I want to calculate
$$
\int_{0}^{\infty} \frac{\arctan{2t}}{e^{2\pi t}+1} \, {\rm d}t
$$
and am wondering if there at all is an analytical expression for the result.
I haven't tried much so far, but expressing the denominator as the geometric series and then substituting $z=2\pi t (n+1)$. Equivalently I get
$$
\sum_{n=0}^{\infty} \frac{{\rm si}\left(\pi(n+1)\right)}{2\pi (n+1)}
$$
where
$$
{\rm si}(x) = -\int_{x}^{\infty} \frac{\sin(t)}{t} \, {\rm d}t
$$
is the shifted ${\rm Si}(x)$ function. The value is approximately $0.03618247146$, in fact I know from comparison with something else that it is exactly $\frac{\ln \pi -1}{4}$, but there should be some (direct) derivation, or?
Thanks for your input if you have any.
Edit: I tried to follow your instructions, but am now stuck.
So I wrote
$$
{\cal L}^{-1} \left(\frac{\psi(t)-\psi(1)}{t-1}\right)(s) =\sum_{n=0}^{\infty} \frac{e^{-sn}}{n+1} = e^s \int_s^\infty \frac{{\rm d}x}{e^{x}-1}
$$
and taking the Laplace-transform
\begin{align}
\psi(t) &= \psi(1) + (t-1)\int_0^\infty {\rm d}s \, e^{-s(t-1)} \int_s^\infty \frac{{\rm d}x}{e^{x}-1} \\
&= \psi(1) + (t-1) \int_0^\infty {\rm d}x \int_{0}^{x} {\rm d}s \, \frac{e^{-s(t-1)}}{e^{x}-1} \\
&= \psi(1) + \int_{0}^{\infty} {\rm d}x \, \frac{1-e^{-x(t-1)}}{e^x-1} \tag{1} \\
&= -\frac{1}{t} + \psi(1) + \int_0^\infty {\rm d}x \, \frac{1-e^{-tx}}{e^x-1} \tag{2} \\
&= \dots \\
&=-\sum_{k=0}^{n} \frac{1}{t+k} + \psi(1) + \int_0^\infty {\rm d}x \, \frac{1-e^{-(t+n)x}}{e^x-1} \\
&\stackrel{(\rm{2})}{=} -\frac{1}{2t} + \ln(t) – \int_0^\infty {\rm d}x \, \left\{ \frac{1}{2} – \frac{1}{x} + \frac{1}{e^x-1}\right\} e^{-tx} \tag{Binet 1}
\end{align}
where $\psi(1)=-\gamma$.
Now we integrate over $t$ from $1$ to $y$ yielding
$$
\ln \Gamma(y) \stackrel{({\rm 1})}{=} \psi(1)\,(y-1) + \int_0^\infty {\rm d}x \, \frac{y-1+\frac{e^{-x(y-1)}-1}{x}}{e^x-1}
$$
but what now? Substitute?
Best Answer
By considering that $\frac{1}{z+1}= \frac{1}{z-1}-\frac{2}{z^2-1}$ and that, by Binet's second $\log\Gamma$ formula, $$ \int_{0}^{+\infty}\frac{\arctan\frac{t}{z}}{e^{2\pi t}-1}\,dt = \frac{1}{2}\left[\log\Gamma(z)-\left(z-\tfrac{1}{2}\right)\log z+z-\log\sqrt{2\pi}\right]\tag{Binet} $$ we get $$ \int_{0}^{+\infty}\frac{\arctan(2t)}{e^{2\pi t}-1}\,dt = \frac{1-\log 2}{4}$$ $$ \int_{0}^{+\infty}\frac{2\arctan(2t)}{e^{4\pi t}-1}\,dt =\int_{0}^{+\infty}\frac{\arctan(t)}{e^{2\pi t}-1}\,dt = \frac{2-\log(2\pi)}{4}$$ and the given integral equals $\color{red}{\frac{-1+\log\pi}{4}}$ as suspected.
I am adding a brief sketch of the proof of $(\text{Binet})$, since the linked Wikipedia page appears to miss it. By applying the (inverse) Laplace transform to the well-known identity $$ \frac{\psi(a)-\psi(b)}{a-b}=\sum_{n\geq 0}\frac{1}{(n+a)(n+b)} $$ one gets an integral representation for the Digamma function. A further step of integration through Fubini's theorem leads to an integral representation for $\log\Gamma$, which can be easily rearranged into $(\text{Binet})$.