Hello,

I have a function $f(x)$ which is even around $x=L/2$. I need to show that the even coefficents (n is even) of the Fourier sine series of $f(x)$ on $0 \leq x \leq L$ are zero.

So the sine series should be $\sum_{n=1}^\infty B_n \sin{\frac{n\pi x}{L}}$ with $B_n=\int_{-L}^{L} f(x)\sin{\frac{n\pi x}{L}}dx$.

I've been trying to manipulate $$ \int_{-L}^{-L/2-\delta} f(x) \sin{\frac{n\pi x}{L}}dx + \int_{-L/2-\delta}^{-L/2+\delta} f(x) \sin{\frac{n\pi x}{L}}dx + \int_{-L/2 +\delta}^0 f(x) \sin{\frac{n\pi x}{L}}dx + \int_{0}^{L/2-\delta} f(x) \sin{\frac{n\pi x}{L}}dx + \int_{L/2+\delta}^{L} f(x) \sin{\frac{n\pi x}{L}}dx. $$ The symmetry would then give $$\int_{-L}^{-L/2-\delta} f(x) \sin{\frac{n\pi x}{L}}dx + \int_{-L/2 +\delta}^0 f(x) \sin{\frac{n\pi x}{L}}dx + \int_{0}^{L/2-\delta} f(x) \sin{\frac{n\pi x}{L}}dx + \int_{L/2+\delta}^{L} f(x) \sin{\frac{n\pi x}{L}}dx.$$

Then I'm not too sure what to do.

I think that $f(x)$ is odd. Since when we talked about getting the sine series the first thing we talked about was getting an odd extension of what ever function. I was hoping to use this ($f(-x)=-f(x)$) to get some more cancellations, but I do not see how this would be related to the index n and it's parity.