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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. ∫_0 ^∞ ((sin^n x)/x^n )dx= y=x^n x=y^(1/n) dx=(y^((1/n)−1) /n)dy (1/n)∫_0 ^∞ ((y^((1/n)−1) ∙siny)/y)dy=(1/n)∫_0 ^∞ y^((1/n)−2) ∙sinydy= (1/((1−(1/n))!))∫_0 ^∞ ζ_t ^n {(((1−(1/n))!)/s^(1+(1−(1/n))) )}(y)ζ_t (sint)(y)dy (1/(n(1−(1/n))!))∫_0 ^∞ (y^(1−(1/n)) /(1+y^2 ))dy=(1/(2n(1−(1/n))!))∫_0 ^∞ (y^((1−(1/(2n)))−1) /(y+1))dy =((Γ(1−(1/(2n)))Γ((1/(2n))))/(2n(1−(1/n))!))=((π∙(1/(sin(π/(2n)))))/(2n(1−(1/n))!)) n=3 ∫_0 ^∞ ((sin^3 x)/x^3 )dx=((π∙2)/(6∙((2/3))!))=(π/(3∙((2/3))!))

$${solution}.{Safojon}.{Toshtemirov}. \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{sin}^{{n}} {x}}{{x}^{{n}} }{dx}=\:\:{y}={x}^{{n}} \:\:{x}={y}^{\frac{\mathrm{1}}{{n}}} \:\:\:{dx}=\frac{{y}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{{n}}{dy} \\ $$$$\frac{\mathrm{1}}{{n}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{y}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \centerdot{siny}}{{y}}{dy}=\frac{\mathrm{1}}{{n}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{y}^{\frac{\mathrm{1}}{{n}}−\mathrm{2}} \centerdot{sinydy}= \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)!}\underset{\mathrm{0}} {\overset{\infty} {\int}}\zeta_{{t}} ^{{n}} \left\{\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)!}{{s}^{\mathrm{1}+\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)} }\right\}\left({y}\right)\zeta_{{t}} \left({sint}\right)\left({y}\right){dy} \\ $$$$\frac{\mathrm{1}}{{n}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)!}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{y}^{\mathrm{1}−\frac{\mathrm{1}}{{n}}} }{\mathrm{1}+{y}^{\mathrm{2}} }{dy}=\frac{\mathrm{1}}{\mathrm{2}{n}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)!}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{y}^{\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right)−\mathrm{1}} }{{y}+\mathrm{1}}{dy} \\ $$$$=\frac{\Gamma\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{n}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}{n}}\right)}{\mathrm{2}{n}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)!}=\frac{\pi\centerdot\frac{\mathrm{1}}{{sin}\frac{\pi}{\mathrm{2}{n}}}}{\mathrm{2}{n}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)!} \\ $$$${n}=\mathrm{3}\:\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{sin}^{\mathrm{3}} {x}}{{x}^{\mathrm{3}} }{dx}=\frac{\pi\centerdot\mathrm{2}}{\mathrm{6}\centerdot\left(\frac{\mathrm{2}}{\mathrm{3}}\right)!}=\frac{\pi}{\mathrm{3}\centerdot\left(\frac{\mathrm{2}}{\mathrm{3}}\right)!} \\ $$

Commented by botir last updated on 20/Apr/22

veru solution laplas

$${veru}\:{solution}\:{laplas} \\ $$

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