Plenty of Integrate by Parts?

We are given the following:

$\displaystyle \int_0^{\infty} \frac{e^{ix}-e^{-ix}}{2ix}\,dx=\frac{\pi}{2}\,\,\,(*)$

We have to find the value of

$\displaystyle \int_0^{\infty} \frac{(e^{ix}-e^{-ix})^9}{2^9ix}\,dx$

Expand the numerator using Binomial theorem to get:

$\displaystyle (e^{ix}-e^{-ix})^9={9 \choose 0}e^{i9x}-{9 \choose 1}e^{i7x}+{9 \choose 2}e^{i5x}-{9 \choose 3}e^{i3x}+{9 \choose 4}e^{ix}-{9 \choose 5}e^{-ix}+{9 \choose 6}e^{-i3x}-{9 \choose 7}e^{-i5x}+{9 \choose 8}e^{-i7x}-{9 \choose 9}e^{-i9x}$

$\Rightarrow \displaystyle (e^{ix}-e^{-ix})^9={9 \choose 0}\left(e^{i9x}-e^{-i9x}\right)-{9 \choose 1}\left(e^{i7x}-e^{-i7x}\right)+{9 \choose 2}\left(e^{i5x}-e^{-i5x}\right)-{9 \choose 3}\left(e^{i3x}-e^{-i3x}\right)+{9 \choose 4}\left(e^{ix}-e^{-ix}\right)$

When the above is substituted in the integral we have to evaluate, we get the following kind of integral:

$\int_0^{\infty} \frac{e^{inx}-e^{-inx}}{x}\,dx$

To evaluate the above, use the substitution $nx=t$ and using $(*)$ , we get, $i\pi$ . Hence,

$\displaystyle \int_0^{\infty} \frac{(e^{ix}-e^{-ix})^9}{2^9ix}\,dx =\frac{1}{2^9i}\left({9 \choose 0}-{9 \choose 1}+{9 \choose 2}-{9 \choose 3}+{9 \choose 4}\right)(i\pi)=\frac{\pi}{2^9}(70)$

$\displaystyle \Rightarrow \int_0^{\infty} \frac{(e^{ix}-e^{-ix})^9}{2^9ix}\,dx=\boxed{\dfrac{35\pi}{256}}$

For odd $n$ we have the Fourier series

$\sin^n x=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^n=-2^{-n}(-1)^{\frac{n-1}{2}}i\sum_{k=0}^{n}(-1)^k{n \choose k}e^{(n-2k)ix}$

$=-2^{-n}(-1)^{\frac{n-1}{2}}i\sum_{k=0}^{\frac{n-1}{2}}(-1)^k{n \choose k}\left(e^{(n-2k)ix}-e^{-(n-2k)ix}\right)$

$=2^{1-n}\sum_{k=0}^{\frac{n-1}{2}}(-1)^{(\frac{n-1}{2}+k)}{n \choose k}\sin((n-2k)x)$

so

$\int_{0}^{\infty}\frac{\sin^n x}{x}dx=\frac{\pi}{2^n}\sum_{k=0}^{\frac{n-1}{2}}(-1)^{(\frac{n-1}{2}+k)}{n \choose k}$

For $n=9$ this comes out to be $\frac{35\pi}{256}$ , so that the answer is $\boxed{291}$

$$ In general $$ $\int_0^\infty\frac{\sin^{2n-1}}{x}dx=\frac{\pi}{2^{2n-1}}\sum_{k=0}^{n-1}(-1)^{k+n-1}\binom{2n-1}{k},\quad n\in\mathbb{N}.$ $$ Sorry, no proof given. This is formula from old college assignment. Anyway, try this: Insane Integral . $$

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$