Prove \(\int_{0}^{1} \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2\)

I stumbled upon the interesting definite integral
\int\limits_{0}^{1} \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2

Here is my proof of this result.

Let \(u=\sin^{-1}(x)\) then integrate by parts,
\int \frac{\sin^{-1}(x)}{x} dx &= \int u \cot(u) du \\
&= u \ln\sin(u) – \int \ln\sin(u) du

\int \ln\sin(u) du &= \int \ln\left(\frac{\mathrm{e}^{iu} – \mathrm{e}^{-iu}}{i2} \right) du \\
&= \int \ln \left(\mathrm{e}^{iu} – \mathrm{e}^{-iu} \right) du \,- \int \ln (i2) du \\
&= \int \ln \left(1 – \mathrm{e}^{-i2u} \right) du + \int \ln \mathrm{e}^{iu} du \,-\, u\ln (i2) \\
&= \int \ln \left(1 – \mathrm{e}^{-i2u} \right) du + \frac{i}{2}u^{2} -u\ln 2 \,-\, ui\frac{\pi}{2}

To evaluate the integral above, let \(y=\mathrm{e}^{-i2u}\)
\int \ln \left(1 – \mathrm{e}^{-i2u} \right) du = \frac{i}{2} \int \frac{\ln (1-y)}{y} dy
= -\frac{i}{2} \mathrm{Li}_{2}(y) = -\frac{i}{2} \mathrm{Li}_{2}\mathrm{e}^{-i2u}

Now we substitute equation \eqref{eq:20161030-3} into equation \eqref{eq:20161030-2}, then substitute that result into equation \eqref{eq:20161030-1}, switch variables back to \(x\), and apply limits,
\int\limits_{0}^{1} \frac{\sin^{-1}(x)}{x} dx
&= \sin^{-1}(x) \ln (x) + \sin^{-1}(x)\left(\ln 2 + i\frac{\pi}{2}\right) \\
&- \frac{i}{2}[\sin^{-1}(x)]^{2} + \frac{i}{2} \mathrm{Li}_{2}\mathrm{e}^{-i2\sin^{-1}(x)} \Big|_{0}^{1} \\
&= \frac{\pi}{2}\ln2

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