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.
 A: 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*}
A: To evaluate $I(1)=\int_0^1 I^\prime(a) da$ we only need $I^\prime(a)$ for $a\in [0,\,1]$, which guarantees $\frac{\pi}{(2a)}>1$. As you've already noted, $$I^\prime(a)=\int_1^{\pi/(2a)}\frac{dx}{x^2}-\sum_{n\ge 0}\left(\int_{\pi/(2a)+2n\pi/a}^{3\pi/(2a)+2n\pi/a}\frac{dx}{x^2}-\int_{3\pi/(2a)+2n\pi/a}^{5\pi/(2a)+2n\pi/a}\frac{dx}{x^2}\right)\\=1-\frac{2a}{\pi}-\frac{2a}{\pi}\sum_{n\ge 0}\left(\frac{1}{1+4n}-\frac{2}{3+4n}+\frac{1}{5+4n}\right).$$We'll come back to that infinite sum over $n$ later, but for now denote it by $\Sigma$ so$$I(1)=\int_0^1\left(1-\frac{2a(1+\Sigma)}{\pi}\right)da=1-\frac{1+\Sigma}{\pi}.$$So now we need to evaluate $\Sigma$; it's clearly $$1-2\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\cdots\right)=1-2\left(1-\frac{\pi}{4}\right)=\frac{\pi}{2}-1,$$so$$I(1)=\tfrac{1}{2}.$$
A: 
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!
