∫_0 ^∞ sinx^p dx ∫_0 ^∞ ((sinx^p )/x^q )dx collected question


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Question Number 124608 by TANMAY PANACEA last updated on 04/Dec/20

$\int_{0}^{\infty} {s i n x}^{p} d x$ $\int_{0}^{\infty} \frac{{s i n x}^{p}}{x^{q}} d x$ $c o l l e c t e d q u e s t i o n$
	Answered by Dwaipayan Shikari last updated on 04/Dec/20

	$\int_{0}^{\infty} {s i n x}^{p} d x$ $= \frac{1}{2 i} \int_{0}^{\infty} e^{{i x}^{p}} - e^{- {i x}^{p}} d x x^{p} = u$ $= \frac{1}{2 i p} \int_{0}^{\infty} {(u)}^{\frac{1 - p}{p}} e^{i u} - \frac{1}{2 i p} \int_{0}^{\infty} u^{\frac{1 - p}{p}} e^{- i u} d t i u = - Ψ i u = φ$ $= - \frac{1}{2 p} \int_{0}^{\infty} {(\frac{- Ψ}{i})}^{\frac{1 - p}{p}} e^{- Ψ} d Ψ + \frac{1}{2 p} \int_{0}^{\infty} {(\frac{φ}{i})}^{\frac{1 - p}{p}} e^{- φ} d φ$ $= \frac{{(i)}^{\frac{1}{p} - 1}}{2 p} Γ (\frac{1}{p}) + \frac{{(- i)}^{\frac{1}{p} - 1}}{2 p} Γ (\frac{1}{p})$ $\frac{1}{2 p} Γ (\frac{1}{p}) (\frac{1}{i} (e^{\frac{π}{2 p} i} - e^{- \frac{π}{2 p} i})) = \frac{Γ (\frac{1}{p}) s i n (\frac{π}{2 p})}{p} = Γ (\frac{1}{p} + 1) s i n (\frac{π}{2 p})$
	Commented by TANMAY PANACEA last updated on 04/Dec/20

	$e x c e l l e n t$
	Answered by mathmax by abdo last updated on 04/Dec/20
	$∫_0 ^∞ ((sin(x^p ))/x^q )dx =−Im(∫_0 ^∞ (e^(−ix^p ) /x^q )dx) but ∫_0 ^∞ (e^(−ix^p ) /x^q )dx =_(ix^p =t) (−i)^p ∫_0 ^∞ (e^(−t) /(((−it)^(1/p) )^q )) (1/p)t^((1/p)−1) dt =(1/p)e^(−((ipπ)/2)) ×∫_0 ^∞ (t^((1/p)−1) /((e^(−((iπ)/2)) )^(q/p) t^(q/p) ))e^(−t) dt =(1/p)e^(−((ipπ)/2)) e^((iqπ)/(2p)) ∫_0 ^∞ t^((1/p)−(q/p)−1) e^(−t) dt =(1/p) e^(i{((qπ)/(2p))−((pπ)/2)}) Γ(((1−q)/p)) (so o<q<1) =(1/p){cos(((qπ)/(2p))−((pπ)/2))+isin(((qπ)/(2p))−((pπ)/2))}Γ(((1−q)/p)) ⇒ ∫_0 ^∞ ((sin(x^p ))/x^q )dx =−(1/p)sin(((qπ)/(2p))−((pπ)/2))Γ(((1−q)/p)) =(1/p)sin(((pπ)/2)−((qπ)/(2p)))×Γ(((1−q)/p))$
	$\int_{0}^{\infty} \frac{\sin (x^{p})}{x^{q}} dx = - Im (\int_{0}^{\infty} \frac{e^{- {ix}^{p}}}{x^{q}} dx) but$ $\int_{0}^{\infty} \frac{e^{- {ix}^{p}}}{x^{q}} dx =_{{ix}^{p} = t} {(- i)}^{p} \int_{0}^{\infty} \frac{e^{- t}}{{({(- it)}^{\frac{1}{p}})}^{q}} \frac{1}{p} t^{\frac{1}{p} - 1} dt$ $= \frac{1}{p} e^{- \frac{ip π}{2}} \times \int_{0}^{\infty} \frac{t^{\frac{1}{p} - 1}}{{(e^{- \frac{i π}{2}})}^{\frac{q}{p}} t^{\frac{q}{p}}} e^{- t} dt = \frac{1}{p} e^{- \frac{ip π}{2}} e^{\frac{iq π}{2 p}} \int_{0}^{\infty} t^{\frac{1}{p} - \frac{q}{p} - 1} e^{- t} dt$ $= \frac{1}{p} e^{i {\frac{q π}{2 p} - \frac{p π}{2}}} Γ (\frac{1 - q}{p}) (so o < q < 1)$ $= \frac{1}{p} {\cos (\frac{q π}{2 p} - \frac{p π}{2}) + isin (\frac{q π}{2 p} - \frac{p π}{2})} Γ (\frac{1 - q}{p}) \Rightarrow$ $\int_{0}^{\infty} \frac{\sin (x^{p})}{x^{q}} dx = - \frac{1}{p} \sin (\frac{q π}{2 p} - \frac{p π}{2}) Γ (\frac{1 - q}{p})$ $= \frac{1}{p} \sin (\frac{p π}{2} - \frac{q π}{2 p}) \times Γ (\frac{1 - q}{p})$
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