Prove $\gamma = \frac{1}{2} + 2 \cdot \int_0^\infty \frac{\sin(\arctan(x))}{(e^{2 \pi x} - 1) \cdot \sqrt{1 + x^2} } dx$

Question

I've found the following integral on the Wikipediapage of the Euler-Mascheroni constant and I want to prove it.

$\gamma = \frac{1}{2} + 2 \cdot \int_0^\infty \frac{\sin(\arctan(x))}{(e^{2 \pi x} - 1) \cdot \sqrt{1 + x^2} } dx$

I know that $\sin(\arctan(x)) = \frac{t}{\sqrt{t^2 + 1}}$. I tried to apply the Abel-Plana- Formula to the first derivate of the digammafunction, but it does not work.

Any help would be appreciated. Thanks in advance.

Answer 1

Use the fact that $$\psi(z)=\log(z)-\frac{1}{2z}-2\int_0^\infty \frac{x}{(x^2+z^2)(e^{2\pi x}-1)}dx$$ Since $\sin(\arctan(x))=\frac{x}{\sqrt{1+x^2}}$, the integral is equal to $$-\psi(1)=\frac{1}{2}+2\int_0^\infty \frac{x}{(x^2+1)(e^{2\pi x}-1)}dx$$ and using $\psi(1)=-\gamma$, the result follows.

EDIT: $\psi$ is the Digamma Function.

Answer 2

Thanks to @Gary I fixed this answer.

Recall Hermite´s integral representation of the Hurwitz zeta function.

$$\zeta(s,u)=\frac{u^{-s}}{2}+\frac{u^{1-s}}{s-1}+2\int_0^{\infty}\frac{\sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{\left( u^2+x^2\right)^{s/2}\left(e^{2 \pi x}-1\right)}dx\qquad$$

Which ca be rewritten as

$$\zeta(s,u)-\frac{u^{1-s}}{s-1}=\frac{u^{-s}}{2}+2\int_0^{\infty}\frac{\sin\left( s \arctan \left( \frac{x}{u}\right)\right)}{\left( u^2+x^2\right)^{s/2}\left(e^{2 \pi x}-1\right)}dx\qquad\tag{1}$$

Letting $u =1$ and taking the limit as $s \to 1$ in $(1)$ we obtain

\begin{align} \frac12+2\int_0^{\infty}\frac{\sin\left( \arctan \left( x\right)\right)}{\sqrt{1+x^2}\left(e^{2 \pi x}-1\right)}dx&=\lim_{s \to 1} \left[\zeta(s,1)-\frac{1}{s-1}\right]\\ &=\lim_{s \to 1} \left[\zeta(s)-\frac{1}{s-1}\right]\\ &=\gamma \qquad \blacksquare \end{align}