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This isn't necessarily true even when $\phi(x)$ and $\psi(x)$ are real-valued functions and $f(x)$ is nonnegative. For example, let $\phi(x) = \cos(x)$ and $\psi(x) = \cos(2x)$. Then one can integrate by parts to get $$\int_0^{\pi} \cos(x)f(x)^n dx = \int_0^{\pi}n\sin(x)f'(x)f(x)^{n-1}dx$$ $$\int_0^{\pi} \cos(2x)f(x)^n dx = {1 \over 2}\int_0^{\pi}n\sin(2x)f'(x)f(x)^{n-1}dx$$ If your statement were true, then by looking at the ratio of the left-hand sides and taking limits as $n$ goes to infinity one should get ${\cos(\xi) \over \cos(2\xi)}$, while looking at the ratio of the right-hand sides and taking limits as $n$ goes to infinity one should get ${\sin(\xi) \over 2\sin(2 \xi)}$, which is generally not the same. (although I'm interpreting the ratio for the right-hand sides in a certain way).