I need to show that $$ \int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3} $$

I have already derived the result $\int_0^\infty \frac{\sin^2x}{x^2} = \frac{\pi}{2}$ using complex analysis, a result which I am supposed to start from. Using a change of variable $ x \mapsto 2x $ :

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

Now using the identity $\sin^2(2x) = 4\sin^2x - 4\sin^4x $, we obtain

$$ \int_0^\infty \frac{\sin^2x - \sin^4x}{x^2}dx = \frac{\pi}{4} $$ $$ \frac{\pi}{2} - \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4} $$ $$ \int_0^\infty \frac{\sin^4x}{x^2}dx = \frac{\pi}{4} $$

But I am now at a loss as to how to make $x^4$ appear at the denominator. Any ideas appreciated.

Important: I must start from $ \int_0^\infty \frac{\sin^2x}{x^2}dx $, and use the change of variable and identity mentioned above

  • $\begingroup$ Check this technique. $\endgroup$ – Mhenni Benghorbal Mar 31 '14 at 22:34
  • $\begingroup$ I reopened this question because a) it is not an exact duplicate and b) REALLY?!? This question and its alleged duplicate are nearly 5 years old! How on earth are they harming anyone? This is an abuse of the duplicate function and it is not the first time I have come across this. Again, stop and think before flagging as duplicate. $\endgroup$ – Ron Gordon Nov 8 '17 at 2:55

You are likely expected to integrate by parts (twice) $$ \begin{eqnarray} \int \frac{\sin^4(x)}{x^4} \mathrm{d}x &=& -\frac{1}{3} \frac{\sin^4(x)}{x^3} + \frac{4}{3} \int \frac{\cos(x) \sin^3(x) }{x^3} \mathrm{d} x \\ &=& -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \frac{3 \cos^2(x) \sin^2(x) - \sin^4(x)}{x^2} \mathrm{d} x \\ &=& -\frac{1}{3} \frac{\sin^4(x)}{x^3} -\frac{2 \cos(x) \sin^3(x)}{3 x^2} + \frac{2}{3} \int \left(\frac{\sin^2(2x)}{x^2} - \frac{\sin^2(x)}{x^2} \right) \mathrm{d}x \end{eqnarray} $$ where the last equality used $$\begin{eqnarray} 3 \cos^2(x) \sin^2(x) - \sin^4(x) &=& 3 \cos^2(x) \sin^2(x) - \sin^2(x) (1-\cos^2(x)) \\ &=& \left(2 \sin(x) \cos(x) \right)^2 - \sin^2(x) = \sin^2(2x) - \sin^2(x) \end{eqnarray} $$ Now $$\begin{eqnarray} \int_0^\infty \frac{\sin^4(x)}{x^4} \mathrm{d}x &=& \frac{2}{3} \int_0^\infty \frac{\sin^2(2x)}{x^2} \mathrm{d} x - \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x \\ &=& \frac{4}{3} \int_0^\infty \frac{\sin^2(y)}{y^2} \mathrm{d} y - \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x \\ &=& \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x = \frac{\pi}{3} \end{eqnarray} $$


Hint: use Parseval/Plancherel theorem on $(\sin{x}/x)^2$.

That is, the FT of $(\sin{x}/x)^2$ is

$$\int_{-\infty}^{\infty} dx \: \frac{\sin^2{x}}{x^2} e^{i k x} = \begin{cases} \\\pi \left (1 - \frac{|k|}{2} \right ) & |k| \le 2 \\ 0 & |k| > 2 \end{cases}$$

Plancherel/Parseval says that

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

$$\therefore \: \int_{0}^{\infty} dx \: \frac{\sin^4{x}}{x^4} = \frac{\pi}{3}$$

  • 1
    $\begingroup$ Thank you, that's quite an elegant way to do it. However, the problem explicitly mentions I must start from $ \int_0^\infty \frac{\sin^2x}{x^2}dx $, and I'm not sure another solution will be accepted. I've changed my original post to reflect that. $\endgroup$ – Jeff Mar 1 '13 at 20:08
  • $\begingroup$ Well, if you or someone figures that out, it'd be a neat trick. Sadly, I have no idea how you'd do that. It's sort of akin to deriving the number $3$ from the number $4$. $\endgroup$ – Ron Gordon Mar 1 '13 at 21:15
  • $\begingroup$ +1 I like this method. Will be using more of it in the future. $\endgroup$ – Sasha Mar 1 '13 at 22:12

I don't know if it's possible to determine the value of $\int_{0}^{\infty} \frac{\sin^{4}(x) }{x^{4}} \, dx$ from the value of $\int_{0}^{\infty} \frac{\sin^{4}(x) }{x^{2}} \, dx $.

But you mentioned that you used complex analysis to evaluate $\int_{0}^{\infty} \frac{\sin^{2}(x)}{x^{2}} \, dx$.

We can also use complex analysis to evaluate $\int_{0}^{\infty} \frac{\sin^{4}(x) }{x^{4}} \, dx$.

Using the trigonometric identity $ \displaystyle \sin^{4} x = \frac{1}{8} \Big(\cos 4x - 4 \cos 2x + 3 \Big)$, we get

$$ \begin{align} \int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} \, dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{4} x}{x^4} \, dx \\ &= \frac{1}{16} \int_{-\infty}^{\infty} \operatorname{Re} \, \frac{e^{4ix}-4e^{2ix}+3}{x^{4}} \ dx \\ &= \frac{1}{16}\int_{-\infty}^{\infty} \operatorname{Re} \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, dx \\ &= \frac{1}{16} \, \operatorname{Re} \, \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, dx. \end{align}$$

So let's integrate the function $$f(z) = \frac{e^{4iz}-4e^{2iz}+3+4iz}{z^{4}}$$ around a contour the consists of the portion of the real axis from $-R$ to $R$, $R>0$, and the upper half of the circle $|z|=R$. To avoid the simple pole at the origin, the contour needs to be indented at the origin.

