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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
∫_0 ^∞ sinx^p dx ∫_0 ^∞ ((sinx^p )/x^q )dx collected question
∫0∞sinxpdx∫0∞sinxpxqdxcollectedquestion
Answered by Dwaipayan Shikari last updated on 04/Dec/20
∫_0 ^∞ sinx^p dx =(1/(2i))∫_0 ^∞ e^(ix^p ) −e^(−ix^p ) dx x^p =u =(1/(2ip))∫_0 ^∞ (u)^((1−p)/p) e^(iu) −(1/(2ip))∫_0 ^∞ u^((1−p)/p) e^(−iu) dt iu=−Ψ iu=ϕ =−(1/(2p))∫_0 ^∞ (((−Ψ)/i))^((1−p)/p) e^(−Ψ) dΨ + (1/(2p))∫_0 ^∞ ((ϕ/i))^((1−p)/p) e^(−ϕ) dϕ =(((i)^((1/p)−1) )/(2p))Γ((1/p))+(((−i)^((1/p)−1) )/(2p))Γ((1/p)) (1/(2p))Γ((1/p))((1/i)(e^((π/(2p))i) −e^(−(π/(2p))i) ))=((Γ((1/p))sin((π/(2p))))/p)=Γ((1/p)+1)sin((π/(2p)))
∫0∞sinxpdx=12i∫0∞eixp−e−ixpdxxp=u=12ip∫0∞(u)1−ppeiu−12ip∫0∞u1−ppe−iudtiu=−Ψiu=φ=−12p∫0∞(−Ψi)1−ppe−ΨdΨ+12p∫0∞(φi)1−ppe−φdφ=(i)1p−12pΓ(1p)+(−i)1p−12pΓ(1p)12pΓ(1p)(1i(eπ2pi−e−π2pi))=Γ(1p)sin(π2p)p=Γ(1p+1)sin(π2p)
Commented by TANMAY PANACEA last updated on 04/Dec/20
excellent
excellent
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))
∫0∞sin(xp)xqdx=−Im(∫0∞e−ixpxqdx)but∫0∞e−ixpxqdx=ixp=t(−i)p∫0∞e−t((−it)1p)q1pt1p−1dt=1pe−ipπ2×∫0∞t1p−1(e−iπ2)qptqpe−tdt=1pe−ipπ2eiqπ2p∫0∞t1p−qp−1e−tdt=1pei{qπ2p−pπ2}Γ(1−qp)(soo<q<1)=1p{cos(qπ2p−pπ2)+isin(qπ2p−pπ2)}Γ(1−qp)⇒∫0∞sin(xp)xqdx=−1psin(qπ2p−pπ2)Γ(1−qp)=1psin(pπ2−qπ2p)×Γ(1−qp)

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