# Evaluation of integral involving $\tanh(ax)$

Is it possible to evaluate in closed form the integral

$$\int_{-\sqrt{x}}^{\sqrt{x}}\frac{r\tanh(ar)}{\sqrt{x-r^{2}}}dr=2\int_{0}^{\sqrt{x}}\frac{r\tanh(ar)}{\sqrt{x-r^{2}}}dr$$

here $a$ is a positive constant, $\tanh(x)$ means the Hyperbolic tangent and $x$ is also positive.

If I apprximate $\tanh(x) \sim 1+ \frac{1}{x}$ I know how to compute the integral of course :)

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That's not an approximtion for $\tanh$, neither at $0$ nor at infinity. At $0$ it diverges whereas $\tanh$ vanishes. –  joriki Jan 28 '13 at 10:21

A way forward, albeit not very pretty, is to express the $\tanh$ term as exponentials:

\begin{align} \tanh{a r} &= \frac{1-e^{-2 a r}}{1+e^{-2 a r}}\\ &= (1-e^{-2 a r}) \sum_{k=0}^{\infty} (-1)^k e^{-2 k a r} \\ &= \sum_{k=0}^{\infty} (-1)^k (e^{-2 k a r}-e^{-2 (k+1)a r}) \end{align}

so that the integral becomes

\begin{align}\int_{0}^{\sqrt{x}}dr \:\frac{r}{\sqrt{x-r^{2}}} \sum_{k=0}^{\infty} (-1)^k (e^{-2 k a r}-e^{-2 (k+1)a r}) &= \sum_{k=0}^{\infty} (-1)^k \int_{0}^{\sqrt{x}}dr \:\frac{r}{\sqrt{x-r^{2}}} (e^{-2 k a r}-e^{-2 (k+1)a r}) \\ \end{align}

It turns out that the integrals may be evaluated in terms of Bessel and Struve functions:

$$\int_{0}^{\sqrt{x}}dr \:\frac{r}{\sqrt{x-r^{2}}} (e^{-2 k a r}-e^{-2 (k+1)a r}) = \frac{1}{2} \pi \sqrt{x} (\pmb{L}_{-1}(2 a \sqrt{x} k)-\pmb{L}_{-1}(2 a \sqrt{x} (k+1))-I_1(2 a \sqrt{x} k)+I_1(2 a \sqrt{x} (k+1)))$$

This is about as far as I am able to go for now. I cannot determine if the series is summable in closed form; Mathematica is not able to do it, but that does not mean it is so. Also I do not have a feel for the rate of convergence of the series, so I am not sure how useful it is compared to a simple numerical evaluation or, as you imply, a power series approximation for small $r$.

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