0-sinx-2n-1-x-dx-pi-2-2n-1-2n-n-

Question Number 152141 by Ar Brandon last updated on 26/Aug/21

\int_{0}^{+ \infty} \frac{{(\sin x)}^{2 n + 1}}{x} d x = \frac{π}{2^{2 n + 1}} (\begin{matrix} 2 n \\ n \end{matrix})

Answered by Olaf_Thorendsen last updated on 26/Aug/21

I_{n} = \int_{0}^{+ \infty} \frac{\sin^{2 n + 1} x}{x} d x

Let f (x) = \sin^{2 n} x

I_{n} = \int_{0}^{+ \infty} \frac{\sin x}{x} f (x) d x

f is a π - periodic function . We can

apply the Lobachevsky - Dirchlet

integral formula :

\int_{0}^{+ \infty} \frac{\sin x}{x} f (x) d x = \int_{0}^{\frac{π}{2}} f (x) d x

\Rightarrow I_{n} = \int_{0}^{\frac{π}{2}} \sin^{2 n} x d x = W_{2 n} = \frac{π}{2} . \frac{(2 n)!}{{(2^{n} n!)}^{2}}

(W_{2 n} : integral of Wallis)

I_{n} = \frac{π}{2^{2 n + 1}} . \frac{(2 n)!}{n! n!} = \frac{π}{2^{2 n + 1}} C_{n}^{2 n}

Commented by Ar Brandon last updated on 26/Aug/21

Gra e x t \c c cias se \overset{}{n o r}!

Commented by puissant last updated on 26/Aug/21

h u m

Answered by Kamel last updated on 26/Aug/21

{s i n}^{2 n} (x) = \underset{k = 0}{\sum^{2 n}} \frac{{(- 1)}^{2 n - k} {(- 1)}^{n}}{2^{2 n}} C_{2 n}^{k} e^{i k x} e^{- i (2 n - k) i x}

= \frac{1}{2^{2 n}} C_{2 n}^{n} + \underset{k = 0}{\sum^{n - 1}} \frac{{(- 1)}^{n + k}}{2^{2 n}} C_{2 n}^{k} e^{i k x} e^{- (2 n - k) i x} + \underset{k = 0}{\sum^{n - 1}} \frac{{(- 1)}^{n + k}}{2^{2 n}} C_{2 n}^{k} e^{i (2 n - k) x} e^{- i k x}

= \frac{1}{2^{2 n}} C_{2 n}^{n} + \underset{k = 0}{\sum^{n - 1}} \frac{{(- 1)}^{n + k}}{2^{2 n}} C_{2 n}^{k} (e^{- (2 n - 2 k) i x} + e^{i (2 n - 2 k) x})

= \frac{1}{2^{2 n}} C_{2 n}^{n} + \underset{k = 0}{\sum^{n - 1}} \frac{{(- 1)}^{n + k}}{2^{2 n - 1}} C_{2 n}^{k} c o s (2 n - 2 k)

∴ \int_{0}^{+ \infty} \frac{{s i n}^{2 n + 1} (x)}{x} d x = \frac{1}{2^{2 n}} C_{2 n}^{n} \int_{0}^{+ \infty} \frac{s i n (x)}{x} d x + \underset{k = 0}{\sum^{n - 1}} \frac{{(- 1)}^{n + k}}{2^{2 n}} C_{2 n}^{k} \int_{0}^{+ \infty} \frac{s i n ((2 n - 2 k + 1) x) - s i n ((2 n - 2 k - 1) x)}{x} d x

W e h a v e : 2 n - 2 k + 1 > 0, 2 n - 2 k - 1 > 0 \forall k = \overset{―}{0, n - 1}, n ⩾ 1

∴ \int_{0}^{+ \infty} \frac{s i n (a x)}{x} d x \overset{a > 0}{=} \frac{π}{2}

S o : \int_{0}^{+ \infty} \frac{{s i n}^{2 n + 1} (x)}{x} d x = \frac{π}{2^{2 n + 1}} (\begin{matrix} 2 n \\ n \end{matrix})

Commented by Ar Brandon last updated on 26/Aug/21

Oh my! Thanks Sir