Evaluate $\int_0^1 x\arcsin\left(\sin\left(\frac 1x\right)\right) dx$ 
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.

Write $I_n = \left[\left(n-\frac{1}{2}\right)\pi, \left(n+\frac{1}{2}\right)\pi\right]$ and notice that the behavior of $\arcsin\sin x$ on $I_n$ depends on the value of $n$. In particular, $\arcsin\sin x$ is linear on each interval $I_n$ with the slope $(-1)^n$.

So, we can compute the integral as
\begin{align*}
\int_{1}^{\infty} \frac{\arcsin\sin x}{x^3} \, \mathrm{d}x
&= \left[ - \frac{\arcsin\sin x}{2x^2} \right]_{1}^{\infty}
 + \int_{1}^{\infty} \frac{(\arcsin\sin x)'}{2x^2} \, \mathrm{d}x \\
&= \frac{1}{2} + \int_{1}^{\infty} \frac{1}{2x^2} \left( \sum_{n=0}^{\infty} (-1)^n \mathbf{1}_{I_n}(x) \right) \, \mathrm{d}x \\
&= \frac{1}{2} + \sum_{n=0}^{\infty} (-1)^n \int_{I_n\cap[1,\infty)} \frac{1}{2x^2} \, \mathrm{d}x \\
&= \frac{1}{2} + \left[ - \frac{1}{2x} \right]_{1}^{\frac{\pi}{2}} + \sum_{n=1}^{\infty} (-1)^n \left[ - \frac{1}{2x} \right]_{\left(n-1/2\right)\pi}^{\left(n+1/2\right)\pi} \\
&= 1 - \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{\left( n - \frac{1}{2} \right) \pi} \\
&= 1 - \frac{1}{2}
 = \frac{1}{2}. 
\end{align*}
Or, more concisely,
\begin{align*}
\int_{1}^{\infty} \frac{\arcsin\sin x}{x^3} \, \mathrm{d}x
&= \left[ - \frac{\arcsin\sin x}{2x^2} \right]_{1}^{\infty}
 + \int_{1}^{\infty} \frac{(\arcsin\sin x)'}{2x^2} \, \mathrm{d}x \\
&= \frac{1}{2} + \left[ -\frac{(\arcsin\sin x)'}{2x} \right]_{1}^{\infty}
 + \int_{1}^{\infty} \frac{(\arcsin\sin x)''}{2x} \, \mathrm{d}x \\
&= 1 + \int_{1}^{\infty} \left( \sum_{n=1}^{\infty} \frac{(-1)^n}{x} \delta\left(x - \frac{2n-1}{2}\pi \right) \right) \, \mathrm{d}x \\
&= 1 + \sum_{n=1}^{\infty} \frac{(-1)^n}{\left(\frac{2n-1}{2}\right)\pi}
= 1 - \frac{1}{2}
= \frac{1}{2}.
\end{align*}
