Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 225 / 4 / ii / Solution

Replacing ${\varphi }_k$ by $-{\varphi }_k$ also replaces $a_k$ by $-a_k$, so both $\mathbb{E}[a_k{b}_l]$ and ${\varphi }_k(s)$ change sign and their product is unchanged. Replacing $u_l$ by $-u_l$ similarly replaces $b_l$ by $-b_l$. Every summand, and hence $\beta$, is independent of all eigenfunction sign choices.