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

$${\int }_0^{+\mathrm{\infty }}\frac{{\left(\sin x\right)}^{2n+1}}{x}dx=\frac{\pi }{2^{2n+1}}\left(\begin{array}{c}2n\\ n\end{array}\right)$$

Answered by Olaf_Thorendsen last updated on 26/Aug/21

$${\mathrm{I}}_n={\int }_0^{+\mathrm{\infty }}\frac{{\sin }^{2n+1}x}{x}dx$$$$\mathrm{Let}f\left(x\right)={\sin }^{2n}x$$$${\mathrm{I}}_n={\int }_0^{+\mathrm{\infty }}\frac{\sin x}{x}f\left(x\right)dx$$$$$$$$f\mathrm{is}\mathrm{a}\pi -\mathrm{periodic}\mathrm{function}.\mathrm{We}\mathrm{can}$$$$\mathrm{apply}\mathrm{the}\mathrm{Lobachevsky}-\mathrm{Dirchlet}$$$$\mathrm{integral}\mathrm{formula}:$$$${\int }_0^{+\mathrm{\infty }}\frac{\sin x}{x}f\left(x\right)dx={\int }_0^{\frac{\pi }{2}}f\left(x\right)dx$$$$\Rightarrow {\mathrm{I}}_n={\int }_0^{\frac{\pi }{2}}{\sin }^{2n}xdx=W_{2n}=\frac{\pi }{2}.\frac{\left(2n\right)!}{{(2^{n}n!)}^2}$$$$({\mathrm{W}}_{2n}:\mathrm{integral}\mathrm{of}\mathrm{Wallis})$$$${\mathrm{I}}_n=\frac{\pi }{2^{2n+1}}.\frac{\left(2n\right)!}{n!n!}=\frac{\pi }{2^{2n+1}}{\mathrm{C}}_n^{2n}$$

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

$$\mathrm{Gra}ext\text{c}c\mathrm{cias}\mathrm{se}\stackrel{}{\mathrm{n}or}!$$

Commented by puissant last updated on 26/Aug/21

$$hum$$

Answered by Kamel last updated on 26/Aug/21

$${sin}^{2n}\left(x\right)=\underset{k=0}{\sum ^{2n}}\frac{{(-1)}^{2n-k}{(-1)}^n}{2^{2n}}{C}_{2n}^k{e}^{ikx}{e}^{-i(2n-k)ix}$$$$=\frac{1}{2^{2n}}{C}_{2n}^n+\underset{k=0}{\sum ^{n-1}}\frac{{(-1)}^{n+k}}{2^{2n}}{C}_{2n}^k{e}^{ikx}{e}^{-(2n-k)ix}+\underset{k=0}{\sum ^{n-1}}\frac{{(-1)}^{n+k}}{2^{2n}}{C}_{2n}^k{e}^{i(2n-k)x}{e}^{-ikx}$$$$=\frac{1}{2^{2n}}{C}_{2n}^n+\underset{k=0}{\sum ^{n-1}}\frac{{(-1)}^{n+k}}{2^{2n}}{C}_{2n}^k(e^{-(2n-2k)ix}+e^{i(2n-2k)x})$$$$=\frac{1}{2^{2n}}{C}_{2n}^n+\underset{k=0}{\sum ^{n-1}}\frac{{(-1)}^{n+k}}{2^{2n-1}}{C}_{2n}^{k}cos(2n-2k)$$$$\therefore {\int }_0^{+\mathrm{\infty }}\frac{{sin}^{2n+1}\left(x\right)}{x}dx=\frac{1}{2^{2n}}{C}_{2n}^n{\int }_0^{+\mathrm{\infty }}\frac{sin\left(x\right)}{x}dx+\underset{k=0}{\sum ^{n-1}}\frac{{(-1)}^{n+k}}{2^{2n}}{C}_{2n}^k{\int }_0^{+\mathrm{\infty }}\frac{sin\left((2n-2k+1)x\right)-sin\left((2n-2k-1)x\right)}{x}dx$$$$Wehave:2n-2k+1>0,2n-2k-1>0\mathrm{\forall }k=\overline{0,n-1},n⩾1$$$$\therefore {\int }_0^{+\mathrm{\infty }}\frac{sin\left(ax\right)}{x}dx\stackrel{a>0}{=}\frac{\pi }{2}$$$$So:{\int }_0^{+\mathrm{\infty }}\frac{{sin}^{2n+1}\left(x\right)}{x}dx=\frac{\pi }{2^{2n+1}}\left(\begin{array}{c}2n\\ n\end{array}\right)$$

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

$$\mathrm{Oh}\mathrm{my}!\mathrm{Thanks}\mathrm{Sir}$$

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