Question-168852

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

s o l u t i o n . S a f o j o n . T o s h t e m i r o v .

\underset{0}{\int^{\infty}} \frac{{s i n}^{n} x}{x^{n}} d x = y = x^{n} x = y^{\frac{1}{n}} d x = \frac{y^{\frac{1}{n} - 1}}{n} d y

\frac{1}{n} \underset{0}{\int^{\infty}} \frac{y^{\frac{1}{n} - 1} \cdot s i n y}{y} d y = \frac{1}{n} \underset{0}{\int^{\infty}} y^{\frac{1}{n} - 2} \cdot s i n y d y =

\frac{1}{(1 - \frac{1}{n})!} \underset{0}{\int^{\infty}} ζ_{t}^{n} {\frac{(1 - \frac{1}{n})!}{s^{1 + (1 - \frac{1}{n})}}} (y) ζ_{t} (s i n t) (y) d y

\frac{1}{n (1 - \frac{1}{n})!} \underset{0}{\int^{\infty}} \frac{y^{1 - \frac{1}{n}}}{1 + y^{2}} d y = \frac{1}{2 n (1 - \frac{1}{n})!} \underset{0}{\int^{\infty}} \frac{y^{(1 - \frac{1}{2 n}) - 1}}{y + 1} d y

= \frac{Γ (1 - \frac{1}{2 n}) Γ (\frac{1}{2 n})}{2 n (1 - \frac{1}{n})!} = \frac{π \cdot \frac{1}{s i n \frac{π}{2 n}}}{2 n (1 - \frac{1}{n})!}

n = 3 \underset{0}{\int^{\infty}} \frac{{s i n}^{3} x}{x^{3}} d x = \frac{π \cdot 2}{6 \cdot (\frac{2}{3})!} = \frac{π}{3 \cdot (\frac{2}{3})!}

Commented by botir last updated on 20/Apr/22

v e r u s o l u t i o n l a p l a s

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