The Problem:
If $f(x) $ is a real funciton, rewrite the integral: $$ \frac{1}{2L} \int_{-L}^{L} |f(x)|^{2} \, dx$$ in terms of the usual Fourier Coefficients, $A_n$ and $B_n$
The attempt: Recall the Fourier Series of $f(x)$ is in the form $$f(x) = \sum_{n=0}^{\infty} A_n \cos{ \frac{n \pi x}{L}} + \sum_{n=0}^{\infty}B_n \sin{\frac{n \pi x}{L}}$$
If we write the cosine and sine terms by exponentials, this would imply that
$$f(x) = \sum_{n=0}^{\infty} A_n \left(\frac{\exp{\frac{n \pi x}{L}}+\exp{\frac{-n \pi x}{L}}}{2}\right) + \sum_{n=0}^{\infty} B_n \left(\frac{\exp{\frac{n \pi x}{L}}-\exp{\frac{-n \pi x}{L}}}{2i}\right)$$
In simplified terms, the function would be
$$f(x) = A_0 + \frac{1}{2} \sum_{n=1}^{\infty} (A_n - iB_n) \exp{\frac{i n \pi x}{L}}+ \frac{1}{2} \sum_{n=1}^{\infty} (A_n + iB_n) \exp{\frac{-i n \pi x}{L}}$$
Now I know that $|f(x)|^{2}$ = $f(x) \overline{f(x)}$ But I am not sure how to find the complex conjugate. Am I on the right track about this?
Thank you for all of your help!