Integral with arctan and e: $\int_{0}^{\infty}\frac{\arctan(x^{3})}{e^{2\pi x}-1}\,\mathrm dx$

Question

Here is an integral I ran across that appears to be tough.

It seems to me I have seen integrals like this before. I have looked around the site, but saw nothing like this.

$$\displaystyle \int_{0}^{\infty}\frac{\arctan(x^{3})}{e^{2\pi x}-1}\,\mathrm dx=\frac{1}{4}\ln(2\pi)-\frac{\pi}{4\sqrt{3}}-\frac{1}{2}\ln(1+e^{-\pi\sqrt{3}})$$

I was wondering if it is possible to do this with residues/branch cuts?.

I have not seen many integrals with inverse trig functions done using residues. Maybe there is a reason for that :)

I realize that $\arctan(z^{3})=\frac{i}{2}\ln\left(\left|\frac{z^{3}-i}{z^{3}+i}\right|\right)-\frac{1}{2}\operatorname{Arg}\left(\frac{z^{3}-i}{z^{3}+i}\right)$.

Then, the zeros of $z^{3}+i$ and $z^{3}-i$ would have to be dealt with, as well as those of $e^{2\pi z}-1=0$.

But, I am not sure how to proceed...if possible.

Besides using contours, any clever method would be nice to see and appreciated.

Thanks all.

Answer 1

you can seehttp://www.artofproblemsolving.com/Forum/viewtopic.php?f=67&t=493626