Evaluate:
$$\int_{0}^{\infty}\dfrac{\sin^3(x-\frac{1}{x} )^5}{x^3} dx$$
I've been stumped by this Integral and cannot think of how to evaluate it. I substituted $\dfrac{1}{x^2}=t \Rightarrow \dfrac{-dt}{2}= \dfrac{dx}{x^3}.$ $$$$However, I can't understand what to do next. $$$$Any help on solving this would be truly appreciated. Many thanks!
On
as @achille hint,I have post it since $$I=\int_{0}^{+\infty}\dfrac{\sin^3{(x-\frac{1}{x})^5}}{x^3}dx=(\int_{0}^{1}+\int_{1}^{\infty})\dfrac{\sin^3{(x-\frac{1}{x})^5}}{x^3}dx=I_{1}+I_{2}$$ since $$I_{2}=-\int_{0}^{1}x\sin^3{(\dfrac{1}{x}-x)^5}dx$$ so $$I=\int_{0}^{1}\left(\dfrac{1}{x^3}-x\right)\sin^3{\left(x-\dfrac{1}{x}\right)^5}dx=\int_{0}^{\infty}=\int_{0}^{1}\left(\dfrac{1}{x}-x\right)\sin^3{\left(x-\dfrac{1}{x}\right)^5}d\left(x-\dfrac{1}{x}\right)$$ let $x-\dfrac{1}{x}=t$,then $$I=-\int_{0}^{+\infty}u\sin^3{u^5}du=-\dfrac{1}{5}\int_{0}^{+\infty}\sin^3{u}u^{-\frac{3}{5}}du=-\dfrac{1}{20}\int_{0}^{+\infty}(3\sin{u}-\sin{3u})u^{-3/5}du$$
and use well known result $$\int_{0}^{+\infty}\sin{x}\cdot x^pdx=\cos{(\pi p/2)}\Gamma{(1+p)},-2<p<0$$ can find it.becasue $$-\dfrac{3}{20}\int_{0}^{+\infty}\sin{u}\cdot u^{-\frac{3}{5}}du=-\dfrac{3}{20}\cos{\dfrac{6\pi}{5}}\Gamma{(\dfrac{2}{5})}$$ we know $$\cos{\dfrac{6\pi}{5}}=-\cos{\dfrac{\pi}{5}}=-\sqrt{1-\sin^2{\dfrac{\pi}{5}}}=-\dfrac{1}{2}\sqrt{\dfrac{5+\sqrt{5}}{2}}$$ and other is simaler
The integral is Riemann improper integrable. To save the argument that whether it converges or not. Let us look at following integral as a function of cutoff $\Lambda$.
$$I(\Lambda) \stackrel{def}{=}\int_{1/\Lambda}^\Lambda \sin^3\left[\left(x - \frac{1}{x}\right)^5\right] \frac{dx}{x^3} = \left(\int_{1/\Lambda}^1 + \int_1^\Lambda\right) \sin^3\left[\left(x - \frac{1}{x}\right)^5\right] \frac{dx}{x^3} $$ Change variable to $\displaystyle\;t = \begin{cases}\frac{1}{x},& x < 1\\x, & x > 1\end{cases}$ and notice for $x \in (0,1)$,
$$\frac{dx}{x^3} = -tdt\quad\text{ and }\quad \sin\left[\left(x - \frac{1}{x}\right)^5\right] = -\sin\left[\left(t - \frac{1}{t}\right)^5\right]$$ We have
$$I(\Lambda) = \int_1^\Lambda \sin^3\left[\left(t - \frac{1}{t}\right)^5\right] \left(\frac{1}{t^2} - t^2\right) \frac{dt}{t}$$
Using the fact $\displaystyle\;\frac{dt}{t} = \frac{d(t - \frac{1}{t})}{t + \frac{1}{t}}\;$, we get
$$I(\Lambda) = -\int_1^\Lambda \sin^3\left[\left(t - \frac{1}{t}\right)^5\right]\left(t - \frac{1}{t}\right) d\left(t - \frac{1}{t}\right)$$
Change variable to $\displaystyle\;u = \left(t - \frac{1}{t}\right)^5$, we find
$$I(\Lambda) = -\frac15 \int_0^{\left(\Lambda - \frac{1}{\Lambda}\right)^5} \sin^3(u) u^{-3/5} du$$
Since the function $\int_0^u \sin^3(t) dt$ is uniformly bounded for $u$ over $(0,\infty)$ and the factor $u^{-3/5}$ is monotonic decreasing to $0$ as $u \to \infty$. By Dirichlet's test on improper integral, following limit exists $$\lim_{X\to\infty} \int_0^X \sin^3(u) u^{-3/5} du$$ This means the original integral exists as a Riemann improper integral and equal to $$\lim_{\Lambda\to\infty} I(\Lambda) = -\frac15\int_0^\infty \sin^3(u) u^{-3/5} du$$
The last integral is something WA understand. If you throw this to WA or just look at math110's answer, the integral we want is equal to
$$\frac{1}{96} (3^{3/5} - 9) \sqrt{10-2\sqrt{5}}\,\Gamma\left(\frac75\right) \approx -0.15356212695352349$$
As a double check, one can ask WA to evaluate the original integral numerically and it reproduces above number. If one want to picky, one can use a different upper bound and lower bound for the cutoff but the final answer remains the same.