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The exercise wants me to prove the identity

$$\pi \coth \pi a= \frac{1}{a}+ \sum_{n=1}^{\infty}\frac{2a}{n^2+a^2}$$

using the Fourier series of $\cosh ax, \; x \in [-\pi, \pi], \; a \neq 0$.

Evaluating the coefficients ($b_n$ is actually zero as $\cosh ax$ is even) we have that:

$\displaystyle {\color{gray} \blacksquare} \;\; a_0 = \frac{1}{\pi}\int_{-\pi}^{\pi}\cosh ax \, {\rm d}x= \frac{2\sinh \pi a}{\pi a}$

$\require{cancel}\displaystyle {\color{gray} \blacksquare} \;\; a_n = \frac{1}{\pi}\int_{-\pi}^{\pi}\cosh ax \cos n x\, {\rm d}x= \frac{2[a \sinh \pi a \cos \pi n+ \cancelto{0}{n \cosh \pi a \sin \pi n}]}{\pi (a^2+n^2)}$

So far so good except one little problem. I get that $\cos \pi n$ in the nominator which is $(-1)^n$ so I get the alternating series not the wanted one. That is , this way I evaluated the series:

$$\cosh ax = \frac{\sinh \pi a}{\pi a}+ \frac{2 a\sinh \pi a}{\pi}\sum_{n=1}^{\infty}\frac{ (-1)^n}{n^2 +a^2} \implies \\ \implies \pi\coth \pi a = \frac{1}{a}+ \sum_{n=1}^{\infty}\frac{2(-1)^n}{n^2+a^2}$$

and not what I want. Calculations of the coefficients were done by Wolfram because they were too tedious to be done by hand.

However, I know that using contour integration using the kernel $\pi \cot \pi z$ that this series indeed evaluates to the LHS. What is wrong here?

Blue
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Tolaso
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1 Answers1

3

Actually I forgot to add the $\cos nx$ in the RHS. Hence:

$$\cosh ax = \frac{\sinh \pi a}{\pi a}+ \frac{2 a\sinh \pi a}{\pi}\sum_{n=1}^{\infty}\frac{ (-1)^n \cos nx}{n^2 +a^2} \overset{x=\pi}{\implies}$$

$$\overset{x=\pi}{\implies} \pi \coth \pi a= \frac{1}{a}+ \sum_{n=1}^{\infty}\frac{2a}{n^2+a^2}$$

and the identity is proven.

@Aretino, thanks for the heads up.

clathratus
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Tolaso
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