Two integrals related to Binet´s second formula

I came across these two integrals here.

$\int_{0}^{\infty}\frac{x}{(1+x^2)(e^{2\pi x-1})}dx=\frac{\gamma}{2}-\frac{1}{4}$

$\int_{0}^{\infty}\frac{x}{(1+x^2)^2(e^{2\pi x-1})}dx=\frac{\pi^2}{24}-\frac{3}{8}$

They are very interesting integrals because through Stirling´s approximation for the gamma function and a more general version of them, it´s possible to proof Binet´s second formula for the Loggamma function. Today, I´ll show the proof of the two results and in a future post, the proof of Binet´s second formula. Lets start by recalling the Laplace transform of sin(xt)

$\int_{0}^{\infty}e^{-st}\sin(xt)dt=\frac{x}{x^2+s^2}$

$\int_{0}^{\infty}\frac{x}{(s^2+x^2)(e^{2\pi x}-1)}dx=\int_{0}^{\infty}\frac{1}{(e^{2\pi x-1})}\int_{0}^{\infty}e^{-st}\sin(xt)dt\,\,dx$

$=\int_{0}^{\infty}e^{-st}\int_{0}^{\infty}\frac{\sin(xt)}{(e^{2\pi x}-1)}dx\,\,dt$

$=\int_{0}^{\infty}e^{-st}\sum_{k=1}^{\infty}\frac{t}{(2 \pi k)^2+t^2}dt$

$=\int_{0}^{\infty}e^{-st}\left\{ -\frac{1}{2t}+\frac{\coth\left(\frac{t}{2}\right)}{4}\right\}dt$

The proof of the inner integral can be found here

Now, let t=2w

$\int_{0}^{\infty}\frac{x}{(1+x^2)(e^{2\pi x}-1)}dx=-\frac{1}{2}\int_{0}^{\infty}e^{-2sw}\left\{ \frac{1}{w}-\coth\left(w\right)\right\}dw$ (1)

Lets now focus in the integral on the right hand side of (1)

$I= \int_{0}^{\infty} e^{-\beta w}\left(\frac{1}{w}-\operatorname{coth}(w)\right) d w$

From the definition of $\operatorname{coth}(w)$ we can rewrite

$\operatorname{coth}(w)=\frac{e^{w}+e^{-w}}{e^{w}-e^{-w}}$

$\operatorname{coth}(w)=\frac{e^{-w}}{e^{-w}} \cdot \frac{e^{w}+e^{-w}}{e^{w}-e^{-w}}$

$\operatorname{coth}(w)=\frac{1+e^{-2 w}}{1-e^{-2 w}}$

Plugging this result in the integral

$I= \int_{0}^{\infty} e^{-\beta w}\left(\frac{1}{w}-\frac{1+e^{-2 w}}{1-e^{-2 w}}\right) d w$

Lets do another substitution, $w=\frac{x}{2}$

$I= \int_{0}^{\infty} e^{-\frac{\beta x}{2 }}\left(\frac{2}{x}-\frac{1+e^{-x}}{1-e^{-x}}\right) \frac{d x}{2}$

$I= \int_{0}^{\infty} e^{-\frac{\beta x}{2 }}\left(\frac{1}{x}-\frac{1}{2} \cdot \frac{1+e^{-x}}{\left(1-e^{-x}\right)}\right) d x$

$I=\int_{0}^{\infty}\left\{\frac{e^{-\frac{\beta x}{2 }}}{x}-\frac{e^{-\frac{\beta x}{2 }}+e^{-\frac{\beta x}{2 }} e^{-x}}{2\left(1-e^{-x}\right)}\right\} d x$

$I= \int_{0}^{\infty}\left\{\frac{e^{-\frac{\beta x}{2 }}+e^{-x}-e^{-x}}{x}-\frac{e^{-\frac{\beta x}{2 }}+e^{-\frac{\beta x}{2 }}-e^{-\frac{\beta x}{2 }}+e^{-\frac{\beta x}{2 }} e^{-x}}{2\left(1-e^{-x}\right)}\right\} d x$

We now rewrite this integral as the sum of three integrals:

$I=\int_{0}^{\infty} \frac{e^{-x}}{x}-\frac{2 e^{-\frac{\beta x}{2 }}}{2\left(1-e^{-x}\right)} d x+\int_{0}^{\infty} \frac{e^{-\frac{\beta x}{2 }}-e^{-\frac{\beta x}{2 }} e^{-x}}{2\left(1-e^{-x}\right)} d x+\int_{0}^{\infty} \frac{e^{-\frac{\beta x}{2 }}-e^{-x}}{x} d x$

$I=\int_{0}^{\infty} \frac{e^{-x}}{x}-\frac{e^{-\frac{\beta x}{2 }}}{\left(1-e^{-x}\right)} d x+\frac{1}{2} \int_{0}^{\infty} e^{-\frac{\beta x}{2 }} d x-\int_{0}^{\infty} \frac{e^{-x}-e^{-\frac{\beta x}{2 }}}{x} d x$

the first integral is the integral representation of the Digamma function $\psi(\frac{\beta}{2})$

the third is Frullani’s integral representation of $\log \left(\frac{\beta}{2}\right)$

The second integral is straightforward:

$\frac{1}{2} \int_{0}^{\infty} e^{-\frac{\beta x}{2 }} d x=\frac{1}{2}\left[-\left.\frac{2}{\beta} e^{-\frac{\beta x}{2 }}\right|_{0} ^{\infty}\right]=\frac{1}{\beta}$

Putting all together we finally get that

$\boxed{I= \int_{0}^{\infty} e^{-{\beta w}}\left(\frac{1}{w}-\operatorname{coth}(w)\right) d w=\frac{1}{\beta}+\psi\Big(\frac{\beta}{2}\Big)-\log \Big(\frac{\beta}{2}\Big)}$ (2)