}{x}\right)^n\,\mathrm{d}x $$ is calculated.

Your integral is that integral for $n=3$.

A Different Way

In a fashion similar to this answer, we will use the equation $$ \frac{\mathrm{d}^2}{\mathrm{d}x^2}\frac{\sin^3(kx)}{k^3} =\frac{9\sin(3kx)-3\sin(kx)}{4k}\tag{1} $$ and the series for $0\lt x\le\pi$, $$ \sum_{k=1}^\infty\frac{\sin(kx)}{k}=\frac{\pi-x}{2}\tag{2} $$ Using $(2)$, we get $$ \begin{align} \sum_{k=1}^\infty\frac{9\sin(3kx)-3\sin(kx)}{4k} &=\frac94\frac{\pi-3x}{2}-\frac34\frac{\pi-x}{2}\\ &=\frac{3\pi}{4}-3x\tag{3} \end{align} $$ Integrating from $0$ twice to back out the derivatives taken in $(1)$ yields $$ \sum_{k=1}^\infty\frac{\sin^3(kx)}{k^3}=\frac{3\pi}{8}x^2-\frac12x^3\tag{4} $$ Set $x=1/n$ and multiply by $n^2$ to get $$ \sum_{k=1}^\infty\frac{\sin^3(k/n)}{k^3/n^3}\frac1n=\frac{3\pi}{8}-\frac1{2n}\tag{5} $$ and $(5)$ is a Riemann sum for $$ \int_0^\infty\frac{\sin^3(x)}{x^3}\,\mathrm{d}x=\frac{3\pi}{8}\tag{6} $$

  • $\begingroup$ where is the wrong in my step. we know that $$\frac{b-a}{n}\sum_{k=0}^{n-1} f\left(a+k\frac{b-a}n\right) \approx \int_a^b f(x)\ dx$$ put a=0 b=1 and take $n\rightarrow \infty$ $$\lim_{n\rightarrow \infty}(\frac{1}{n}\sum_{k=0}^{n-1} f\left(k\frac{1}n\right)) = \int_0^1 f(x)\ dx$$ $$f(x)=(\frac {\sin x} {x} )^3$$ $$\lim_{n\rightarrow \infty}(\sum_{k=0}^{\infty} \frac{1}{n} (\frac {\sin \frac{k}{n}} {\frac{k}{n}} )^3 = \int_0^1 (\frac {\sin x} {x} )^3\ dx\neq \int_0^\infty (\frac {\sin x} {x} )^3\ dx$$ $\endgroup$ – mnsh Jul 27 '13 at 15:32
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    $\begingroup$ @hmedan.mnsh: $$ \lim_{n\to\infty}\sum_{k=0}^\infty\frac1n\left(\frac {\sin\left(\frac{k}{n}\right)} {\frac{k}{n}}\right)^3\ne\lim_{n\to\infty}\sum_{k=0}^n\frac1n\left(\frac {\sin\left(\frac{k}{n}\right)} {\frac{k}{n}}\right)^3 $$ $\endgroup$ – robjohn Jul 27 '13 at 18:09
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    $\begingroup$ @hmedan.mnsh: You are correct in the formula for the Riemann Sum on a finite interval, but there are conditions under which the same idea can be extended to infinite intervals (improper Riemann Sums). The condition that applies most easily here is that $$ \sum_{k=0}^\infty\sup_{x\in[k,k+1]}|f'(x)|\lt\infty $$ $\endgroup$ – robjohn Jul 27 '13 at 18:45

Well, there's a lot of creative answers but it seems that no one bothered to put here the "follow your nose" one, so I'll add it here for completeness.

