When I used the formula $$\frac{e^{2a\pi x}}{e^{2\pi x}-1}=\frac1{2\pi x}+\frac1\pi\sum_{k\ge1}\frac1{x^2+k^2}\big(x\cos2ak\pi-k\sin2ak\pi\big)\tag{$\Re(a)\le1$}$$ to evaluate $$\int_0^{\infty}\left(\frac{e^{2a\pi x}}{e^{2\pi x}-1}-\frac1{2\pi x}\right)\frac{dx}{x^2+b^2}\:,\tag1$$I got \begin{align}&\frac1\pi\int_0^\infty\sum_{k\ge1}\bigg(\frac{x\cos2ak\pi}{(x^2+k^2)(x^2+b^2)}-\frac{k\sin2ak\pi}{(x^2+k^2)(x^2+b^2)}\bigg)dx\\[2ex] &=\frac1\pi\int_0^\infty\sum_{k\ge1}\frac{x\cos2ak\pi}{b^2-k^2}\bigg(\frac1{x^2+k^2}-\frac1{x^2+b^2}\bigg)dx\\[2ex] &\quad-\frac1\pi\int_0^\infty\sum_{k\ge1}\frac{k\sin2ak\pi}{b^2-k^2}\bigg(\frac1{x^2+k^2}-\frac1{x^2+b^2}\bigg)dx\\[3ex] &=\frac1\pi\sum_{k\ge1}\frac{\cos2ak\pi}{b^2-k^2}\cdot\frac12\ln\frac{x^2+k^2}{x^2+b^2}_0^\infty-\frac1\pi\sum_{k\ge1}\frac{k\sin2ak\pi}{b^2-k^2}\bigg(\frac1k\tan^{-1}\frac xk-\frac1b\tan^{-1}\frac xb\bigg)_0^\infty\\[3ex] &=\frac1{2\pi}\sum_{k\ge1}\frac{\cos2ak\pi}{b^2-k^2}\bigg(\ln\bigg[1+\frac{k^2-b^2}{\infty^2+b^2}\bigg]-\ln\,\frac{k^2}{b^2}\bigg)-\frac12\sum_{k\ge1}\frac{k\sin2ak\pi}{b^2-k^2}\bigg(\frac1k-\frac1b\bigg)\\[3ex] &=\frac1{2\pi}\sum_{k\ge1}\frac{\cos2ak\pi}{k^2-b^2}\big(\ln k^2-\ln b^2\big)-\frac1{2b}\sum_{k\ge1}\frac{\sin2ak\pi}{k+b}\tag{$*$} \end{align}But I see three problems with this formulation:
- While the integral is convergent for positive $b^2$, the sums in $(*)$ have singularities when $b$ is a nonzero integer (those in the left sum are removable, but the right sum's (if $b\lt0$) are not);
- The integral is even in $b$, but the right sum isn't; and
- The sums are periodic in $a$, but not so the integral (the integral definition, however, can be manipulated easily to cover $\Re(a)\gt1$, but it produces the additional integral $\int_0^\infty e^{2a\pi x}\!/(x^2+b^2)\;dx$, which I think requires use of $\operatorname{Ei}$).
I can't figure out how to regularize the expression $(*)$ to match values of the corresponding integral $(1)$.
[Edit]
After evaluating $\int_0^\infty\frac{dx}{x^2+b^2}$ for complex $b$, I was able to make $(*)$ even...but it is now very unwieldy (below, $b$ is still real): \begin{align}\int_0^{\infty}\left(\frac{e^{2a\pi x}}{e^{2\pi x}-1}-\frac1{2\pi x}\right)\frac{dx}{x^2+b^2}=&\frac1\pi\sum_{k\ge1}\frac{\cos2ak\pi}{k^2-b^2}\,\ln k+\frac12\sum_{k\ge1}\frac{\sin2ak\pi}{k^2-b^2}\\ &-\frac{\ln b^2}{4b}\bigg(\frac1b+\pi\sin2ab\pi\bigg)+\frac\pi{4b}\operatorname{sgn}(b)\cos2ab\pi\\[1ex] &+\frac\pi{4b}\cot b\pi\big[\cos2ab\pi\cdot\ln b^2-\operatorname{sgn}(b)\sin2ab\pi\big] \end{align} This new expression has no periodicity issues, like $(*)$, but it obfuscates the singularities.
I was able to rewrite some of the expression by separating the $x$-odd and $x$-even parts of my first formula.