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I am trying to find a simple expression for the integral

$\hskip 3cm{\displaystyle 2a\,\text{sin}\,\alpha\int\limits_0^\infty \big(1 - \text{tanh}\,\pi x\,\text{tanh}\,\alpha x}\big)\, dx, \quad a \in \mathbb{R},\, \alpha \in[0, 2\pi)$

The above integral is a simplified version of

$\hskip .3cm{\displaystyle (a\,\text{sin}\,\alpha + b\,\text{sin}\,\beta) \times \int\limits_0^\infty\bigg(1 - 2\,\text{tanh}\,\pi x\,\frac{\text{sinh}\,\alpha x \, \text{sinh}\,\beta x}{\text{sinh} \,(\alpha + \beta)x}\bigg)dx}, \quad a,b\in\mathbb{R}, \, \alpha, \beta \in [0,2\pi)$

when $a = b$ and $\alpha = \beta$. I would ideally like a solution to the latter integral, however, a soultion to the former one is also very helpful. I have tried playing around with trigonometric properties but it seems like I just end up with an expression thats even harder to solve. Either numerical or analytic solutions would be appreciated!

These expressions are derived when trying to solve for the capacitance of two overlapping spheres (picture for reference). enter image description here

$\textbf{EDIT:}$ I guess the appropriate question would be "If I wanted to numerically approximate this integral, what method should I look at?". Off the top of my head, I would just choose a well known quadrature rule and apply a while loop. In this while loop I check if the value being returned is smaller than some threshold, and if it is terminate and return the approximation. Would this be a valid method?

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  • $\begingroup$ If $\alpha$ were a rational multiple of $\pi$ one could use this substitution, which rationalizes the integrand. Then the rational function can be effectively integrated. $\endgroup$ – OR. Jun 30 '17 at 12:53
  • $\begingroup$ Probably the same thing. When $\alpha = \pi m/n$, Maple evaluates your integral as a sum of logarithmic terms, one term for each root of a certain polynomial of degree $n-1$. $\endgroup$ – GEdgar Jun 30 '17 at 13:25
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$$\begin{eqnarray*}\int_{0}^{+\infty}\left(1-\tanh(\pi x)\tanh(\alpha x)\right)\,dx &=&2\int_{0}^{+\infty}\frac{e^{2ax}+e^{2\pi x}}{(1+e^{2ax})(1+e^{2\pi x})}\,dx\\&=&\int_{0}^{+\infty}\frac{\cosh((\pi-a)x)}{\cosh(ax)\cosh(\pi x)}\,dx\end{eqnarray*}$$ and $$ \int_{0}^{+\infty}\frac{dx}{1+e^{2\pi x}}=\frac{\log(2)}{2\pi},\qquad \int_{0}^{+\infty}\frac{dx}{1+e^{2ax}}=\frac{\log(2)}{2a} $$ so the evaluation of the first integral boils down to the evaluation of $$\begin{eqnarray*} \int_{0}^{+\infty}\frac{dx}{(1+e^{2\pi x})(1+e^{2ax})}&=&\int_{1}^{+\infty}\frac{dz}{z(1+z^{2\pi})(1+z^{2a})} \\&\stackrel{z\mapsto z^{-1}}{=}&\int_{0}^{1}\frac{dz}{z(1+z^{-2\pi})(1+z^{-2a})}\\&\stackrel{x\mapsto w^{\frac{1}{2\pi}}}{=}&\frac{1}{2\pi}\int_{0}^{1}\frac{dw}{(1+w)(1+w^{a/\pi})}\end{eqnarray*}$$ that is related with Lerch zeta function.
So, bad news: a simple expression for the first integral does not exist, in general.
However, good news: accurate numerical approximations for the last integral are simple to achieve, and if $\frac{a}{\pi}=\frac{p}{q}\in\mathbb{Q}$ a closed form for the last integral does exist by setting $w=v^q$ and applying partial fraction decomposition.

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  • $\begingroup$ Could you point me towards a resource on numerical approximations for this integral? $\endgroup$ – David Jul 3 '17 at 1:54

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