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:
$$\int_{0}^\infty \frac{\sin^{-1}(\sin(x))}{x}dx=2\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)^{2}}=2G$$

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*}

Here is another approach if you are happy to use results which I will quote and give links to their proofs elsewhere on this site. I agree it is not as self-contained as @Sangchul Lee's answer, so it really depends on what you are after.

I will make use of three results. They are:

1. The Maclaurin series expansion for the inverse sine function of $$\arcsin (x) = \sum_{n = 0}^\infty \frac{x^{2n + 1}}{2^{2n} (2n + 1)} \binom{2n}{n}, \quad |x| \leqslant 1.$$
2. The improper integral $$ \int_0^\infty \frac{\sin^{2n + 1}(x)}{x} \, dx = \frac{\pi}{2^{2n + 1}} \binom{2n}{n}.$$ Here $n = 0,1,2,\ldots$
3. A series involving the square of the central binomial coefficient $$\sum_{n = 0}^\infty \binom{2n}{n}^2\frac{1}{2^{4n}(2n + 1)} = \frac{4\mathbf{G}}{\pi}.$$

If $x$ is replaced with $\sin x$ in the first result we have $$\arcsin(\sin x) = \sum_{n = 0}^\infty \frac{\sin^{2n + 1} x}{2^{2n}(2n + 1)} \binom{2n}{n}, \quad \text{for all $x$}.$$

The interchange made between the summation and the integration signs based on invoking Fubini's theorem in this case is not permissible. A salvaged answer using a regularisation suggested by @Sangchul Lee is given in the Update Answer below.

Replacing this term in our integral, after interchanging the summation and integration signs which is permissible due to $\require{enclose}\enclose{horizontalstrike}{\text{Fubini's theorem}}$, we have \begin{align*} \int_0^\infty \frac{\arcsin(\sin x)}{x} \, dx &= \sum_{n = 0}^\infty \frac{1}{2^{2n}(2n + 1)} \binom{2n}{n} \int_0^\infty \frac{\sin^{2n + 1} x}{x} \, dx\\ &= \frac{\pi}{2} \sum_{n = 0}^\infty \binom{2n}{n}^2 \frac{1}{2^{4n}(2n + 1)} \qquad \text{(from the second result)}\\ &= \frac{\pi}{2} \cdot \frac{4 \mathbf{G}}{\pi} = 2\mathbf{G}, \qquad \text{(from the third result)} \end{align*} as required to show.

Updated Answer

Here is a salvaged answer based on a regularisation of the demoninator $x$ by $x^{1 + s}$ for $0 < s < 1$.

Since $$\int_{0}^{\infty}\sum_{n=0}^{\infty} \left| \frac{1}{2^{2n}(2n+1)}\binom{2n}{n}\frac{\sin^{2n+1}x}{x^{1+s}}\right|\,\mathrm{d}x < \infty,$$ the interchange between the summation and the integration signs is justified by Fubini's theorem. Thus $$\int_0^\infty \frac{\arcsin(\sin x)}{x} \, dx = \sum_{n = 0}^\infty \frac{1}{2^{2n}(2n + 1)} \binom{2n}{n} \lim_{s \to 0^+} \int_0^\infty \frac{\sin^{2n + 1} x}{x^{1 + s}} \, dx.$$

For the integral that has appeared, using $$\sin^{2n+1}(x) = \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1} \sin\left((2k+1)x\right),$$ one has \begin{align*} \int_0^\infty \frac{\sin^{2n + 1}(x)}{x^{1 + s}} \, dx &= \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1}\int_0^\infty \frac{\sin\left((2k+1)x\right)}{x^{1 + s}} \, dx\\ &= \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1} (2k + 1)^s\int_0^\infty \frac{\sin\left(x\right)}{x^{1 + s}} \, dx. \end{align*} Note in the second line a substitution of $(2k + 1)x \mapsto x$ has been made. A value for the integral that has appeared can be found (see here). It is $$\int_0^\infty \frac{\sin x}{x^{1 + s}} \, dx = -\Gamma(-s) \sin \left (\frac{s\pi}{2} \right ), \quad 0 < s < 1.$$ One therefore has $$\int_0^\infty \frac{\sin^{2n + 1}(x)}{x^{1 + s}} \, dx = -\Gamma (-s) \sin \left (\frac{s \pi}{2} \right ) \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1} (2k + 1)^s.$$ Thus \begin{align*} \lim_{s \to 0^+} \int_0^\infty \frac{\sin^{2n + 1}(x)}{x^{1 + s}} \, dx &= \lim_{s \to 0^+} \left [-\Gamma (-s) \sin \left (\frac{s \pi}{2} \right ) \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1} (2k + 1)^s \right ]\\ &= \lim_{s \to 0^+} \left (-\Gamma (-s) \sin \left (\frac{s\pi}{2} \right ) \right ) \sum_{k=0}^n \frac{(-1)^k }{4^n} \binom{2n+1}{n+k+1} \lim_{s \to 0^+} (2k + 1)^s. \end{align*} Since $k = 0,1,2,\ldots$, then $\lim_{s \to 0^+} (2k + 1)^s = 1$ and it can be shown that $$\lim_{s \to 0^+} \left (-\Gamma (-s) \sin \left (\frac{s\pi}{2} \right ) \right ) = \frac{\pi}{2},$$ giving $$\lim_{s \to 0^+} \int_0^\infty \frac{\sin^{2n + 1}(x)}{x^{1 + s}} \, dx = \frac{\pi}{2^{2n + 1}} \sum_{k = 0}^n (-1)^k \binom{2n+1}{n+k+1}.$$ The binomial sum can be found using a telescoping trick. The result is: $$\sum_{k = 0}^n (-1)^k \binom{2n+1}{n+k+1} = \binom{2n}{n}.$$ So finally $$\lim_{s \to 0^+} \int_0^\infty \frac{\sin^{2n + 1}(x)}{x^{1 + s}} \, dx = \frac{\pi}{2^{2n + 1}} \binom{2n}{n},$$ which is a regularised version of the second result I initially gave and the final result for the integral then follows as before.

Okay, I believe that salvages my initial result but it is hardly straight forward nor a "neat" way of proving your result.

Conclusion

Stick with @Sangchul Lee's solution.