Recall the identity: $$\sin^3x = \frac{3 \sin x - \sin(3x)}{4}.$$

So we have: $$\int_{-\infty}^{+\infty} \left(\frac{\sin x}{x}\right)^3\,{\rm d}x = \int_{-\infty}^{\infty} \frac{3 \sin x - \sin (3x)}{4x^3}\,{\rm d}x. $$Let $0 < r < R$. Consider $\gamma_1$ the line segment joining $r$ to $R$, $\gamma_2$ the circular arc oriented counterclockwise joining $R$ to $-R$, $\gamma_3$ the line segment joining $-R$ to $-r$ and $\gamma_4$ the circular arc oriented clockwise joining $-r$ to $r$. Consider $\gamma = \gamma_1 \ast \gamma_2 \ast \gamma_3 \ast \gamma_4$ the concatenation. Sketch:

enter image description here

Consider the function: $$ f(z) = \frac{3e^{iz}-e^{3iz}}{4z^3}. $$The only singularity is $z = 0$ (triple pole). Expanding in Laurent: $$ f(z) = \frac{1}{4z^3}\left(3\sum_{n \geq 0}\frac{i^nz^n}{n!} - \sum_{n \geq 0}\frac{3^ni^nz^n}{n!}\right) = \sum_{n \geq 0} \left(\frac{(3-3^n)i^n}{4\cdot n!}\right)z^{n-3}. $$ Since $f$ is holomorphic in $\gamma$ and inside it, by Cauchy-Goursat we get: $$ \oint_{\gamma} f(z)\,{\rm d}z = \int_{\gamma_1} f(z)\,{\rm d}z + \int_{\gamma_2} f(z)\,{\rm d}z + \int_{\gamma_3}f(z)\,{\rm d}z + \int_{\gamma_4}f(z)\,{\rm d}z = 0.$$ Let's analyze everything sistematically.

Parametrizing $\gamma_1(x) = x$, with $r \leq x \leq R$, we have: $$ \int_{\gamma_1}f(z)\,{\rm d}z = \int_r^R \frac{3e^{ix}-e^{3ix}}{4x^3}\,{\rm d}x. $$For $\gamma_2$, we have that its lenght is $\pi R$, $|e^{iz}| = e^{{\rm Re}(iz)} = e^{-{\rm Im}(z)} < 1$, and similarly $|e^{3iz}| < 1$, since for all $z$ in $\gamma_2$ we have ${\rm Im}(z) > 0$. Hence: $$ \left|\int_{\gamma_2} f(z)\,{\rm d}z\right| \leq \frac{\pi R(3+4)}{4 R^3} = \frac{\pi}{R^2} \stackrel{R\, \to \,+\infty}{\longrightarrow} 0. $$

Parametrizing $\gamma_3^-(x) = -x$, with $r \leq x \leq R$, we have: $$ \int_{\gamma_3}f(z)\,{\rm d}z =- \int_r^R \frac{3e^{-ix}-e^{-3ix}}{4x^3}\,{\rm d}x, $$once the signs in $(-x^3) = -x^3$ and ${\rm d}z = -{\rm d}x$ cancel each other. Notice here that: $$ \int_{\gamma_1}f(z)\,{\rm d}z + \int_{\gamma_3}f(z)\,{\rm d}z = 2i\int_r^R \left(\frac{\sin x}{x}\right)^3\,{\rm d}x.$$

For $\gamma_4$, we have: $$\begin{align} \int_{\gamma_4}f(z)\,{\rm d}z &= \int_{\gamma_4} \sum_{n \geq 0}\left(\frac{(3-3^n)i^n}{4 \cdot n!}\right)z^{n-3}\,{\rm d}z \\ &= \int_{\gamma_4}\frac{1}{2z^3}\,{\rm d}z + \int_{\gamma_4}\frac{3}{4z}\,{\rm d}z + \int_{\gamma_4} \sum_{n \geq 3} \left(\frac{(3-3^n)i^n}{4 \cdot n!}\right)z^{n-3}\,{\rm d}z \end{align}$$

