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If $T>2$ and exist $p$ such that $2(p-1)\leq T<2p$ show that $\int_{0}^{T}|e^{i n\pi t}|^2dx\leq p\int_{0}^{2}|e^{i n \pi t}|^2dt.$

Hint: Use 2-periodicity of $e^{i n\pi t}$

I have this: $\int_{0}^{T}|e^{i n \pi t}|^2dt\leq \int_{0}^{2p} |e^{i n \pi t}|^2dt=p\int_{0}^{2}|e^{i n\pi sp}|^2ds$ with the change $s=t/p$

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  • $\begingroup$ Doesn't $\int_0^2 e^{in\pi t} \, dt$ equal zero? $\endgroup$ – angryavian Jan 15 at 19:04
  • $\begingroup$ Sorry. Now, I fixed it. $\endgroup$ – eraldcoil Jan 15 at 19:19
  • $\begingroup$ This is just silly; $|e^{in\pi t}|=1$. $\endgroup$ – David C. Ullrich Jan 15 at 23:42
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I have it! $\int_{0}^{2p} |e^{i n\pi it}|^2dt=\int_{0}^{2} |e^{i n\pi it}|^2dt+\cdots+ \int_{2(p-1)}^{2p} |e^{i n\pi it}|^2dt=p\int_{0}^{2p} |e^{i n\pi it}|^2dt$

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