Evaluate the improper integral $\int _1^\infty x^{-10/9} \coth(x) \,\mathrm dx$

Question

Could someone help me to evaluate this integral please: $$\int _1^\infty \dfrac{1}{x^{10/9} \tanh(x)} \,\mathrm dx$$

I tried using change variable method in order to change the integral bound.

Answer 1

By considering the logarithmic derivative of the Weierstrass product for the $\sinh$ function: $$ I=\int_{1}^{+\infty}\frac{\coth(x)}{x^{10/9}}\,dx = 2\sum_{n\geq 1}\int_{1}^{+\infty}\frac{dx}{x^{1/9}(\pi^2 n^2+x^2)}=2\sum_{n\geq 1}\int_{0}^{1}\frac{x^{1/9}dx}{1+\pi^2 n^2x^2}\tag{1}$$ hence: $$ I = 18\sum_{n\geq 1}\int_{0}^{1}\frac{x^9\,dx}{1+\pi^2 n^2 x^{18}}\tag{2} $$

where the last integrals can be computed through partial fraction decomposition.
The eighteenth roots of unity are clearly involved.