We have: $$ \int_{\gamma_4} \frac{1}{2z^3}\,{\rm d}z = -\frac{1}{4z^2}\Bigg|_{-r}^{r} = 0, \quad \int_{\gamma_4} \frac{3}{4z}\,{\rm d}z = \frac{3}{4}\int_{\gamma_4} -i \frac{{\rm d}z}{-iz} = \frac{3}{4}\int_0^\pi -i\,{\rm d}t = -\frac{3\pi i}{4}, $$ and: $$ \left|\int_{\gamma_4} \sum_{n \geq 3}\frac{(3-3^n)i^n}{4\cdot n!}z^{n-3}\right| \leq \pi r \sum_{n \geq 3}\frac{3^n-3}{4\cdot n!}r^{n-3} = \sum_{n \geq 3}\frac{\pi(3^n-3)}{4 \cdot n!}r^{n-2} \stackrel{r \to 0}{\longrightarrow} 0. $$

Making first $r \to 0$, and then $R \to +\infty$, in $\oint_\gamma f = 0$, we get: $$ 2i\int_0^{+\infty}\left(\frac{\sin x}{x}\right)^3\,{\rm d}x - \frac{3\pi i}{4} = 0 \implies \int_0^{+\infty}\left(\frac{\sin x}{x}\right)^3\,{\rm d}x = \frac{3\pi}{8}, $$as desired.


Related technique. You can use the Laplace transform technique. Recalling the Laplace transform

$$F(s)= \int_{0}^{\infty} f(x) e^{-sx}dx. $$

Taking $ f(x) = \frac{\sin(x)^3}{x^3} $ gives

$$ F(s)= \frac{\pi \,{s}^{2}}{8}+\frac{3\,\pi}{8}- \frac{3( {s}^{2}-1) }{8}\,\arctan \left( s \right) +\frac{( {s}^{2}-9)}{8}\,\arctan \left( \frac{s}{3} \right) $$ $$+\frac{3s}{8}\, \left( -\ln \left( {s}^{2}+9 \right) +\ln \left( {s} ^{2}+1 \right) \right). $$

Taking the limit as $s\to 0$ gives the desired result $\frac{3\pi}{8}$.

Another Laplace transform approach: Referring to the problem, we can use the following relation

$$ \begin{align} \int_0^\infty F(u)g(u) \, du & = \int_0^\infty f(u)G(u) \, du \\[6pt] L[f(t)] & = F(s) \\[6pt] L[g(t)] & = G(s)\end{align} $$


$$ G(u)=\frac{1}{u^3} \implies g(u)=\frac{u^2}{2!}, $$


$$ f(u)= \sin(u)^3 \implies F(u) = {\frac {6}{ \left( {u}^{2}+1 \right) \left( {u}^{2}+9 \right) }}. $$


$$ \int_0^\infty \frac{\sin^3 x}{x^3} \, dx = \frac{6}{2}\int_0^\infty \frac{u^2}{\left( {u}^{2}+1 \right) \left( {u}^{2}+9 \right)} \, du = \frac{3\pi}{8}$$.

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    $\begingroup$ thanks but can you show me step by step the laplace transform of $\frac{\sin(x)^3}{x^3}$ $\endgroup$ – mnsh Jul 27 '13 at 14:41
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    $\begingroup$ @hmedan.mnsh: Mhenni has some explaining to do. Your expression is right (save for the sign, but never mind for now). Integrating this three times gives an expression that goes to zero as $s \to 0$. The integration constants provide a quadratic in $s$ which goes to the constant term from the last integration as $s \to 0$. Thus, the integration constant is...the original integral! In other words, as far as I can see, this method is completely useless in deriving the desired result. This is not the first time that Mhenni has provided a "hint" that was poorly thought out. $\endgroup$ – Ron Gordon Jul 28 '13 at 2:05
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    $\begingroup$ @MhenniBenghorbal: you owe the OP an explanation of how you got the $3 \pi/8$ term. I worked the integral out according to your technique specified in the link, and the limit as $s \to 0$ of the result produces an integration constant whose value is...the result we seek. Please explain how you got this term, because I cannot; as far as I can see, this method simply is of no use in deriving the result. $\endgroup$ – Ron Gordon Jul 28 '13 at 2:11
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    $\begingroup$ @MhenniBenghorbal Iam up voter not down voter and the way is very beautiful and for that iam going to learn more and more about laplace transform $\endgroup$ – mnsh Jul 28 '13 at 12:15
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    $\begingroup$ @MhenniBenghorbal: Then prove me wrong. How did you derive this Laplace transform without needing to evaluate the original integral we were supposed to solve? I did this and came to that the the only way to get your result was to write $3 \pi/8$ for the result of the integral without deriving it. I would love to be proven wrong, but you seem to not want to do this. By the way, the downvote is not about the right answer: the OP could just go to Wolfram|Alpha for that. It is about a useful technique, which this does not seem to be. Please show us something. $\endgroup$ – Ron Gordon Jul 28 '13 at 14:12

