How to calculate the improper integral $\int_0^{\infty}\frac{(\sin x)^4}{x^3} \,{\rm d}x$?

Question

According to Wolfram Alpha,

$$ \int_0^{\infty}\frac{(\sin x)^4}{x^3} \,{\rm d}x = \log{2} $$

but I can't find a way to prove it.

Answer 1

Well, we are trying to find the following integral:

$$\mathcal{I}:=\int_0^\infty\frac{\sin^4 x}{x^3}\space\text{d}x\tag1$$

Using the 'evaluating integrals over the positive real axis' property of the Laplace transform, we can write:

$$\mathcal{I}=\int_0^\infty\mathscr{L}_x\left[\sin^4 x\right]_{\left(\sigma\right)}\cdot\mathscr{L}_x^{-1}\left[\frac{1}{x^3}\right]_{\left(\sigma\right)}\space\text{d}\sigma\tag2$$

Which gives:

$$\mathcal{I}=\int_0^\infty\frac{1}{\sigma}\cdot\frac{24}{\sigma^4+20\sigma^2+64}\cdot\frac{\sigma^2}{2}\space\text{d}\sigma=12\int_0^\infty\frac{\sigma}{\sigma^4+20\sigma^2+64}\space\text{d}\sigma\tag3$$

Now, substitute $\text{u}=\sigma^2$:

$$\mathcal{I}=6\int_0^\infty\frac{1}{\text{u}^2+20\text{u}+64}\space\text{du}=6\int_0^\infty\frac{1}{\left(\text{u}+10\right)^2-36}\space\text{du}\tag4$$

Now, substitute $\text{p}=\text{u}+10$:

$$\mathcal{I}=6\int_{10}^\infty\frac{1}{\text{p}^2-36}\space\text{dp}=-\frac{1}{6}\int_{10}^\infty\frac{1}{1-\left(\frac{\text{p}}{6}\right)^2}\space\text{dp}\tag5$$

Now, substitute $\text{w}=\frac{\text{p}}{6}$:

$$\mathcal{I}=-\int_\frac{5}{3}^\infty\frac{1}{1-\text{w}^2}\space\text{dw}\tag6$$

I'll let you conclude.

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EDIT, solving it more generally you could use:

When $\text{n}\in\mathbb{N}$ we know that:

$$\mathscr{L}_x\left[\sin^{2\text{n}}\left(x\right)\right]_{\left(\sigma\right)}=\frac{\left(2\text{n}\right)!}{\sigma}\prod_{\text{k}\space=\space1}^\text{n}\frac{1}{\sigma^2+4\text{k}^2}\tag7$$

And:

$$\mathscr{L}_x^{-1}\left[\frac{1}{x^\text{p}}\right]_{\left(\sigma\right)}=\frac{\sigma^{\text{p}-1}}{\Gamma\left(\text{p}\right)}\tag8$$

So, we get:

$$\mathcal{I}_\text{n}\left(\text{p}\right):=\int_0^\infty\frac{\sin^{2\text{n}}\left(x\right)}{x^\text{p}}\space\text{d}x=\frac{\left(2\text{n}\right)!}{\Gamma\left(\text{p}\right)}\int_0^\infty\sigma^{\text{p}-2}\prod_{\text{k}\space=\space1}^\text{n}\frac{1}{\sigma^2+4\text{k}^2}\space\text{d}\sigma\tag9$$

Answer 2

Another technique: since $\frac{1}{x^3}=\frac12\int_0^\infty y^2e^{-xy}dy$, by Fubini's theorem the integral is$$\begin{align}&\color{white}{=}\frac{1}{2(2i)^4}\int_0^\infty y^2\left(\int_0^\infty(e^{4ix}-4e^{2ix}+6-4e^{-2ix}+e^{-4ix})e^{-xy}dx\right)dy\\&=\frac{1}{32}\int_0^\infty y^2\left(\frac{2y}{y^2+16}-\frac{8y}{y^2+4}+\frac{6}{y}\right)dy,\end{align}$$which I leave you to simplify to $\int_0^\infty\left(\frac{y}{y^2+4}-\frac{y}{y^2+16}\right)dy$, which is easily shown to be $\ln2$.