Prove $1-2\int_{0}^{\infty}{\tan^{-1}(x)\over e^{x\pi}-1}dx=\ln(2)$

Question

I want to prove the following:

$$J=1-2\int_{0}^{\infty}{\tan^{-1}(x)\over e^{x\pi}-1}dx=\color{red}{\ln(2)}.$$

My attempt: we know that $$\tan^{-1}(x)=x-{x^3\over 3}+{x^5\over 5}-\cdots$$

Plugging in, we obtain, $$J=1-2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}\int_{0}^{\infty}{x^{2n-1}\over e^{x\pi}-1}dx$$

Recall that, $${\pi^{2s}\over \Gamma(2s)}\int_{0}^{\infty}{x^{2s-1}\over e^{x\pi}-1}dx=\zeta(2s)$$

Finally, we have,$$J=1-2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2n-1}{\Gamma(2n)\zeta(2n)\over \pi^{2n}}$$

I need help on evaluating the sum $J$. Some help please?

Answer 1

I waited a lifetime for this moment!

It is enough to use Binet's second $\log\Gamma$ formula with $z=\color{red}{\frac{1}{2}}$.

Notice that you cannot simply plug in the Taylor series for the arctangent function, since the radius of convergence of its power series at $x=0$ is finite (it is $1$), so that way you get a divergent series. But Binet's second $\log\Gamma$ formula is a not so difficult consequence of Frullani's theorem or the Abel-Plana formula, since $\arctan(x)=\text{Im}\log(1+ix)$.

Answer 2

We have $$I=\int_{0}^{\infty}\frac{\arctan\left(x\right)}{e^{\pi x}-1}dx=\sum_{k\geq1}\int_{0}^{\infty}e^{-k\pi x}\arctan(x)dx $$ $$=\frac{1}{\pi}\sum_{k\geq1}\frac{1}{k}\int_{0}^{\infty}\frac{e^{-k\pi x}}{1+x^{2}}dx\tag{1} $$ and this is the Laplace transform of $\frac{1}{1+x^{2}}$ at $s=\pi k$. This integral can be expressed in terms of the Sine integral function and Cosine integral function (it follows from the Laplace transform of the sine function), but in our case the integral is $$I=\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k}\left(\frac{\textrm{Si}\left(k\pi\right)}{\pi}-\frac{1}{2}\right)\tag{2} $$ $$=\frac{\log\left(2\right)}{2}-\frac{1}{\pi}\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k}\textrm{Si}\left(k\pi\right) $$and now we can observe from the Taylor series of the $\textrm{Si}\left(k\pi\right) $ function that $$\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k}\textrm{Si}\left(k\pi\right)=\sum_{k\geq1}\frac{\left(-1\right)^{k}}{k}\sum_{m\geq1}\left(-1\right)^{m}\frac{\left(k\pi\right)^{2m-1}}{\left(2m-1\right)\left(2m-1\right)!} $$ $$=\sum_{m\geq1}\left(-1\right)^{m}\frac{\pi^{2m-1}}{\left(2m-1\right)\left(2m-1\right)!}\zeta\left(-2m+2\right) $$ $$=\zeta\left(0\right)\pi=-\frac{\pi}{2} $$ since $\zeta\left(-2m\right)=0,\,\forall m\geq1$ and this conclude the proof.

Note. A way for proving $(2)$ from $(1)$ is the following: $$ \begin{align} \int_0^\infty\frac{e^{-sx}}{1+x^2}dx &=\int_0^\infty\int_0^\infty e^{-sx}e^{-xy}\sin(y)dydx\\ &=\int_0^\infty\frac{\sin(y)}{s+y}dy\\ &=\int_s^\infty\frac{\sin(y)\cos(s)-\cos(y)\sin(s)}{y}dy\\ &=\textrm{Ci}(s)\sin(s)+\frac{\pi\cos(s)}{2}-\textrm{Si}(s)\cos(s). \end{align} $$