Evaluating $\int_{0}^{\infty} \left( \text{coth} (x) - x \text{csch}^2 (x) \right) \left( \ln \left( \frac{4 \pi^2}{x^2} + 1 \right) \right) \, dx$

Question

How can the following improper integral be evaluated?

$$\int_{0}^{\infty} \left( \text{coth} (x) - x \text{csch}^2 (x) \right) \left( \ln \left( \frac{4 \pi^2}{x^2} + 1 \right) \right) \, dx$$

or alternatively:

$$\int_{0}^{\infty} \frac{x \text{coth} (x) - 1}{x^2 (2 \pi + i x)} \, dx$$ Note: I am only really interested in the imaginary component of the second integral.

I've attempted multiple methods, all of which seeming unsuccessful, however, I believe contour integration may be the solution to the second integral above, which would also easily allow me to get the integral I'm interested in.

Answer 1

Partial solution

We can present the second integral as a series (imaginary part of the integral forms a series) $$I= \int_{0}^{\infty} \frac{x \text{coth} (x) - 1}{x^2 (2 \pi + i x)} \, dx=\int_{0}^{\infty} \frac{(x \text{coth} (x) - 1)(2\pi-ix)}{x^2 ((2 \pi)^2 + x^2)} \, dx$$ $$x\coth x=1+2\sum_{k=1}^{\infty}\frac{x^2}{\pi^2k^2+x^2}$$

$$I=2\int_{0}^{\infty}dx\frac{2\pi-ix}{(2\pi)^2+x^2}\sum_{k=1}^{\infty}\frac{1}{\pi^2k^2+x^2}=\Re I+i\Im I=I_1+I_2$$ $$I_1=2\pi\int_{0}^{\infty} \frac{x \text{coth} (x) - 1}{x^2 (4 \pi^2 + x^2)}dx=4\pi\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{1}{(2\pi)^2+x^2}\frac{1}{\pi^2k^2+x^2}dx$$$$=4\pi\sum_{k=1}^{\infty}\int_{0}^{\infty}\Bigl(\frac{1}{(2\pi)^2+x^2}-\frac{1}{\pi^2k^2+x^2}\Bigr)\frac{dx}{\pi^2k^2-4\pi^2}$$ $$=4\pi\sum_{k=1}^{\infty}\frac{1}{\pi^2k^2-4\pi^2}\Bigl(\frac{1}{2\pi}-\frac{1}{\pi k}\Bigr)\frac{\pi}{2}=2\pi^2\sum_{k=1}^{\infty}\frac{1}{2\pi^2k(2\pi+\pi k)}=\frac{1}{\pi}\sum_{k=1}^{\infty}\frac{1}{k(k+2)}$$ $$I_1=\frac{1}{2\pi}\sum_{k=1}^{\infty}\biggl(\frac{1}{k}-\frac{1}{k+2}\biggr)=\frac{3}{4\pi}$$ $$I_2=-2i\sum_{k=1}^{\infty}\int_{0}^{\infty}\frac{1}{(2\pi)^2+x^2}\,\frac{x}{\pi^2k^2+x^2}dx$$ For $k\neq2$ $$I_2=-2i\sum_{k\neq2}^{\infty}\lim_{N\to\infty}\int_{0}^{N}\Bigl(\frac{1}{(2\pi)^2+x^2}-\frac{1}{\pi^2k^2+x^2}\Bigr)\frac{xdx}{\pi^2k^2-4\pi^2}$$ $$=-2i\sum_{k\neq2}^{\infty}\lim_{N\to\infty}\int_{0}^{N}\Bigl(\frac{d(x^2)}{(2\pi)^2(1+\frac{x^2}{(2\pi)^2})}-\frac{d(x^2)}{(\pi k)^2(1+\frac{x^2}{(\pi k)^2})}\Bigr)\frac{1}{\pi^2k^2-4\pi^2}$$ $$=-\frac{4i}{\pi^2}\sum_{k\neq2}^{\infty}\frac{\log\bigl(\frac{k}{2}\bigr)}{k^2-4}$$ For $k=2$ $$I_2=-2i\int_{0}^{\infty}\frac{xdx}{((2\pi)^2+x^2)^2}=-i\int_{0}^{\infty}\frac{dt}{((2\pi)^2+t)^2}=-\frac{i}{4\pi^2}$$

$$I=I_1+I_2=\frac{3}{4\pi}-\frac{4i}{3\pi^2}\log2-\frac{i}{4\pi^2}-\frac{4i}{\pi^2}\sum_{k>2}^{\infty}\frac{\log\bigl(\frac{k}{2}\bigr)}{k^2-4}$$

Answer 2

For large $x$: $$\frac{x\coth(x)-1}{x^2(2\pi+ix)}\approx\frac{x}{x^3}=\frac{1}{x^2}$$ and for small $x$: $$\frac{x\coth(x)-1}{x^2(2\pi+ix)}\approx\frac{1}{2\pi(e+e^{-1})}$$ So by comparison the integral converges. However we cannot split the top of this integral up as the two diverge for $x\to0$ unless u take a limit for the lower bound