Catalan constant's integral representation Catalan constant is known to have a rich source of integral identities, here is the formula I found:
$$ \int_0^\infty \frac{\sin^{-1}(\sin(x))}{x} \,dx \ =2G.$$
This can be proved by analyzing the function and using the identity:
$$ \frac{2G}{\pi}-\frac{1}{2}=\ln\left(\prod_{n=1}^\infty \frac{(4n-1)^{4n-1}}{(4n-3)^{2n-1}(4n+1)^{2n}}\right).$$
Is there any "neat" way to prove the formula directly without using the identity? It's really hard for me to prove the identity.
Update
I found my own way to prove this integral indentity:
Using Fourier series of triangle wave, with a little change in the coefficient, I got this equation:
$$\sin^{-1}(\sin(x))=\frac{4}{\pi}\sum_{n=0}^\infty (-1)^{n}\frac{\sin((2n+1)x)}{(2n+1)^{2}}$$
Divide both sides by $x$ and integrate:
$$\int_{0}^\infty \frac{\sin^{-1}(\sin(x))}{x}dx=\frac{4}{\pi}\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)^{2}}\int_{0}^\infty \frac{\sin((2n+1)x)}{x}dx$$
Using the fact that $\int_{0}^\infty \frac{\sin((2n+1)x)}{x}dx=\int_{0}^\infty \frac{\sin(x)}{x}dx=\frac{\pi}{2}$, we got:
$$\int_{0}^\infty \frac{\sin^{-1}(\sin(x))}{x}dx=2\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)^{2}}=2G$$
 A: Note that
\begin{align*}
\int_{0}^{\infty} \frac{\arcsin(\sin x)}{x} \, \mathrm{d}x
&= \frac{1}{2}\int_{-\infty}^{\infty} \frac{\arcsin(\sin x)}{x} \, \mathrm{d}x \\
&= \lim_{N\to\infty} \frac{1}{2}\int_{-N\pi}^{(N+1)\pi} \frac{\arcsin(\sin x)}{x} \, \mathrm{d}x  \\
&= \lim_{N\to\infty} \frac{1}{2}\int_{0}^{\pi} \left( \sum_{n=-N}^{N} \frac{(-1)^n}{x + n \pi} \right) \arcsin(\sin x) \, \mathrm{d}x.
\end{align*}
Using the identity $ \lim_{N\to\infty} \sum_{n=-N}^{N} \frac{(-1)^n}{x + n \pi} = \frac{1}{\sin x}$, we get
\begin{align*}
\int_{0}^{\infty} \frac{\arcsin(\sin x)}{x} \, \mathrm{d}x
&= \frac{1}{2}\int_{0}^{\pi} \frac{\arcsin(\sin x)}{\sin x} \, \mathrm{d}x \\
&= \int_{0}^{\pi/2} \frac{x}{\sin x} \, \mathrm{d}x.
\end{align*}
Substituting $t=\tan(x/2)$, the last integral reduces to
\begin{align*}
\int_{0}^{\pi/2} \frac{x}{\sin x} \, \mathrm{d}x
&= \int_{0}^{\pi/2} \frac{x\sec^2(x/2)}{2\tan(x/2)} \, \mathrm{d}x \\
&= 2\int_{0}^{1} \frac{\arctan(t)}{t} \, \mathrm{d}t \\
&= 2\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} \int_{0}^{1} t^{2n} \, \mathrm{d}t \\
&= 2\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2} \\
&= 2G.
\end{align*}
