Is it possible to choose a function to zero an integral?

Question

Let $I(\omega,L)=\int_{0}^{L}xf(x,L)\cos(\omega x/L)dx$. I was wondering whether or not it was possible to find a nontrivial $f(x,L)$ (which can't depend on $\omega$) such that $I(\omega,L) = 0$. I tried messing around with integration by parts a lot, but it never really got me anywhere. I think it probably isn't possible, because it looks like the $\omega$ is always going to be there, but maybe some sort of shrewd cancellation would do it?

Answer 1

It's convenient to extend $f$ to $[-L,L]$ as an even function, so that its Fourier expansion on $[-L,L]$ contains only cosines: $f(x)=\frac12 a_0+\sum_{n=1}^\infty a_n \cos nx$. Here $a_n=\frac{2}{L}I(\pi n,L)$, so if these integrals all vanish, $f$ must be zero. See Fourier cosine series.