Evaluate $\int_0^\infty\frac {\sin^4(x)} {x^4} \operatorname dx$

Question

How to evaluate the definite integral:

$$ \int \limits_0^\infty\frac {\sin^4(x)} {x^4} \operatorname dx $$

Also provide the reference to various theorems to be to used to evaluate it! Thank you.

Answer 1

Recalling the result

$$ \int_{0}^{\infty} G(u)f(u) du = \int_{0}^{\infty} g(u)F(u) du, $$

where $F(u)$ and $G(u)$ are the Laplace transform of $f$ and $g$. Now, applying this to our problem gives

$$ \int_0^\infty\frac {\sin^4(x)} {x^4} dx = 4\int_{0}^{\infty} {\frac {x^2}{ \left( {x}^{2}+4 \right) \left( {x}^{2}+16 \right) } }=\frac{\pi}{3}\,.$$

Note:

1) We used the following Laplace transforms

$$ \mathcal{L} (\sin(x)^4) = {\frac {24}{s \left( {s}^{2}+4 \right) \left( {s}^{2}+16 \right) } },$$

$$ \mathcal{L} ( \frac{x^3}{6} )= \frac{1}{s^4}. $$

2) Use the partial fraction to evaluate the integral

$$ {\frac {x^2}{ \left( {x}^{2}+4 \right) \left( {x}^{2}+16 \right) } }= \frac{4}{3}\, \frac{1}{ {x}^{2}+16 } - \frac{1}{3}\, \frac{1}{ {x}^{2}+4 }.$$

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{\infty}{\sin^4\pars{x} \over x^4}\,\dd x:\ {\large ?}}$.

Let's $\ds{{\cal I}\pars{\mu} \equiv \half\int_{-\infty}^{\infty} {\sin^4\pars{\mu x} \over x^4}\,\dd x\ \mbox{with}\ \mu > 0}$ and such that $\ds{\int_{0}^{\infty}{\sin^4\pars{x} \over x^4}\,\dd x = {\cal I}\pars{1}}$. Also, $\ds{{\cal I}\pars{0^{+}} = 0}$.

\begin{align} {\cal I}'\pars{\mu}&=2\int_{-\infty}^{\infty}{\sin^{3}\pars{\mu x}\cos\pars{\mu x} \over x^3}\,\dd x =\int_{-\infty}^{\infty}{\sin^{2}\pars{\mu x}\sin\pars{2\mu x} \over x^{3}}\,\dd x \\[3mm]&=\half\int_{-\infty}^{\infty} {\sin\pars{2\mu x} - \sin\pars{2\mu x}\cos\pars{2\mu x} \over x^{3}}\,\dd x ={1 \over 4}\int_{-\infty}^{\infty}{2\sin\pars{2\mu x} - \sin\pars{4\mu x} \over x^{3}} \,\dd x\\[3mm]&\mbox{with}\ {\cal I}'\pars{0^{+}} = 0 \end{align}

\begin{align} {\cal I}''\pars{\mu}&=\int_{-\infty}^{\infty} {\cos\pars{2\mu x} - \cos\pars{4\mu x} \over x^{2}}\,\dd x\,, \qquad{\cal I}''\pars{0^{+}} = 0 \end{align}

\begin{align} {\cal I}'''\pars{\mu}&=\int_{-\infty}^{\infty} {-2\sin\pars{2\mu x} + 4\sin\pars{4\mu x} \over x}\,\dd x\ =-2\int_{-\infty}^{\infty}{\sin\pars{x} \over x}\,\dd x + 4\int_{-\infty}^{\infty}{\sin\pars{x} \over x}\,\dd x \\[3mm]&=2\pi\quad\mbox{where we used the well know result}\ \int_{-\infty}^{\infty}{\sin\pars{x} \over x}\,\dd x = \pi \end{align}

Then, $$ {\cal I}''\pars{\mu} = 2\pi\mu\,,\quad{\cal I}'\pars{\mu} = \pi\mu^{2}\quad \mbox{and}\quad{\cal I}\pars{\mu} = {1 \over 3}\,\pi\mu^{3} $$ $$\color{#00f}{\large% \int_{0}^{\infty}{\sin^{4}\pars{x} \over x^{4}}\,\dd x = {1 \over 3}\,\pi} $$

Answer 3

We can use a version of Parseval's equality:

$$\int_{-\infty}^{\infty} dx \, f(x)^2 = \frac1{2 \pi} \int_{-\infty}^{\infty} dk \, F(k)^2$$

where

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

Now, we may use the well-known FT:

$$f(x) = \frac{\sin^2{x}}{x^2} \implies F(k) = \pi \left (1-\frac{|k|}{2} \right ) $$

(This result may be easily derived using, e.g., the convolution theorem on the even more basic FT of $\sin{x}/x$.)

Thus the integral is $1/2$ of

$$\frac1{2 \pi} \pi^2 \int_{-2}^2 dk \, \left (1-\frac{|k|}{2} \right )^2 = \pi \int_0^2 \left ( 1-k+\frac14 k^2\right ) = \frac{2 \pi}{3}$$

or $\pi/3$.

Answer 4

NOTE: This is a sketch of what a solution may look like, but it is what it is and nothing more - a sketch.

I would do it 'a la residue'.

First, note that:

$$\sin^4{x} = \Re \frac{e^{4it}-4e^{2it}+3}{8}$$

Then note that if we choose a contour integral over the semicircle in the complex plane - that is on $\Gamma_{R}$ as $R \rightarrow \infty$ - we are bounded by $\frac{1}{R^3} \rightarrow 0$ as $R \rightarrow \infty$. We note that the only pole is at $z=0$ and so this problem reduces to $2\cdot \int_{0}^{\infty} \frac{\sin^4(x)}{x^4} = -\pi \cdot i \cdot Res_{z=0} \frac{\sin^4(z)}{z^4}$. Now the third derivative is $\frac{-64ie^{4it} + 32ie^{2it}}{8}$. This gives a residue of $1/3! \cdot \frac{-32i}{8} = -2i/3$.

Hence, we conclude that $\int_{0}^{\infty} \frac{\sin^4(x)}{x^4} dx = 1/2\cdot \pi i \cdot -2i/3 = \pi/3$.