$\int_1^\infty \operatorname{sech} x \cdot \ln x \ dx$ The integral cant be expressed in "standard mathematical functions" -Wolfram
I'm asked to determine if its convergent or divergent (I can do that via comparison theorem) and its convergent (and wolfram agrees) but them I'm asked to evaluate because it's convergent. Am I missing something? Is there a way to evaluate if I cant express the integral?
 A: It is convergent for sure since:
$$ 0\leq \int_{1}^{+\infty}\frac{2\log x}{e^{x}+e^{-x}}\,dx\leq \int_{1}^{+\infty}2(x-1)e^{-x}\,dx = \frac{2}{e}.$$
The upper bound can be improved up to $\frac{\sqrt{\pi}}{e}$ if we exploit $\log x\leq\sqrt{x-1}$ for any $x\geq 1$.
A: One method to attempt to evaluate the integral is as follows. 
Using 
\begin{align}
sech(x) = 2 \, \sum_{k=0}^{\infty} (-1)^{k} \, e^{-(2k+1) \, x}
\end{align}
then
\begin{align}
I(x) &= \int_{1}^{\infty} \frac{t^{x-1} \, dt}{\cosh(t)} \\
&= 2 \, \sum_{k=0}^{\infty} (-1)^{k} \, \int_{1}^{\infty} t^{x-1} \, e^{-(2k+1) t} \, dt \\
&= 2 \, \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^{x}} \, \int_{2k+1}^{\infty} e^{-u} \, u^{x-1} \, du \mbox{  where  } u = (2k+1) t \\
&= 2 \sum_{k=0}^{\infty} \frac{(-1)^{k} \, \Gamma(x, 2k+1)}{(2k+1)^{x}},
\end{align}
where $\Gamma(x,\alpha)$ is one of the incomplete gamma functions. Now taking the derivative of both sides with respect to $x$ and setting $x=1$ leads to the desired integral being
\begin{align}
\int_{1}^{\infty} \frac{\ln(t) \, dt}{\cosh(t)} = 2 \, \sum_{k=0}^{\infty} \frac{(-1)^{k} \, \Gamma'(1,2k+1)}{2k+1} - 2 \, \sum_{k=0}^{\infty} \frac{(-1)^{k} \, \ln(2k+1) \, \Gamma(1,2k+1)}{2k+1}.
\end{align}
