# Showing an inequality of a complex exponential function

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 $$s=t/p$$

• Doesn't $\int_0^2 e^{in\pi t} \, dt$ equal zero? – angryavian Jan 15 at 19:04
• Sorry. Now, I fixed it. – eraldcoil Jan 15 at 19:19
• This is just silly; $|e^{in\pi t}|=1$. – David C. Ullrich Jan 15 at 23:42

I have it! $$\int_{0}^{2p} |e^{i n\pi it}|^2dt=\int_{0}^{2} |e^{i n\pi it}|^2dt+\cdots+ \int_{2(p-1)}^{2p} |e^{i n\pi it}|^2dt=p\int_{0}^{2p} |e^{i n\pi it}|^2dt$$