Explanation of $\int_0^{2\pi}\sin^{100}x\,dx$.

Question

I was browsing this thread when I came across this answer. I can neither make heads nor tail of it. Can someone help me understand it?

This I understand: $$\int_0^{2\pi}e^{ikx}\,dx=\left\{\begin{array}{cl}0&k\ne0\\ 2\pi&k=0\end{array}\right.,$$

But how does he get this? Where does the summation come from?

$$\int_0^{2\pi}\sin^{100}x\,dx=\frac1{2^{100}}\sum_{k=0}^{100}\binom{100}{k}\int_0^{2\pi}e^{ikx}(-1)^{100-k}e^{-i(100-k)x}\,dx=\frac{\binom{100}{50}}{2^{100}}2\pi.$$

EDIT: The answer is so obvious...I can't believe I didn't notice it! I kept expanding $\sin^{100}(x)$ as $\frac{1}{2i} (e^{100ix} + e^{-100ix})$ which got me nowhere.

Answer 1

We have

$$\sin x=\frac1{2i}(e^{ix}-e^{-ix})$$

so by the binomial formula we get $$\frac1{2^{100}}\sum_{k=0}^{100}(-1)^{100-k}\binom{100}ke^{kix}e^{-ix(100-k)}$$ so $$\int_0^{2\pi}\sin^{100}xdx=\frac1{2^{100}}\sum_{k=0}^{100}(-1)^{k}\binom{100}k\underbrace{\int_0^{2\pi}e^{(2k-100)ix}dx}_{=2\pi\;\text{only for}\; k=50}=\frac{2\pi}{2^{100}}\binom{100}{50}$$

Answer 2

Hint: Using the binomial theorem

$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}\implies \sin^{100}x=\frac1{2^{100}}\sum_{k=0}^{100}\binom{100}ke^{kix}e^{-ix(100-k)}\;\ldots$$

Answer 3

Write $$ \sin x = \frac{e^{ix} - e^{-ix}}{2i} $$ Then $$ \sin^{100} x = \frac{(a + b)^{100}}{2^{100}} $$ where $a = e^{ix}$ and $b = -e^{-ix}$.

Now use the binomial theorem: $$ (a + b)^{100} = \sum_{k=0}^{100}\left( \begin{array}{c} 100 \\ k \end{array} \right) a^k b^{100-k} $$ and the first summation step in the formula you were puzzled about you give falls out. And then in that integral, only the term where the factor in front of $x$ in the exponent gives a non-zero integral. So only the term with $k - (100-k) = 0$, or $k=50$, is non-zero. That is where the $\left( \begin{array}{c} 100 \\ 50 \end{array} \right)$ comes from.