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a in all descriptions is $\sqrt{2}-1$

I recently tried to change $\frac{1}{\sin{x}}$ to $x$ in a simple Feynman integral problem: $\int_0^{\frac{\pi}{2}}\frac{\arctan(\sin{x})}{\sin{x}}\mathrm{d}x$, but I found that such a change would make the problem too complicated to be solved by Feynman integrals. I tried the calculator, and the result is an infinite series:$-2\sum_{n=1}^{\infty}\frac{(-1)^{n}a^{2n-1}}{(2n-1)^{3}}$ In other words,I found that $$\int_0^{\frac{\pi}{2}}{x\arctan(\sin{x})}\mathrm{d}x=-2\sum_{n=1}^{\infty}\frac{(-1)^{n}a^{2n-1}}{(2n-1)^{3}}$$ but I can't prove this equation.

I tried to expand the integral, but unfortunately I couldn't prove it this way. I have also tried to use the residue theorem, but it is clear that this does not conform to several forms of using the residue theorem. Can someone please prove this equation? Thank you!

spacedog
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1 Answers1

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We can use the second approach that Sangchul Lee used in this answer.

For all real values of $x$, the function $\arctan (\sin x)$ can be represented by the Fourier series $$\arctan(\sin x) =2 \sum_{n=1}^{\infty} \frac{\sin\left((2n-1)x\right)}{2n-1} \left(\sqrt{2}-1\right)^{2n-1}.$$

So we have

$ \begin{align} \int_{0}^{\pi/2} x \arctan (\sin x) \, \mathrm dx &= 2\int_{0}^{\pi/2} x \sum_{n=1}^{\infty} \frac{\sin\left((2n-1)x\right)}{2n-1} \left(\sqrt{2}-1\right)^{2n-1} \, \mathrm dx\\ &= 2\sum_{n=1}^{\infty} \frac{\left(\sqrt{2}-1 \right)^{2n-1}}{2n-1} \int_{0}^{\pi /2} x \sin \left((2n-1)x \right) \, \mathrm dx \\ &= \small 2 \sum_{n=1}^{\infty} \frac{\left(\sqrt{2}-1 \right)^{2n-1}}{2n-1} \left(-\frac{x \cos \left((2n-1)x \right)}{2n-1} \bigg|_{0}^{\pi/2} + \int_{0}^{\pi/2} \frac{\cos \left((2n-1)x \right)}{2n-1} \, \mathrm dx\right) \\ &= 2\sum_{n=1}^{\infty} \frac{\left(\sqrt{2}-1 \right)^{2n-1}}{2n-1} \left(0 + \int_{0}^{\pi/2} \frac{\cos \left((2n-1)x \right)}{2n-1} \, \mathrm dx\right) \\ &= 2\sum_{n=1}^{\infty} \frac{\left(\sqrt{2}-1 \right)^{2n-1}}{2n-1} \frac{\sin \left((2n-1)x \right)}{(2n-1)^{2}} \bigg|_{0}^{\pi/2} \\ &=-2\sum_{n=1}^{\infty} \frac{\left(\sqrt{2}-1 \right)^{2n-1} (-1)^{n}}{(2n-1)^{3}}, \end{align}$

which is the imaginary part of $2\operatorname{Li}_{3}\left((\sqrt{2}-1)i \right).$

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