$$\lim_{n\to\infty}\int_0^1 f(x,\sin nt)dt=\frac{1}{2\pi}\int_0^{2\pi}f(x,\sin y)d y$$ for continuous $f$?
It sounds like Riemann's lemma. But I am unable to do it.
2026-05-16 07:34:59.1778916899
$\lim_{n\to\infty}\int_0^1 f(x,\sin nt)dt=\frac{1}{2\pi}\int_0^{2\pi}f(x,\sin y)d y$ for continuous $f$
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