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Evaluate$$\int_{-1}^1\frac{\sin(x)}{\arcsin(x)} \, dx$$

Since $\frac{\sin(x)}{\arcsin(x)}$ is an even function, my first step was to simplify it to:

$$2\int_0^1\frac{\sin(x)}{\arcsin(x)} \, dx$$

And after trying IBP, I quickly realised that the indefinite integral is non-elementary as verified by WolframAlpha. The usual method I use to continue for these types of problems is by differentiating under the integral sign, however I fail to see how it could help here.

My only other idea is that using contour integration could help, but i'm not to good at it and can't use it very well.

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  • $\begingroup$ The Maclaurin series doesn't look hopeful, so far as getting a closed form solution goes. $\endgroup$ – saulspatz Apr 10 '18 at 16:06
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Well, your integral equals

$$\begin{eqnarray*} 2 \int_{0}^{\pi/2}\frac{\sin\sin\theta\cos\theta}{\theta}\,d\theta&=&4\sum_{m\geq 0}J_{2m+1}(1)\int_{0}^{\pi/2}\frac{\sin((2m+1)\theta)\cos\theta}{\theta}\,d\theta\\&=&2\sum_{m\geq 0}J_{2m+1}(1)\left[\text{Si}(m\pi)+\text{Si}((m+1)\pi)\right]\end{eqnarray*}$$ which is a rapidly convergent series involving Bessel functions of the first kind and the sine integral function $\text{Si}(s)=\int_{0}^{s}\frac{\sin u}{u}\,du =\frac{\pi}{2}+O\left(\frac{1}{s}\right)$. I wouldn't bet on the existence of a simpler representation.

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