Compute $\int_{0}^{\infty} \frac{\sin^4x}{x^4}dx$

Question

Compute $$ \int_{0}^{\infty} \frac{\sin^4x}{x^4}dx\;\;. $$

Well, I'm not sure how to approach this. I thought about using Plancheral, and using $f(x) = \frac{\sin^2x}{x^2}$, and finding the fourier transform of $f(x)$. Any other better approach?

Answer 1

I think you sould use integration by parts to do it more easily in the following way:

$$ \int \frac{\sin^4(x)}{x^4}dx = -\frac{1}{3} \frac{\sin^4(x)}{x^3} + \frac{4}{3} \int \frac{\cos(x) \sin^3(x) }{x^3} dx $$ $$ = -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \frac{3 \cos^2(x) \sin^2(x) - \sin^4(x)}{x^2} dx$$

$$= -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \left(\frac{\sin^2(2x)}{x^2} - \frac{\sin^2(x)}{x^2} \right) dx$$

Last line I have used $\sin^4(x) = \sin^2(x)(1-cos^2(x)) $ and simplify it.

In general we know that $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis.

Also put $2x = y$ and find the integral foe the other part.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[15px,#ffd]{\int_{0}^{\infty}{\sin^{4}\pars{x} \over x^{4}}\,\dd x} = {1 \over 2}\,\lim_{N \to \infty}\int_{-N\pi}^{N\pi} {\sin^{4}\pars{x} \over x^{4}}\,\dd x \\[5mm] = &\ {1 \over 2}\,\lim_{N \to \infty}\bracks{% \int_{-N\pi}^{-\pars{N - 1}\pi}{\sin^{4}\pars{x} \over x^{4}}\,\dd x + \cdots + \int_{\pars{N - 1}\pi}^{N\pi}{\sin^{4}\pars{x} \over x^{4}}\,\dd x} \\[5mm] = &\ {1 \over 2}\int_{0}^{\pi} \sin^{4}\pars{x}\sum_{k = -\infty}^{\infty}{1 \over \pars{x + k\pi}^{4}}\,\dd x \\[5mm] = &\ {1 \over 2}\int_{0}^{\pi} \sin^{4}\pars{x} \,{2\cot^{2}\pars{x}\csc^{2}\pars{x} + \csc^{4}\pars{x} \over 3}\,\dd x \\[5mm] = &\ {1 \over 3}\int_{0}^{\pi}\cos^{2}\pars{x}\,\dd x + {1 \over 6}\int_{0}^{\pi}\dd x = {1 \over 6}\,\pi + {1 \over 6}\,\pi = \bbx{\ds{\pi \over 3}} \\ & \end{align}

Answer 3

Using Plancherel:

$$ \begin{aligned} \int_{0}^{+\infty} \frac{ \sin^4 (x)}{x^4} \ \mathrm dx &= \frac{1}{2} \int_{-\infty}^{+\infty} \left (\frac{ \sin^2 (x)}{x^2} \right )^2 \ \mathrm dx \\&= \frac{\pi}{2} \int_{-\infty}^{+\infty} | f(\omega) |^2 \ \mathrm d\omega \end{aligned} $$

Where:

$$ f(\omega) := \begin{cases} 1 - |\omega|, & |\omega| < 1 \\ 0, & \text{otherwise} \end{cases} $$

Therefore:

$$ \int_{0}^{+\infty} \frac{ \sin^4 (x)}{x^4} \ \mathrm dx = \frac{\pi}{2} \int_{-1}^{1} (1-|\omega|)^2 \ \mathrm d\omega = \frac{\pi}{3} $$

