Analytic solution to given integration Is there an analytic solution to the following integral:
$\int_B^\infty \mathrm{d}\Lambda \frac{ e^{\frac{\pi  (\Lambda +1)}{c}}}{ \left(e^{\frac{2 \pi  \Lambda }{c}}+e^{\frac{2 \pi }{c}}\right)}  \tan ^{-1}\left(\frac{2 \Lambda -2}{c}\right)$
where $c\in \mathbb{R^+}$.
Mathematica shows that the integral converges even for any random value of $B\in (-\infty,\infty)$.
 A: $$I=\int \frac{e^{\frac{\pi  (x+1)}{c}} }{e^{\frac{2 \pi  x}{c}}+e^{\frac{2 \pi }{c}}}\,\tan ^{-1}\left(\frac{2 x-2}{c}\right)\,dx$$
$$x=1+\frac{c }{2}t\implies I=\frac{c}{4} \int \tan ^{-1}(t)\,\, \text{sech}\left(\frac{\pi }{2}t\right)\, dt=\frac c{2\pi}\int \tan ^{-1}\left(\frac{2 y}{\pi }\right) \text{sech}(y)\,dy $$ I do not see how to have a closed form but this integral looks like a distribution function.
However, we can write
$$I=\frac c \pi \int \frac {\tan ^{-1}\left(\frac{2 y}{\pi }\right) }   {e^y+e^{-y} }\,dy=\frac c \pi \sum_{n=0}^\infty (-1)^n \int e^{-(2 n+1) y} \tan ^{-1}\left(\frac{2 y}{\pi }\right) \,dy$$
$$J_n= \int_b^\infty e^{-(2 n+1) y} \tan ^{-1}\left(\frac{2 y}{\pi }\right) \,dy$$
$$J_n=\frac {(-1)^n } {2(2n+1) }\Bigg[\text{Ei}\left(\frac{i}{2}  (2 n+1) (2 i
   b-\pi)\right)+\text{Ei}\left(\frac{i}{2}  (2 n+1) (2 i b+\pi )\right)+2 (-1)^n
   e^{-(2 n+1) b} \tan ^{-1}\left(\frac{2 b}{\pi }\right)  \Bigg]$$
$$K_n=\int_0^\infty (-1)^n
   e^{-(2 n+1) y} \tan ^{-1}\left(\frac{2 y}{\pi }\right)\,dy=\frac 1{2n+1}\text{Ci}\left(  (2 n+1)\frac{\pi }{2}\right)$$
