I am trying to integrate $$\int_{0}^{\infty}\left(\frac{\tanh(x)}{x^2}-\frac{\mathrm{sech}^2(x)}{x}\right)\cdot \frac{\mathrm dx}{\sinh(x)}\tag1$$ and to simplify a little bit, $$\int_{0}^{\infty}\frac{\tanh(x)}{x^2}\cdot \frac{\mathrm dx}{\sinh(x)}-\int_{0}^{\infty}\frac{\mathrm{sech}(x)}{x}\cdot \frac{\mathrm dx}{\sinh(x)}$$

$$\int_{0}^{\infty}\frac{1}{x^2\cosh(x)} \mathrm dx-\int_{0}^{\infty}\frac{2}{x}\cdot \frac{\mathrm dx}{\cosh(x)\sinh(2x)}$$

I don't think I can use series here to help me.

Can anyone help to evaluate $(1)?$

  • 1
    $\begingroup$ I don’t think splitting the original integral into two integrals like that will be useful since the last two integrals don’t seem to converge. $\endgroup$ Dec 19, 2019 at 13:12
  • $\begingroup$ The answer seems to be $\ln 2$ according to WA. $\endgroup$
    – Toby Mak
    Dec 19, 2019 at 13:38
  • $\begingroup$ Agree with @AlexMacedo ... writing a convergent integral as the difference of two divergent integrals is not helpful. $\endgroup$
    – GEdgar
    Dec 19, 2019 at 14:13

1 Answer 1


Write $$ I \equiv \int \limits_0^\infty \left[\frac{\tanh(x)}{x^2} - \frac{1}{x \cosh^2 (x)}\right] \, \frac{\mathrm{d} x}{\sinh(x)} = \int \limits_0^\infty \left[\frac{1}{x^2 \cosh(x)} - \frac{1}{x \sinh(x)} + \frac{\sinh(x)}{x \cosh^2 (x)}\right] \, \mathrm{d} x $$ and integrate by parts to find \begin{align} I &= \left[\frac{1}{x} \left(1 - \frac{1}{\cosh(x)}\right)\right]_{x=0}^{x=\infty} + \int \limits_0^\infty \left[\frac{1}{x^2 \cosh(x)} - \frac{1}{x \sinh(x)} + \frac{1}{x^2}\left(1 - \frac{1}{\cosh(x)}\right)\right] \, \mathrm{d} x \\ &= \int \limits_0^\infty \left[\frac{1}{x^2} - \frac{1}{x \sinh(x)} \right] \, \mathrm{d} x \, . \end{align} The remaining integral has already been evaluated here and in a comment to this question (using Frullani integrals or the pole expansion of $\operatorname{csch}$) and we obtain $I = \ln(2)$.


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