Fourier series and a known identity

[Tolaso J Kos]

Let \( f(x) =e^{ax} , \;\; x \in [-\pi, \pi)\) . Show that the Fourier series of \( f \) converges in \( [-\pi, \pi) \) to \( f \) and at \( x = \pi \) to \( \displaystyle \frac{e^{a\pi}+e^{-a\pi}}{2} \). Deduce that:


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

[Riemann]

The function is expanded (fair and square) in a Fourier series in the given interval. Let us evaluate the coefficients:

• $\displaystyle a_0= \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \, {\rm d}x = \frac{2\sinh \pi a}{ \pi a}$
• $\begin{aligned}
  a_n&= \frac{1}{\pi}\int_{-\pi}^{\pi} f(x) \cos nx \, {\rm d}x \\
  &=\frac{1}{\pi}\int_{-\pi}^{\pi} e^{ax} \cos nx \, {\rm d}x \\
  &=\frac{1}{\pi} \mathfrak{Re}\left ( \int_{-\pi}^{\pi} e^{ax} e^{in x } \, {\rm d}x \right ) \\
  &=\frac{2 a (-1)^n \sinh \pi a}{ \pi (a^2+n^2)}
  \end{aligned}$
• $\displaystyle b_n = \frac{1}{\pi} \mathfrak{Im}\left ( \int_{-\pi}^{\pi} e^{ax} e^{in x} \, {\rm d}x \right )= - \frac{2 (-1)^n n \sinh \pi a }{\pi \left ( a^2+n^2 \right )}$

Now, we note that at $x=\pi$ the series converges at half of the value at the endpoints of the interval. Recalling the formula

$$f(x)=\frac{a_0}{2} + \sum_{n=1}^{\infty} [a_n \cos nx + b_n \sin x]$$

we have the result posted above.