Question on average value of $\sin^{100}(x)$

Question

I was looking at this answer but ran into some problems while working through it. Using $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$ :

$$\begin{align}\frac{1}{2\pi}\int_{0}^{2\pi}{\sin^{100}(x)\, dx}&=\frac{1}{2\pi}\int_{0}^{2\pi}{(e^{ix}-e^{-ix})^{100}(2i)^{-100}\, dx}\\&=\frac{1}{2^{101}\pi}\int_{0}^{2\pi}{\sum_{k=0}^{100}{100\choose{k}}e^{ikx}(-1)^{100-k}e^{-ix(100-k)}\, dx}\\&=\frac{1}{2^{101}\pi}\sum_{k=0}^{100}{100\choose{k}}(-1)^{100-k}\int_{0}^{2\pi}{e^{ix(2k-100)}\, dx}\end{align}$$

Edit : Assuming the interchange is allowed at all

The exponent inside the integral ($2k-100$) will be nonzero unless $k=50$, but from the answer if $k\neq0$ then the integral goes to $0$, no? How should this be evaluated?

Yes, the integral of $e^{inx}$ over $[0,2\pi]$ vanishes for $n\neq 0,$ as the antiderivative is equal $(in)^{-1}e^{inx}.$ — Ryszard Szwarc, Sep 04 '23 at 15:58
@Luthier415Hz not necessarily because $\sin^{100}(x) \geq 0$ — AnthonyML, Sep 04 '23 at 16:04
If the integral is calculated over $[0,2\pi]$ would it not make more sense to divide by $2\pi$ to get the average ? — Peter, Sep 04 '23 at 16:07
@Peter sorry it was a typo, the average could be done with $\frac{\pi}{2}$ but for this method it's easier with $2\pi$ — AnthonyML, Sep 04 '23 at 16:17

score 3 · Accepted Answer · answered Sep 04 '23 at 16:27

3

If k is a non-zero integer than $$\int_0^{2\pi} e^{ikx} \, dx =\frac{1}{ik}\left[ e^{ikx} \right]_0^{2\pi}$$ $$=\frac{1}{ik}\left( e^{2ik\pi}-e^0 \right)$$ $$=\frac{1}{ik}\left( 1-1\right)$$ $$=0$$

answered Sep 04 '23 at 16:27

Blitzer

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Question on average value of $\sin^{100}(x)$

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