TITLE: Evaluate $\int_0^1 x\arcsin\left(\sin\left(\frac 1x\right)\right) dx$
QUESTION [6 upvotes]: Evaluate $$\int_0^1 x\arcsin\left(\sin\left(\frac 1x\right)\right) dx=\int_1^{\infty} \frac {\arcsin(\sin x)}{x^3}dx$$

I tried to use Feynman's trick in the second one as 
$$I(a)=\int_1^{\infty} \frac {\arcsin(\sin ax)}{x^3}dx$$
We have $I(0)=0$
Now differentiating both sides leads me to an absurd integral as $$I'(a)=\int_1^{\infty} \frac {\cos ax }{x^2\vert \cos ax \vert}dx$$
I am now unable proceed further. 
Also If I keep in mind that I need value of integral at $a=1$ and so break the integral in intervals as $$\left(1,\frac {\pi}{2}\right) ;\left(\frac {\pi}{2},\frac {3\pi}{2}\right) ; \left(\frac {3\pi}{2},\frac {5\pi}{2}\right) $$
and so on then this gives me the value of integral as 0. But I suspect it isn't correct because one online software suggests it's value to be $\frac {1}{2}$
Moreover the signum function is always reminding me of the Grandi's series which also in some definition equals $1/2$
Any help would be greatly appreciated.

REPLY [6 votes]: I thought it might be instructive to present a way forward that does not use integration by parts, but rather evaluates the integral directly after enforcing the substitution $x\mapsto 1/x$.  To that end, we proceed.


Note that we have
$$\begin{align}
\int_1^\infty \frac{\arcsin(\sin(x))}{x^3}\,dx&=\int_1^{\pi/2}\frac{\arcsin(\sin(x))}{x^3}\,dx+\sum_{n=1}^\infty \int_{-\pi/2+n\pi}^{\pi/2+n\pi}\frac{\arcsin(\sin(x))}{x^3}\,dx\\\\
&=\int_1^{\pi/2}\frac{1}{x^2}\,dx+\sum_{n=1}^\infty (-1)^n\int_{-\pi/2}^{\pi/2}\frac{x}{(x+n\pi)^3}\,dx\\\\
&=\left(1-\frac2\pi\right)+\frac2\pi\sum_{n=1}^\infty (-1)^n\left(\frac{1}{2n-1}-\frac{1}{2n+1}+\frac{n}{(2n+1)^2}-\frac{n}{(2n-1)^2}\right)\\\\
&=\left(1-\frac2\pi\right)+\frac1\pi\sum_{n=1}^\infty (-1)^n\left(\frac{1}{2n-1}-\frac{1}{2n+1}-\frac{1}{(2n+1)^2}-\frac{1}{(2n-1)^2}\right)\\\\
&=\left(1-\frac2\pi\right)+\frac1\pi\left(-\frac\pi4+1-\frac\pi4+1\right)\\\\
&=\frac12
\end{align}$$
as expected!