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I expected the following formula to hold:

$\int^{2n\pi}_0\cos(\sin t+t/n)dt=0$, for ${}^\forall n\in\mathbb{N},\ n\geq2$

But I can't prove it. Could you please tell me.

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    $\begingroup$ I'm not quite sure why people are voting to close this. Is it a well-known result? Is it obvious? I think the poster is saying that he found that this seems to be true numerically, and is asking for a proof or reference. I checked it numerically up to $n=10$. $\endgroup$ – Joe Silverman Jul 31 at 18:34
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    $\begingroup$ I voted to close because I would have expected an explanation of why it might be expected to hold. More generally the question lacks context. Also it's evidently true for even $n$ by comparing $t$ and $t+n\pi$. $\endgroup$ – Anthony Quas Jul 31 at 18:45
  • $\begingroup$ This is a question for MSE. $\endgroup$ – user64494 Jul 31 at 19:37
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You can rewrite the integral as $$ \int_0^{2\pi} \left(\sum_{j=0}^{n-1}\cos\Big(\sin t+\tfrac tn+2\pi \tfrac jn\Big)\right)\,dt. $$ But $\sum_{j=0}^{n-1}\cos\big(a+2\pi \tfrac jn\big)=0$ for all $a$.

In particular, the equality holds if $\sin t$ is replaced by any $2\pi$-periodic function.

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    $\begingroup$ Nice. And for $n=1$ you sum is $\cos(a)$, not 0, which explains why the integral is non-zero for $n=1$. I guess there will be a difference of opinion as to whether this belongs on MO or MSE, but I will say it's a clever trick that I enjoyed seeing. $\endgroup$ – Joe Silverman Jul 31 at 20:30
  • $\begingroup$ Thank you very much. You are smart. $\endgroup$ – Matsuno Aug 1 at 2:57

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