Answer 4

I would use Parseval, which states that the integral of the square of the a function is related to the integral of the square of the FT of that function. In this case, the function is $(\sin{x}/x)^2$, which has FT equal to $\pi (1-|k|/2)$ for $|k| \le 2$ and $0$ otherwise. The integral here is equal to

$$\frac1{2 \pi} \int_0^2 dk \, \pi^2 \left (1-\frac{k}{2} \right )^2 = \frac{\pi}{3}$$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[15px,#ffd]{\int_{0}^{\infty}{\sin^{4}\pars{x} \over x^{4}}\,\dd x} \\ = &\ \int_{0}^{\infty}\ \overbrace{\cos\pars{4x} - 4\cos\pars{2x} + 3 \over 8}^{\ds{\sin^{4}\pars{x}}}\ \overbrace{{1 \over 6}\int_{0}^{\infty}t^{3}\expo{-xt}\dd t} ^{\ds{1 \over x^{4}}}\ \dd x \\[5mm] = &\ {1 \over 48}\int_{0}^{\infty}t^{3}\Re\int_{0}^{\infty} \bracks{\expo{-\pars{t - 4\ic}x} - 4\expo{-\pars{t - 2\ic}x} + 3\expo{-tx}}\dd x\,\dd t \\[5mm] = &\ {1 \over 48}\int_{0}^{\infty} \bracks{{t^{4} \over t^{2} + 16} - {4t^{4} \over t^{2} + 4} + 3t^{2}}\dd x\,\dd t \\[1cm] = &\ {1 \over 48}\int_{0}^{\infty} \left[\pars{t^{4} + 16t^{2}} - 16\pars{t^{2} + 16} + 256 \over t^{2} + 16\right. \\[2mm] & \left.\phantom{{1 \over 48}\int_{0}^{\infty}\,\,\,\,\,\,\,}- {4\pars{t^{4} + 4t^{2}} - 16\pars{t^{2} + 4} + 64\over t^{2} + 4} + 3t^{2}\right]\dd x\,\dd t \\[1cm] = &\ {1 \over 48}\int_{0}^{\infty} \pars{{256 \over t^{2} + 16} - {64 \over t^{2} + 4}}\dd t \\[5mm] = &\ {1 \over 48}\,{1 \over 16}\,256 \times 4\int_{0}^{\infty}{\dd t \over t^{2} + 1} - {1 \over 48}\,{1 \over 4}\,64 \times 2\int_{0}^{\infty} {\dd t \over t^{2} + 1} \\[5mm] = &\ \pars{{4 \over 3} - {2 \over 3}}\,{\pi \over 2} = \bbx{\large{\pi \over 3}} \end{align}

Answer 6

$${\int_0^\infty \left( \frac {\sin x}{x} \right)^p dx = \frac{\pi}{2^p(p-1)!} \left(p^{p-1} - { p \choose 1 } (p-2)^{p-1} + { p \choose 2 } (p-4)^{p-1} - \cdots \right)}$$

Here's a nice reference,

$$\int_0^\infty \frac {\sin x}{x} dx = \frac{\pi}{2}$$

$$\int_0^\infty \left( \frac {\sin x}{x} \right)^2 dx = \frac{\pi}{2}$$

$$\int_0^\infty \left( \frac {\sin x}{x} \right)^3 dx = \frac{3\pi}{8}$$

$$\boxed{\int_0^\infty \left( \frac {\sin x}{x} \right)^4 dx = \frac{\pi}{2^4(3)!} \left( 4^{3} - { 4 \choose 1 } (2)^{3} \right)=\frac{\pi}{3}}$$

$$\int_0^\infty \left( \frac {\sin x}{x} \right)^5 dx = \frac{115\pi}{384}$$

$$\int_0^\infty \left( \frac {\sin x}{x} \right)^6 dx = \frac{11\pi}{40}$$

$$\int_0^\infty \left( \frac {\sin x}{x} \right)^7 dx = \frac{5887\pi}{23040}$$

$$\int_0^\infty \left( \frac {\sin x}{x} \right)^8 dx = \frac{151\pi}{630}$$

$$\int_0^\infty \left( \frac {\sin x}{x} \right)^9 dx = \frac{259723 \pi}{1146880}$$

Answer 7

$$I=\int_0^\infty\left(\frac{\sin x}x\right)^4dx=\frac12\int_{-\infty}^\infty\left(\frac{\sin x}x\right)^4dx=\Re \int_{-\infty}^\infty\frac{e^{4ix}-4e^{2ix}+3}{16x^4}dx $$ Now, by Residue theorem with upper semi-circle contour with a jump at zero and "half-contour" at the singularity trick, $$I=\pi i \left(-\frac i3\right)=\frac\pi 3$$