What to do about it?

Consider: $I(a)=\int_{0}^{\infty}\dfrac{x^{a}}{e^{2\pi x}-1}dx$ Rewrite it as: $I(a)=\int_{0}^{\infty}\dfrac{x^{a}e^{-2\pi x}}{1-e^{-2\pi x}}dx$ Using the Geometric Infinite Series and setting $t = 2\pi x$ : $I(a) =\dfrac{1}{(2\pi)^{a+1}}\sum_{r=1}^{\infty}\dfrac{1}{r^{a+1}}\int_{0}^{\infty}t^{a}e^{-t}dt$ which is by definitions of Zeta and Gamma Functions: $I(a) =\dfrac{\Gamma(a+1)\zeta(a+1)}{(2\pi)^{a+1}}$ Now, the required integral can be split up as: $\ln(2\pi)I(1) + I'(1)$ Using the fact that $\zeta'(-1)=\dfrac{1}{12}-\ln(A)$ ---------------(1)
we can get the integral easily to be $\dfrac{(\ln(2\pi)+1)}{24}-\dfrac{1}{2}\ln(A)$ Note that the solution is not complete and the derivation of (1) is needed