Define $\displaystyle{% {\cal F}\left(\mu\right) \equiv \int_{-\infty}^{\infty}{\sin^{3}\left(\mu x\right) \over x^{3}}\,{\rm d}x\,, \quad ? = {1 \over 2}\,{\cal F}\left(1\right)}$

\begin{align} {\cal F}'\left(\mu\right) &= \int_{-\infty}^{\infty} {3\sin^{2}\left(\mu x\right)\cos\left(\mu x\right)x \over x^{3}}\,{\rm d}x = {3 \over 2}\int_{-\infty}^{\infty} {\cos\left(\mu x\right) - \cos\left(2\mu x\right)\cos\left(\mu x\right) \over x^{2}}\,{\rm d}x \\[3mm]&= {3 \over 2}\int_{-\infty}^{\infty} {\cos\left(\mu x\right) - \left\lbrack\cos\left(3\mu x\right) + \cos\left(\mu x\right)\right\rbrack/2 \over x^{2}}\,{\rm d}x = {3 \over 4}\int_{-\infty}^{\infty} {\cos\left(\mu x\right) - \cos\left(3\mu x\right) \over x^{2}}\,{\rm d}x \end{align} \begin{align} -&------------------------------ \end{align} \begin{align} {\cal F}''\left(\mu\right) &= {3 \over 4}\int_{-\infty}^{\infty} {-\sin\left(\mu x\right)x + \sin\left(3\mu x\right)\left(3x\right) \over x^{2}} \,{\rm d}x = {3 \over 2}\,{\rm sgn}\left(\mu\right)\int_{-\infty}^{\infty}{\sin\left(x\right) \over x} \,{\rm d}x \\[3mm]&= {3 \over 2}\,{\rm sgn}\left(\mu\right) \int_{-\infty}^{\infty} \left({1 \over 2}\int_{-1}^{1}{\rm e}^{{\rm i}kx}\,{\rm d}k\right) \,{\rm d}x = {3 \over 2}\,\pi\,{\rm sgn}\left(\mu\right) \int_{-1}^{1}\left(\int_{-\infty}^{\infty}{\rm e}^{{\rm i}kx} \,{{\rm d}x \over 2\pi}\right){\rm d}k \\[3mm]&= {3 \over 2}\,\pi\,{\rm sgn}\left(\mu\right) \int_{-1}^{1}\delta\left(k\right){\rm d}k = {3 \over 2}\,\pi\,{\rm sgn}\left(\mu\right) = {3 \over 2}\,\pi\,{{\rm d}\left\vert \mu\right\vert \over {\rm d}\mu} \end{align} \begin{align} -&------------------------------ \end{align} \begin{align} {\cal F}'\left(\mu\right) & = {3 \over 2}\,\pi\,\left\vert\mu\right\vert\ \Longrightarrow\ {\cal F}\left(\mu\right) = {3 \over 2}\,\pi\int_{0}^{\mu}\left\vert\mu'\right\vert\,{\rm d}\mu' = {3 \over 2}\,\pi\,{\rm sgn}\left(\mu\right)\int_{0}^{\mu}\mu'\,{\rm d}\mu' = {3 \over 4}\,\pi\,\left\vert\mu\right\vert\mu \end{align}

$$ \begin{array}{|c|}\hline\\ \color{#ff0000}{\large\quad% \int_{0}^{\infty}{\sin^{3}\left(x\right) \over x^{3}}\,{\rm d}x = {1 \over 2}\,{\cal F}\left(1\right) = {3\pi \over 8}\quad} \\ \\ \hline \end{array} $$


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