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I am performing asymptotic expansion of the following integral:

$$ \mathcal{I}(\beta) = \int_{\mathbb{R}} \tanh\left(\beta x\right) x \exp\left(-\frac{x^2}{2}\right) \mathrm{d} x, $$

for some parameter $\beta > 0$ using

$$ \tanh(x) = \sum_{n=1}^{\infty} \frac{2^{2n}\left(2^{2n} - 1\right) \mathcal{B}_{2n}}{(2n)!} x^{2n-1}, $$ for $\left\lvert x \right \rvert < \pi / 2$, where $\mathcal{B}_{2n}$ is the $2n$-th Bernoulli number. Assuming we may interchange the order of the summation and integral, we have $$ \mathcal{I}(\beta) = \sqrt{2\pi} \sum_{n=1}^{\infty} \frac{2^n \left(2^{2n} - 1\right) \mathcal{B}_{2n}}{n!} \beta^{2n-1}, $$ by considering Gaussian moments.

Noted the resulting series is similar to that of the $\tanh(x)$, yet I failed to identify if it can be represented by some elementary functions or special integral functions. Or if it is possible that the integral $\mathcal{I}(\beta)$ can be evaluated in some other ways?

Thanks in advance.

Jiayao Zhang
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  • Note that a partial integration leads to$$I(\beta)=\beta\int_{\mathbb R}{e^{-x^2/2}\over\cosh^2(\beta x)}>dx\ .$$ Maybe you could do something with the residue theorem. – Christian Blatter May 25 '18 at 08:34

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