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Question Number 168852 by bagjagugum123 last updated on 19/Apr/22

Commented by CAIMAN last updated on 19/Apr/22

3π/8

Commented by safojontoshtemirov last updated on 20/Apr/22

$$solution.Safojon.Toshtemirov.$$$$\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{{sin}^{n}x}{x^n}dx=y=x^{n}x=y^{\frac{1}{n}}dx=\frac{y^{\frac{1}{n}-1}}{n}dy$$$$\frac{1}{n}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{y^{\frac{1}{n}-1}\cdot siny}{y}dy=\frac{1}{n}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{y}^{\frac{1}{n}-2}\cdot sinydy=$$$$\frac{1}{(1-\frac{1}{n})!}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}{\zeta }_t^n\left\{\frac{(1-\frac{1}{n})!}{s^{1+(1-\frac{1}{n})}}\right\}\left(y\right){\zeta }_t\left(sint\right)\left(y\right)dy$$$$\frac{1}{n(1-\frac{1}{n})!}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{y^{1-\frac{1}{n}}}{1+y^2}dy=\frac{1}{2n(1-\frac{1}{n})!}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{y^{(1-\frac{1}{2n})-1}}{y+1}dy$$$$=\frac{\mathrm{\Gamma }(1-\frac{1}{2n})\mathrm{\Gamma }\left(\frac{1}{2n}\right)}{2n(1-\frac{1}{n})!}=\frac{\pi \cdot \frac{1}{sin\frac{\pi }{2n}}}{2n(1-\frac{1}{n})!}$$$$n=3\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{{sin}^{3}x}{x^3}dx=\frac{\pi \cdot 2}{6\cdot \left(\frac{2}{3}\right)!}=\frac{\pi }{3\cdot \left(\frac{2}{3}\right)!}$$

Commented by botir last updated on 20/Apr/22

$$verusolutionlaplas$$

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