A series from a physics problem: $\sum_{k = 0}^{\infty}\frac{2}{2k + 1}e^{-(2k + 1)\pi x/a}\sin\left( \frac{(2k + 1)\pi y}{a}\right)$

Question

How might we show that $$\sum_{k = 0}^{\infty}\frac{2}{2k + 1}e^{-(2k + 1)\pi x/a}\sin\left( \frac{(2k + 1)\pi y}{a}\right) = \tan^{-1}\left( \frac{\sin(\pi y/a)}{\sinh(\pi x/a)} \right) $$ where $x, y$ are independent variables and $a$ is a parameter?

This arose in the solution to Laplace's equation. My text asserted this without proof, so I'm curious.

Answer 1

Expanding on Mhenni Benghorbal's answer, the Maclaurin series of the inverse hyperbolic tangent function is $$\sum_{k=0}^{\infty} \frac{z^{2k+1}}{2k+1} = \text{artanh} (z) \, , \ |z| <1.$$

So assuming $x, a >0$ and $y \in \mathbb{R}$, we get

$$ \begin{align} &\sum_{k = 0}^{\infty}\frac{2}{2k + 1}e^{-(2k + 1)\pi x/a}\sin\left( \frac{(2k + 1)\pi y}{a}\right) \\ &= \text{Im}\sum_{k = 0}^{\infty}\frac{2}{2k + 1}e^{-(2k + 1)\pi x/a} e^{i(2k+1) \pi y /a} \\ &= 2 \ \text{Im} \ \text{artanh} \left(e^{- \pi x /a} e^{- i \pi y/a} \right) \\ &= \ \text{Im} \left( \text{Log} (1+e^{- \pi x/a} e^{i \pi y/a}) - \text{Log}(1-e^{-\pi x /a} e^{i \pi y /a})\right) \\&= \arctan \left( \frac{e^{- \pi x/a}\sin (\frac{\pi y}{a})}{1 + e^{- \pi x/a} \cos (\frac{\pi y}{a})} \right) - \arctan\left( \frac{-e^{- \pi x /a} \sin (\frac{\pi y}{a})}{1 - e^{- \pi x /a}\cos (\frac{\pi y}{a})} \right) \tag{1} \\ &= \arctan \left( \frac{\sin (\frac{\pi y}{a})}{e^{\pi x /a} +\cos (\frac{\pi y}{a})} \right) + \arctan\left( \frac{\sin (\frac{\pi y}{a})}{e^{\pi x /a} - \cos (\frac{\pi y}{a})} \right) \\ &= \arctan \left(\frac{2 e^{\pi x/a} \sin (\frac{\pi y}{a})}{e^{2 \pi x/a} - \cos^{2} (\frac{\pi y}{a}) - \sin^{2} (\frac{\pi y}{a})} \right) \tag{2}\\ &= \arctan \left( \frac{2e^{\pi x/a} \sin (\frac{\pi y}{a})}{e^{2 \pi x/a}-1} \right) \\ &= \arctan \left( \frac{\sin (\frac{\pi y}{a})}{\sinh (\frac{\pi x}{a})}\right). \end{align}$$

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$(1)$ The points $1+e^{- \pi x/a} e^{i \pi y/a}$ and $ 1-e^{- \pi x/a} e^{i \pi y/a}$ are in the right half-plane.

$(2)$ Since $$ \left|\frac{\sin (\frac{\pi y}{a})}{e^{\pi x/a} + \cos (\frac{\pi y}{a})} \left(\frac{\sin (\frac{\pi y}{a})}{e^{\pi x/a} - \cos (\frac{\pi y}{a})} \right) \right| = \frac{1-\cos^{2} (\frac{\pi y}{a})}{e^{2 \pi x/a} - \cos^{2} (\frac{\pi y}{a})} < 1, $$ we can use the formula $$\arctan(x+y) = \arctan \left(\frac{x+y}{1-xy} \right).$$

Answer 2

Hints: You need the identities

$$ \sin t = \frac{ e^{it} - e^{-it}}{2i} $$

and

$$ \sum_{k=0}^{\infty} t^{2k} = \frac{1}{1-t^2} .$$