Show that $\int_0^\infty \frac{x\log(1+x^2)}{e^{2\pi x}+1}dx=\frac{19}{24} - \frac{23}{24}\log 2 - \frac12\log A$ Any idea on how to prove the following integral 
$$\int_{0}^{\infty} {x\log(1+x^2)\over e^{2\pi x}+1}dx =\require{cancel} \cancel{\frac{17}{24} - \frac{23}{24}\log 2 + \frac{1}{2}\log A}={\frac{19}{24} - \frac{23}{24}\log 2 - \frac{1}{2}\log A }$$
Where $A$ is the Glaisher–Kinkelin constant. 
We define 
$$A= \lim_{n \to \infty}\frac{H(n)}{n^{n^2/2+n/2+1/12}e^{-n^2/4}}$$
Where $$H(n) = \prod^{n}_{k=1} k^k $$
I would start by 
$$F(z) = \int^\infty_0 \frac{2xz}{(x^2+z^2)(e^{2\pi x}+1)} \, dx$$
I know that 
$$\frac{2t}{e^{2\pi t}-1} =\frac{1}{\pi}-t+\frac{2t^2}{\pi}\sum_{k=1}^\infty\frac{1}{k^2+t^2} $$
But I can't find a similar one for 
$$\frac{1}{e^{2\pi t}+1}$$
 A: The following approach uses the Abel-Plana formula, which was mentioned in a comment by the user tired.
By applying the Abel-Plana formula to $f(\frac{x}{2})$ and then subtracting the result from the Abel-Plana formula applied to $f(x)$, we get $$\begin{align} &\sum_{n=0}^{\infty} f(n) - \sum_{n=0}^{\infty} f \left(\frac{n}{2} \right) \\ &= \int_{0}^{\infty} f(x) \, dx - \int_{0}^{\infty} f \left(\frac{x}{2} \right) \, dx + i \int_{0}^{\infty} \frac{f(ix) - f(-ix)}{e^{2 \pi x}-1} \, dx -  i \int_{0}^{\infty} \frac{f\left(i \frac{x}{2}\right) - f\left(-i\frac{x}{2}\right)}{e^{2 \pi x}-1} \, dx \\ &= \int_{0}^{\infty} f(x) \, dx - 2 \int_{0}^{\infty} f(u) \, du + i \int_{0}^{\infty} \frac{f(ix) - f(-ix)}{e^{2 \pi x}-1} \, dx - 2 i \int_{0}^{\infty} \frac{f(iu) - f(-iu)}{e^{4 \pi u}-1} \, du  \\ &= -\int_{0}^{\infty} f(x) \,  dx + i \int_{0}^{\infty} \frac{f(ix) - f(-ix)}{e^{2 \pi x}+1} \, dx. \end{align} $$
Let's apply the above formula to the function $$ f(x)= \frac{1}{(1+x)^{s}}  , \quad \text{Re}(s) >1.$$
Doing so, we get $$\sum_{n=0}^{\infty} \frac{1}{(1+n)^{s}} - 2^{s} \sum_{n=0}^{\infty}\frac{1}{(2+n)^{s}} = \frac{1}{1-s} + 2 \int_{0}^{\infty} \frac{\sin \left(s \arctan x \right)}{(1+x^{2})^{s/2}(e^{2 \pi x}+1)} \, dx,$$ which implies that
$$2 \int_{0}^{\infty} \frac{\sin \left(s \arctan x \right)}{(1+x^{2})^{s/2}(e^{2 \pi x}+1)} \, dx =  (1-2^{s}) \zeta(s) + 2^{s} + \frac{1}{s-1}. \tag{1}$$
By analytic continuation, $(1)$ should hold for all complex values of $s$. (For $s=1$, the right side of the equation should be treated as a limit.)
Now if we differentiate under the integral sign and then let $s=-1$, we get
$$\int_{0}^{\infty} \frac{x \log(1+x^{2})}{e^{2 \pi x}+1} \, dx + 2 \int_{0}^{\infty} \frac{\arctan x}{e^{2 \pi x}+1} \, dx = - \frac{\log 2}{2} \underbrace{\zeta(-1)}_{- \frac{1}{12}}+\frac{\zeta'(-1)}{2} + \frac{\log 2}{2}-\frac{1}{4}. $$
But using Binet's second formula for the log gamma function, we have $$ \begin{align} 2 \int_{0}^{\infty} \frac{\arctan x}{e^{2 \pi x}+1} \, dx &= 2 \int_{0}^{\infty} \frac{\arctan x}{e^{2 \pi x}-1} \, dx - 4 \int_{0}^{\infty} \frac{\arctan x}{e^{4 \pi x}-1} \, dx \\ &=2 \int_{0}^{\infty} \frac{\arctan x}{e^{2 \pi x}-1} \, dx -2 \int_{0}^{\infty} \frac{\arctan \left(\frac{u}{2}\right)}{e^{2 \pi u}-1} \, du \\ &= 1- \frac{\log (2 \pi)}{2} + \frac{3}{2} \log 2 -2 + \frac{\log(2 \pi)}{2}  \\ &= \frac{3 \log 2}{2} -1 .\end{align}$$
Therefore, $$ \begin{align}\int_{0}^{\infty} \frac{x \log(1+x^{2})}{e^{2 \pi x}+1} \, dx &= \frac{\log 2}{24} + \frac{\zeta'(-1)}{2} + \frac{\log 2}{2} - \frac{1}{4} + 1 - \frac{3 \log 2}{2} \\ &= \frac{\zeta'(-1)}{2} - \frac{23}{24} \log 2 + \frac{3}{4}. \end{align}$$
