# Is there a nice way to find this integral $\int_0^1\frac{ \arcsin x}{x} \mathrm{d}x$?

$$\int_0^1\frac{ \arcsin x}{x}\,\mathrm dx$$

I was looking in my calculus text by chance when I saw this example , the solution is written also but it uses very tricky methods for me ! I wonder If there is a nice way to find this integral.

The idea of the solution in the text is in brief , Assume $y=\sin(x)$ and use definition of improper integral and some properties of definite integral to get $-\lim_{\varepsilon \rightarrow0^+} \int_\epsilon^{\frac{\pi}{2}} \ln(\cos(y-\varepsilon))\,dy$ then it uses that fact that $a= (a+a) \times \frac{1}{2}$ where $a$ here is the integral , and then there is a step which I can't understand till the moment( but understand most of the rest of the steps ) . and it keeps going to use more and more tricks to get the final result , $\frac{\pi}{2} \ln2$.

Now, I try to understand this method , If I couldn't I will ask for help, but in the moment, I wonder if there is a good way to find this integral.

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$$I = \int_0^1 \frac{\arcsin x}{x}\,dx = \int_{0}^{\pi/2}\frac{x\,dx}{\sin x}\cos x = x\log\sin x \bigg|_0^{\pi/2}-\int_0^{\pi/2} \log\sin x\,dx$$

This last integral is well known to equal $-\frac{\pi}{2}\log 2$. Thus

$$I = \frac{\pi}{2}\log 2$$

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Using integration by parts with $u=\arcsin(x)$ yields

$$\int_0^1 \frac{\arcsin x}{x}\,dx = -\int_0^1 \frac{\ln(x)}{\sqrt{1-x^2}}\,dx=I .$$

Now, consider the integral

$$F = \int_0^1 \frac{x^\alpha}{\sqrt{1-x^2}}\,dx = \frac{\sqrt{\pi}}{2}{\frac {\Gamma \left( \frac{\alpha}{2}+\frac{1}{2} \right) }{ \Gamma \left( \frac{\alpha}{2}+1 \right) }},$$

which can be evaluated using the $\beta$ function (subs $x^2=t$ in F). Now, $I$ follows from $F$ as

$$I = \lim_{\alpha\to 0}\frac{dF}{d\alpha}= \frac{\pi \ln(2)}{2}.$$

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