Could someone point me to a proof of Parseval's Theorem? That is,
$$\sum_{n=-\infty}^{\infty} a_n\bar b_n = \int_{-\pi}^{\pi} A(x)\bar B(x) dx$$
where $A(x)=\sum_{n=-\infty}^{\infty} a_ne^ {inx}$ and similarly for $B(x)$. All the proofs I've found online deal merely with the case $A(x)=B(x)$.
Let $$A(x)=\sum_{n=-\infty}^{\infty}a_{n}e^{inx}$$ $$B(x)=\sum_{m=-\infty}^{\infty}b_{m}e^{imx}$$ then for a fixed $n$ we know $$\int_{-\pi}^{\pi}a_{n}e^{inx}\overline{B(x)}\ dx = \sum_{m=-\infty}^{\infty}a_{n}\overline{b_{m}}\int_{-\pi}^{\pi}e^{inx}e^{-imx}\ dx= 2\pi a_n\overline{b_{n}}$$ where $$\int_{-\pi}^{\pi}e^{i(n-m)x}dx= \begin{cases} 0&,~~~~n\neq m\\ 2\pi&,~~~~n=m\\ \end{cases} $$ then $$\int_{-\pi}^{\pi} A(x)\overline{B(x)}\ dx=\sum_{n=-\infty}^{\infty} \int_{-\pi}^{\pi}a_{n}e^{inx}\overline{B(x)}\ dx = 2\pi\sum_{n=-\infty}^{\infty} a_n\overline{b_{n}}$$