I am looking at a certain derivation for the full Fourier series of $\phi(x)=\cosh{x}$ on the interval $(-l,l)$. The derivation uses the definition of hyperbolic cosine and the full Fourier series for $e^x$ as follows
$$\begin{align} \cosh{x}&=\frac{e^x+e^{-x}}{2} \\ &=\frac{1}{2}\left(\sum_{n=-\infty}^\infty (-1)^n \frac{l+in\pi}{l^2+n^2\pi^2}e^{\frac{in\pi x}{l}} \sinh{l}+\sum_{n=-\infty}^\infty (-1)^n \frac{l+in\pi}{l^2+n^2\pi^2}e^{-\frac{in\pi x}{l}} \sinh{l}\right)\\&= \frac{1}{2}\left(\sum_{n=-\infty}^\infty (-1)^n \frac{l+in\pi}{l^2+n^2\pi^2}e^{\frac{in\pi x}{l}} \sinh{l}+\sum_{n=-\infty}^\infty (-1)^{-n} \frac{l-in\pi}{l^2+n^2\pi^2}e^{\frac{in\pi x}{l}} \sinh{l}\right) \end{align}$$
My confusion is in seeing how the second sum on the middle line is apparently equal to the second sum on the last line. I have thought very long about this and may be missing something simple, but would really appreciate any help in seeing why this is true. Thanks in advance.
This is similar to writing a full fourier series in complex form.
Follow this example:
$f(x) = \frac{a_0}{2}+\sum_{n=-\infty}^\infty (a_n cos(nx) b_n sin(nx)$
$= \frac{a_0}{2} + \sum_{n=-\infty}^\infty a_n \frac{e^{inx}+e^{-inx}}{2} + b_n \frac{e^{inx} -e^{-inx}}{2i}$
$=\frac{a_0}{2} + \sum_{n=-\infty}^\infty \frac{a_n-ib_n}{2} e^{inx} + \sum_{n=-\infty}^\infty \frac{a_n+ib_n}{2} e^{-inx}$
$=\sum_{n=-\infty}^\infty c_n e^{inx}$
Also, remember that $\cos(n\pi) =(-1)^n$ and De Moivre's Theorem $e^{ix}= \cos(x) + i \sin(x)$
so $(-1)^n e^{\frac{-in \pi x}{l}} = (-1)^n e^{\frac{-in \pi x}{l}} \frac{e^{\frac{in \pi x}{l}} }{e^{\frac{in \pi x}{l}} } $
Using this idea, we can apply it to the 2nd line posed in your question and should see the results follow.
Hope this helps.