calculate Fourier expansion

Question

Let $f(θ) = e^{bθ},\quad\ θ\in[−π, π]$ ,

Calculate Fourier expansion and then demostrate from this result the following expression:

$$\sum_{n=1}^\infty \frac{(-1)^n}{b^2+n^2} = \frac \pi {2b\sinh b\pi} - \frac 1 {2b^2}. $$

my attempt

coefficient expressions

Answer 1

$$c_n=\frac1\pi\int_{-\pi}^\pi e^{b\theta}e^{-in\theta}d\theta=\frac1\pi\int_{-\pi}^\pi e^{(b-in)\theta}d\theta=\ldots\text{(fill in details...)}$$

$$=\ldots=$$

$$=\frac1{\pi(b-in)}\left(e^{b\pi}(-1)^n-e^{-b\pi}(-1)^n\right)=\frac{2(-1)^n}{\pi(b-in)}\sinh b\pi$$

and from here that the desired Fourier series is (fill in details)

$$\sum_{n=-\infty}^\infty(-1)^n\frac{2(b+in)}{\pi(b^2+n^2)}\,\sinh(b\pi)\,\, e^{in\theta}$$

and this series equals $\;e^{b\theta}\;$ at any point of continuity, and it is its arithmetic mean at the function's jump discontinuities...

For example, with $\;\theta=0\;$ we get

$$e^0=1=\sum_{n=-\infty}^\infty(-1)^n\frac{2(b+in)}{\pi(b^2+n^2)}\,\sinh(b\pi)\,\, =\frac2\pi\sin h(b\pi)\sum_{n=-\infty}^\infty (-1)^n\frac{b+in}{b^2+n^2}$$

...and now end the exercise.