Letting the radius of the indentation go to zero and $R \to \infty$ (and applying Jordan's lemma along with the estimation lemma), we get

$$ \operatorname{PV} \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \, dx- i \pi \ \text{Res}[f(z),0] = 0,$$


$$ \operatorname{Res}[f(z),0] = \lim_{z \to 0} \frac{e^{4iz}-4e^{2iz}+3+4iz}{z^{3}} = \frac{16 \pi}{3} . $$


$$ \int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} \, dx = \frac{\pi}{3} .$$

Technically, it wasn't necessary to add $4ix$ to the numerator.

For reasons explained here, the Cauchy principal value of $ \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3}{x^{4}} \, dx $ exists even though $\frac{e^{4iz}-4e^{2iz}+3}{z^{4}}$ has a pole of order $3$ at the origin.

  • 1
    $\begingroup$ @Urgent The Cauchy principal value of the integral. The integral $ \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3}{x^{4}} \ dx$ doesn't' exist in the traditional sense due to a singularity at $x=0$, but it does exist as a Cauchy principal value integral. The Cauchy principal value is a very useful tool for evaluating integrals using contour integration. en.wikipedia.org/wiki/Cauchy_principal_value $\endgroup$ – Random Variable Apr 15 '17 at 14:54
  • $\begingroup$ I meant that the integral $\int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3+{\color{red}{4ix}}}{x^{4}} \, dx $ exists as a Cauchy principal value integral. $\endgroup$ – Random Variable Apr 15 '17 at 22:49

HINT: Use the relation

$$\int_0^\infty \left(\frac{\sin x}{x}\right)^n \mathrm{d}x = \frac{\pi}{2^n (n-1)!} \sum_{k=0}^{\lfloor n/2\rfloor} (-1)^k {n \choose k} (n-2k)^{n-1}$$ You may find a proof here A sine integral $\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$.


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

$\large\tt \mbox{METHOD}\ 0:$ Let's $\ds{{\cal F}\pars{\mu} \equiv \int_{0}^{\infty}{\sin^{4}\pars{\mu x} \over x^{4}}\,\dd x}$ such that $\ds{\int_{0}^{\infty}{\sin^{4}\pars{\mu x} \over x^{4}}\,\dd x ={\cal F}\pars{1}}$. \begin{align} \color{#c00000}{{\cal F}'\pars{\mu}} &=\int_{0}^{\infty}{4\sin^{3}\pars{\mu x}\cos\pars{\mu x} \over x^{3}}\,\dd x =\int_{0}^{\infty}{\bracks{1 - \cos\pars{2\mu x}}\sin\pars{2\mu x} \over x^{3}}\,\dd x \\[3mm]&=\half\int_{0}^{\infty}{2\sin\pars{2\mu x} - \sin\pars{4\mu x} \over x^{3}}\,\dd x \quad\mbox{with}\ {\cal F}\pars{0} = 0\\[5mm]&\mbox{} \end{align} \begin{align} \color{#c00000}{{\cal F}''\pars{\mu}}&= 2\int_{0}^{\infty}{\cos\pars{2\mu x} - \cos\pars{4\mu x} \over x^{2}}\,\dd x \quad\mbox{with}\ {\cal F}'\pars{0} = 0\\[5mm]&\mbox{} \end{align} \begin{align} \color{#c00000}{{\cal F}'''\pars{\mu}}&= 2\int_{0}^{\infty}{-2\sin\pars{2\mu x} + 4\sin\pars{4\mu x} \over x}\,\dd x =4\sgn\pars{\mu}\ \overbrace{\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x} ^{\ds{=\ {\pi \over 2}\,. \mbox{See below}}} = 2\pi\sgn\pars{\mu} \\[3mm]&\mbox{with}\ {\cal F}''\pars{0} = 0 \end{align} $$ \mbox{With}\ \mu > 0\,,\ {\cal F}'''\pars{\mu} = 2\pi\ \imp\ {\cal F}''\pars{\mu} =2\pi\mu\ \imp\ {\cal F}'\pars{\mu}=\pi\mu^{2}\ \imp\ {\cal F}\pars{\mu} = {\pi \over 3}\,\mu^{3} $$ $$ \color{#00f}{\large\int_{0}^{\infty}{\sin^{4}\pars{x} \over x^{4}}\,\dd x ={\cal F}\pars{1} = {\pi \over 3}\large} $$ $-----------------------------------------$
Also \begin{align} \color{#c00000}{\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x}&= \half\int_{-\infty}^{\infty}{\sin\pars{x} \over x}\,\dd x =\half\int_{-\infty}^{\infty}\ \overbrace{\bracks{\half\int_{-1}^{1}\expo{\ic kx}\,\dd k}} ^{\ds{=\ {\sin\pars{x} \over x}}}\ \dd x \\[3mm]&={\pi \over 2}\int_{-1}^{1}\ \overbrace{\bracks{\int_{-\infty}^{\infty}\expo{\ic kx}\,{\dd x \over 2\pi}}} ^{\ds{=\ \delta\pars{k}}}\ \dd k ={\pi \over 2}\int_{-1}^{1}\delta\pars{k}\,\dd k =\color{#c00000}{\pi \over 2} \end{align} $\ds{\delta\pars{k}}$ is the Dirac Delta Function.


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