find ∫_0 ^∞ cos(x^n )dx snd ∫


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Question Number 124530 by Bird last updated on 03/Dec/20

$f i n d \int_{0}^{\infty} c o s (x^{n}) d x s n d \int_{0}^{\infty} s i n (x^{n}) d x$
Commented by Dwaipayan Shikari last updated on 04/Dec/20

$\int_{0}^{\infty} s i n (x^{n}) d x$ $= \frac{1}{2 i} \int_{0}^{\infty} e^{{i x}^{n}} - e^{- {i x}^{n}} d x$ $= \frac{1}{2 i n} \int_{0}^{\infty} u^{\frac{1 - n}{n}} e^{i u} d u - \frac{1}{2 i n} \int t^{\frac{1 - n}{n}} e^{- i u} d u x^{n} = u i u = - Λ, i u = Ψ$ $= - \frac{1}{2 i n} \int_{0}^{\infty} {(\frac{Λ}{- i})}^{\frac{1 - n}{n}} e^{- Λ} d Λ - \frac{{(- i)}^{\frac{1 - n}{n}}}{2 i n} \int_{0}^{\infty} Ψ^{\frac{1 - n}{n}} e^{- Ψ} d Ψ$ $= Γ (\frac{1}{n} + 1) (- \frac{{(i)}^{\frac{1 - n}{n}}}{2 i n} - \frac{{(- i)}^{\frac{1 - n}{n}}}{2 i n}) = \frac{1}{2 n} Γ (\frac{1}{n} + 1) (- \frac{e^{\frac{π}{2} (\frac{1}{n} - 1) i}}{i} - \frac{e^{- \frac{π}{2} (\frac{1}{n} - 1) i}}{i})$ $= \frac{1}{2 n} Γ (\frac{1}{n} + 1) (- 2 s i n \frac{π}{2} (\frac{1}{n} - 1)) = \frac{1}{n} Γ (\frac{1 + n}{n}) c o s (\frac{π}{2} - \frac{π}{2 n})$ $= \frac{1}{n} Γ (\frac{1 + n}{n}) s i n (\frac{π}{2 n})$
	Answered by mathmax by abdo last updated on 04/Dec/20
	$let I =∫_0 ^∞ cos(x^n )dx and J =∫_0 ^∞ sin(x^n )dx ⇒ I−iJ =∫_0 ^∞ e^(−ix^n ) dx changement ix^n =t give x^n =−it ⇒ x=(−it)^(1/n) =(−i)^(1/n) t^(1/n) =(e^(−((iπ)/2)) )^(1/n) t^(1/n) =e^(−((iπ)/(2n))) t^(1/n) ⇒ I−iJ =e^(−((iπ)/(2n))) ∫_0 ^∞ e^(−t) (1/n)t^((1/n)−1) dt =(1/n) e^(−((iπ)/(2n))) ∫_0 ^∞ t^((1/n)−1) e^(−t) dt =(1/n)Γ((1/n)){cos((π/(2n)))−isin((π/(2n)))} ⇒ ∫_0 ^∞ cos(x^n )dx =(1/n)Γ((1/n))cos((π/(2n))) and ∫_0 ^∞ sin(x^n )dx =(1/n)Γ((1/n))sin((π/(2n)))$
	$let I = \int_{0}^{\infty} \cos (x^{n}) dx and J = \int_{0}^{\infty} \sin (x^{n}) dx \Rightarrow$ $I - iJ = \int_{0}^{\infty} e^{- {ix}^{n}} dx changement {ix}^{n} = t give x^{n} = - it \Rightarrow$ $x = {(- it)}^{\frac{1}{n}} = {(- i)}^{\frac{1}{n}} t^{\frac{1}{n}} = {(e^{- \frac{i π}{2}})}^{\frac{1}{n}} t^{\frac{1}{n}} = e^{- \frac{i π}{2 n}} t^{\frac{1}{n}} \Rightarrow$ $I - iJ = e^{- \frac{i π}{2 n}} \int_{0}^{\infty} e^{- t} \frac{1}{n} t^{\frac{1}{n} - 1} dt = \frac{1}{n} e^{- \frac{i π}{2 n}} \int_{0}^{\infty} t^{\frac{1}{n} - 1} e^{- t} dt$ $= \frac{1}{n} Γ (\frac{1}{n}) {\cos (\frac{π}{2 n}) - isin (\frac{π}{2 n})} \Rightarrow$ $\int_{0}^{\infty} \cos (x^{n}) dx = \frac{1}{n} Γ (\frac{1}{n}) \cos (\frac{π}{2 n}) and$ $\int_{0}^{\infty} \sin (x^{n}) dx = \frac{1}{n} Γ (\frac{1}{n}) \sin (\frac{π}{2 n})$
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