# Prove $\int_0^\infty \frac{\sin^4x}{x^4}dx = \frac{\pi}{3}$

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

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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 $$\begin{eqnarray} \int_0^\infty \frac{\sin^4(x)}{x^4} \mathrm{d}x &=& \frac{2}{3} \int_0^\infty \frac{\sin^2(2x)}{x^2} \mathrm{d} x - \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x \\ &=& \frac{4}{3} \int_0^\infty \frac{\sin^2(y)}{y^2} \mathrm{d} y - \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x \\ &=& \frac{2}{3} \int_0^\infty \frac{\sin^2(x)}{x^2} \mathrm{d}x = \frac{\pi}{3} \end{eqnarray}$$

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Very nice way (+1) –  Chris's sis Mar 3 '13 at 9:17

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 $\int_0^{\infty} \left(\frac{\sin x }{x }\right)^n\,\mathrm{d}x$.

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Thank you for that solution. –  Jeff Mar 1 '13 at 20:10
@Jeff: welcome! –  Chris's sis Mar 1 '13 at 20:52

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

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

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

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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. –  Jeff Mar 1 '13 at 20:08
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 $4$. –  Ron Gordon Mar 1 '13 at 21:15
+1 I like this method. Will be using more of it in the future. –  Sasha Mar 1 '13 at 22:12

Here's another idea.

$\sin^{4} x = \frac{1}{8} \Big(\cos 4x - 4 \cos 2x + 3 \Big)$

To derive the above identity you could use the complex exponential form of the sine function and expand using the binomial theorem.

Then $\displaystyle \int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} \ dx = \frac{1}{2} \int_{-\infty}^{\infty} \frac{\sin^{4} x}{x^4} \ dx = \frac{1}{16} \text{Re} \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3}{x^{4}} \ dx$.

However we can't let $\displaystyle f(z) = \frac{e^{4iz}-4e^{2iz}+3}{z^{4}}$ since it has a pole of order 3 at the origin.

Instead we'll let $\displaystyle f(z) = \frac{e^{4iz}-4e^{2iz}+3+4iz}{z^{4}}$ which has a simple pole at the origin.

And notice that $\displaystyle \text{Re} \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3}{x^{4}} \ dx = \text{Re} \int_{-\infty}^{\infty} \frac{e^{4ix}-4e^{2ix}+3+4ix}{x^{4}} \ dx$.

Now just integrate around a closed half cirlce in the upper half complex plane with a small half circle indentation about the origin.

There are no poles inside of the contour, but in the limit the indentation will contribute $- i \pi \text{Res} [f,0] = - \frac{16 \pi}{3}$.

So $\displaystyle \text{PV}\int_{0}^{\infty} \frac{\sin^{4} x}{x^{4}} \ dx = \text{Re} \ \frac{1}{16} \left(\frac{16 \pi}{3} \right) = \frac{\pi}{3}$.

And we can drop the Cauchy principal value label since the integral converges in the normal sense.

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