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show that

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

using different ways

thanks for all

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A different way to what? What have you tried? –  Daniel Rust Jul 27 '13 at 1:16
    
@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 –  hammood Jul 27 '13 at 1:19
7  
This question asks a more general question, $$\int_0^\infty\left(\frac{\sin(x)}{x}\right)^n\,\mathrm{d}x$$ –  robjohn Jul 27 '13 at 4:15

5 Answers 5

up vote 7 down vote accepted

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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2  
+1 for using the parametrization trick. –  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

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

Then

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

Therefore

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

ADDENDUM

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.

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thanks but sorry I don't know alot about this theorem –  hammood Jul 27 '13 at 1:29
    
Fundamental for Fourier transforms; allows evaluation of a wide range of integrals that are otherwise difficult. See en.wikipedia.org/wiki/… –  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 $$ \int_0^\infty\frac{\sin^3(x)}{x^3}\,\mathrm{d}x=\frac{3\pi}{8}\tag{6} $$

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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 $n\rightarrow \infty$ $$\lim_{n\rightarrow \infty}(\frac{1}{n}\sum_{k=0}^{n-1} f\left(k\frac{1}n\right)) = \int_0^1 f(x)\ dx$$ $$f(x)=(\frac {\sin x} {x} )^3$$ $$\lim_{n\rightarrow \infty}(\sum_{k=0}^{\infty} \frac{1}{n} (\frac {\sin \frac{k}{n}} {\frac{k}{n}} )^3 = \int_0^1 (\frac {\sin x} {x} )^3\ dx\neq \int_0^\infty (\frac {\sin x} {x} )^3\ dx$$ –  hammood Jul 27 '13 at 15:32
2  
@hmedan.mnsh: $$ \lim_{n\to\infty}\sum_{k=0}^\infty\frac1n\left(\frac {\sin\left(\frac{k}{n}\right)} {\frac{k}{n}}\right)^3\ne\lim_{n\to\infty}\sum_{k=0}^n\frac1n\left(\frac {\sin\left(\frac{k}{n}\right)} {\frac{k}{n}}\right)^3 $$ –  robjohn Jul 27 '13 at 18:09
1  
@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 $$ \sum_{k=0}^\infty\sup_{x\in[k,k+1]}|f'(x)|\lt\infty $$ –  robjohn Jul 27 '13 at 18:45

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

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

Let

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

and

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

Now,

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

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1  
thanks but can you show me step by step the laplace transform of $\frac{\sin(x)^3}{x^3}$ –  hammood Jul 27 '13 at 14:41
    
@hmedan.mnsh: The link has the technique. –  Mhenni Benghorbal Jul 27 '13 at 15:48
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@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 –  hammood Jul 28 '13 at 12:15
2  
@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. –  Ron Gordon Jul 28 '13 at 14:12
1  
(+1) The second technique is amazing and makes the whole integral brilliantly simple! –  TenaliRaman Aug 4 '13 at 22:38

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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