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.

  • 1
    $\begingroup$ I don't downvote because the question is not bad. However in general you have to provide some thought(inside the post, not just title) yourself for every question you post on this site. $\endgroup$
    – user175968
    Jun 4, 2016 at 3:52
  • $\begingroup$ have you proved that this integral exists? $\endgroup$ Jun 4, 2016 at 5:42

1 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.


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