normalization of a non-orthogonal fourier series

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I'm learning about the 1-dimensional diffusion/heat equation with certain boundary conditions, and currently I'm stuck at the last step, which is normalization at the initial time t=0.

The problem is to find the function $\phi(s,s',t)$, with the spatial variable $s\in (0,1)$, some fixed position $s'\in (0,1)$, and for all time $t\geq 0$. It must satisfy the heat equation

$\frac{\partial \phi (s,s',t)}{\partial t} = \frac{\partial^2 \phi(s,s',t)}{\partial s^2}$

under the boundary conditions

$\frac{\partial \phi(s,s',t)}{\partial s}\Big |_{s=1} = -2\mu \phi(1,s',t)$

$\frac{\partial \phi(s,s',t)}{\partial s}\Big |_{s=0} = 2\mu \phi(0,s',t)$

and the initial shape

$\phi(s,s',0) = e^{-2\mu |s-s'|}$

There is a fixed real constant $\mu > 0$. So far I could work out the solution up to the normalization step:

$e^{-2\mu |s-s'|} = \sum_{n=1}^{\infty} A_n \cos [\alpha_n (2s-1)] + B_n \sin [\beta_n (2s-1)]$

The Fourier frequencies are the solutions of these equations:

$\alpha_n \tan \alpha_n = \mu\\ \beta_n \cot \beta_n = -\mu$

In the Doi and Edwards article http://www.physics.emory.edu/faculty/roth/polymercourse/historical/Doi-Edwards_JCS-FT78.pdf Equation 4.37, the exact amplitudes are given as

$A_n = \frac{ \mu}{\mu^2 + \alpha_n^2 + \mu }\cos(\alpha_n (2s'-1))\\ B_n = \frac{ \mu}{\mu^2 + \beta_n^2 + \mu }\sin(\beta_n (2s'-1))$

without explanation and I was wondering how can one derive these expressions. If the frequencies where the usual $\alpha_n = \pi n$ etc., we could simply use the orthogonality of the trigonometric functions, but in the current situation I have no clue how to proceed.