I am trying to find a closed-form expression for the following integral
$$ \int_0^\infty \ln(x) \operatorname{sech}(x)^n dx $$
There are specific values that I would like to generate (Table of Integrals , Series and Products 4.371)
In the Book there are for even n closed forms but I can’t find a general formula for it. I just know that at the beginning we have a $$ -K*\ln(4 / \pi) + Zeta(p)+…..$$
$\newcommand{\d}{\,\mathrm{d}}\newcommand{\sech}{\operatorname{sech}}\newcommand{\csch}{\operatorname{csch}}\newcommand{\res}{\operatorname{Res}}$The work of Blagouchine is relevant here.
$$I_n=\frac{1}{4}\lim_{a\to0^+}\int_{-\infty}^{\infty}\ln(x^2+a^2)\sech^n(x)\d x\\=\ln(2\pi)\cdot\int_0^\infty\sech^n(x)\d x+\pi\Im(\res_{z=\pi i/2}+\res_{z=3\pi i/2})\left(\sech^n(z)\log\Gamma\left(\frac{z}{2\pi i}\right)\right)$$
Consider that $\sech(z)=-i\csch(z-\pi i/2)=-i\sum_{n=0}^\infty\frac{2(1-2^{2n-1})B_{2n}}{(2n)!}\cdot(z-\pi i/2)^{2n-1}$ and $\sech(z)=i\csch(z-3\pi i/2)=i\sum_{n=0}^\infty\frac{2(1-2^{2n-1})B_{2n}}{(2n)!}\cdot(z-3\pi i/2)^{2n-1}$ when $z$ is close to the respective poles.
You can then use: $$\log\Gamma\left(\frac{z}{2\pi i}\right)=\log\Gamma\left(1/4\right)+\frac{z-\pi i/2}{2\pi i}\psi\left(1/4\right)-\frac{(z-\pi i/2)^2}{8\pi^2}\psi^{(1)}\left(1/4\right)+\cdots$$For $z$ near to $\pi i/2$, and a similar expression for $z$ near to $3\pi i/2$, and then in multiplication combine these and sum to compute these residues. The integral of $\sech^n$ is also computable via the recurrence: $$\int\sech^n(x)\d x=\frac{1}{n-1}\operatorname{tanh}(x)\sech^{n-2}(x)+\frac{n-2}{n-1}\int\sech^{n-2}(x)\d x$$
In this way, for any particular $n$, you have a "straightforward" algorithm to follow to compute your desired integral in terms of (sometimes known, at other times their sums or differences will be deducible from the reflection or multiplication theorems) polygamma values and Bernoulli numbers. This is reasonable to carry out by hand for small $n$. However this approach does not seem to produce a simple closed form.
Example: for $n=2$ the integral of $\sech^2$ is easily $1$. By squaring the series I wrote (including the prefactors $\pm i$, which both square to $-1$) we see the Laurent series for $\sech^2$ contains only a $-(z-a)^{-2}$ term at each pole $a$. The residues are then $(-1)$ of the first order terms in the log-gamma expansion, which are $\psi(1/4)/2\pi i,\psi(3/4)/2\pi i$.
$-1/i=i$ allows us to easily read off the imaginary parts and we find: $$\begin{align}I_2&=\ln(2\pi)+\pi\cdot\frac{\psi(1/4)+\psi(3/4)}{2\pi}\\&=\ln(2\pi)+\pi\cdot\frac{2(\psi(1/2)-\ln2)}{2\pi}\\&=\psi(1/2)+\ln(\pi)\\&=\ln\frac{\pi}{4}-\gamma\end{align}$$Using the Gauss multiplication formula. This agrees with the table of values you show us. It's worth noting their "$C$" is not Catalan's constant but it is notation for $\gamma$ the Euler-Mascheroni constant.
