Recently I've come across three infinite integrals of very similar forms:
$\displaystyle \int_{0}^{\infty}{\dfrac{\ln(1+x^2)}{e^{2πnx}-1}}\ \mathrm{d}x\\ $
$\displaystyle\int_{0}^{\infty}{\dfrac{\ln(1+x^2)\arctan x}{e^{2πnx}-1}\ \mathrm{d}x}$
$\displaystyle\int_{0}^{\infty}{\dfrac{\arctan x}{e^{2πnx}-1}\ \mathrm{d}x}$
which $n∈\mathbb{R}_{+}$
The first thing that occurred to me was that maybe the Abel Plana formula is the final solution.
As expected, I used Abel Plana's formula to solve the second and third integral:
First of all, the Hurwitz zeta function is represented by Abel Plana
$\displaystyle \begin{aligned}\zeta(s,a)&=\dfrac{1}{2a^s}+\int_{0}^{\infty}{\dfrac{1}{(x+a)^s} \ \mathrm{d}x}+\dfrac{2}{a^{s-1}}\int_{0}^{\infty}{\dfrac{\sin\left(s\arctan x\right)}{(1+x^2)^{^{\large\frac{s}{2}}}\left(e^{2aπx}-1\right)}\ \mathrm{d}x}\\\\&=\dfrac{a^{-s}}{2}+\dfrac{a^{1-s}}{s-1}+2a^{1-s}\int_{0}^{\infty}{\dfrac{\sin\left(s\arctan x\right)}{(1+x^2)^{^{\large\frac{s}{2}}}\left(e^{2aπx}-1\right)}\ \mathrm{d}x}\end{aligned}\\$
And then take the first and second derivative of both sides with respect to s:
$\displaystyle \begin{aligned}\dfrac{\partial(\zeta(s,a))}{\partial s}&=-\dfrac{a^{-s}\ln a}{2}+\dfrac{a^{1-s}\ln a}{1-s}-\dfrac{a^{1-s}}{(1-s)^2}-2a^{1-s}\ln a\int_{0}^{\infty}{\dfrac{\sin\left(s\arctan x\right)}{(1+x^2)^{^{\large\frac{s}{2}}}\left(e^{2aπx}-1\right)}\ \mathrm{d}x}+2a^{1-s}\int_{0}^{\infty}{\dfrac{\arctan x\cos\left(s\arctan x\right)-{\large\frac{\ln(1+x^2)}{2}}\sin\left(s\arctan x\right)}{(1+x^2)^{^{\large\frac{s}{2}}}\left(e^{2aπx}-1\right)}\ \mathrm{d}x}\end{aligned}\\$
$\displaystyle \begin{aligned}\dfrac{\partial^2(\zeta(s,a))}{\partial s^2}&=\dfrac{a^{-s}\ln^2a}{2}-\dfrac{a^{1-s}\ln^2a}{1-s}+\dfrac{2a^{1-s}\ln a}{(1-s)^2}-\dfrac{2a^{1-s}}{(1-s)^3}-2a^{1-s}\int_{0}^{\infty}{\dfrac{\ln(x^2+1)\arctan x\cos(s\arctan x)}{(x^2+1)^{\large\frac{s}{2}}(e^{2aπx}-1)}\ \mathrm{d}x}+\mathcal{O}(s)\end{aligned}\\$
let s = $0$ and it solves.
However, for the first integral, I could not find a suitable function object, which was obtained by Abel Plana's formula and derivative operation.
Can the first integral be solved using Abel Plana's formula?
If so, which function object is the most suitable?