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Question Number 144180 by Willson last updated on 22/Jun/21
Prove that ∫^( +∞) _0 ((sh(𝛂t))/(sh(t)))dt = (𝛑/2)tan(((𝛑𝛂)/2))
Provethat∫0+∞sh(αt)sh(t)dt=π2tan(πα2)
Answered by mindispower last updated on 22/Jun/21
=∫_0 ^∞ ((e^(at) −e^(−at) )/(e^t −e^(−t) ))dt= e^(−t) =u ⇔∫_0 ^1 ((u^(−a) −u^a )/((1/u)−u)).(1/u)du =∫_0 ^1 ((u^(−a) −u^a )/(1−u^2 ))du =∫_0 ^1 ((x^((−a)/2) −x^(a/2) )/(1−x)).(x^(−(1/2)) /2) =(1/2)∫_0 ^1 ((x^(−((a+1)/2)) −x^((a−1)/2) )/(1−x))dx Ψ(x+1)=−γ+∫_0 ^1 ((1−t^x )/(1−t))dt we get(1/2)( Ψ(((a+1)/2))−Ψ(((1−a)/2)))=(1/2)(Ψ(1−((1−a)/2))−Ψ(((1−a)/2))) Ψ(1−z)−Ψ(z)=πcot(πz) we get (1/2)(πcot(((π(1−a))/2)))=(π/2)cot((π/2)−((aπ)/2)) =(π/2)tan(((aπ)/2))
=∫0∞eat−e−atet−e−tdt=e−t=u⇔∫01u−a−ua1u−u.1udu=∫01u−a−ua1−u2du=∫01x−a2−xa21−x.x−122=12∫01x−a+12−xa−121−xdxΨ(x+1)=−γ+∫011−tx1−tdtweget12(Ψ(a+12)−Ψ(1−a2))=12(Ψ(1−1−a2)−Ψ(1−a2))Ψ(1−z)−Ψ(z)=πcot(πz)weget12(πcot(π(1−a)2))=π2cot(π2−aπ2)=π2tan(aπ2)
Answered by mathmax by abdo last updated on 23/Jun/21
Φ=∫_0 ^∞ ((sh(αt))/(sh(t)))dt ⇒Φ=∫_0 ^∞ ((e^(αt) −e^(−αt) )/(e^t −e^(−t) ))dt =_(e^(−t) =x) −∫_0 ^1 ((x^(−α) −x^α )/(x^(−1) −x))(−(dx/x)) =∫_0 ^1 ((x^(−α) −x^α )/(1−x^2 ))dx =_(x^2 =y) ∫_0 ^1 ((y^(−(α/2)) −y^(α/2) )/(1−y))(dy/(2(√y))) =(1/2)∫_0 ^1 ((y^(−((α+1)/2)) −y^((α−1)/2) )/(1−y))dy =−(1/2)∫_0 ^1 ((1−y^(−((α+1)/2)) )/(1−y))dy +(1/2)∫_0 ^1 ((1−y^((α−1)/2) )/(1−y))dy we know Ψ(z+1)=−γ +∫_0 ^1 ((1−y^z )/(1−y))dy ⇒ ∫_0 ^1 ((1−y^(−((α+1)/2)) )/(1−y))dy=Ψ(1−((α+1)/2))+γ=Ψ(((1−α)/2))+γ ∫_0 ^1 ((1−y^((α−1)/2) )/(1−y))dy =Ψ(1+((α−1)/2))+γ=Ψ(((α+1)/2)) ⇒ Φ=(1/2){Ψ(((α+1)/2))−Ψ(((1−α)/2))} we have also Ψ(1−z)−Ψ(z)=πcotan(πz) ⇒ Ψ(((1+α)/2))−Ψ(((1−α)/2))=Ψ(1−((1−α)/2))−Ψ(((1−α)/2)) =πcotan(π(((1−α)/2)))=πcotan((π/2)−((πα)/2))=πtan(((πα)/2))
Φ=∫0∞sh(αt)sh(t)dt⇒Φ=∫0∞eαt−e−αtet−e−tdt=e−t=x−∫01x−α−xαx−1−x(−dxx)=∫01x−α−xα1−x2dx=x2=y∫01y−α2−yα21−ydy2y=12∫01y−α+12−yα−121−ydy=−12∫011−y−α+121−ydy+12∫011−yα−121−ydyweknowΨ(z+1)=−γ+∫011−yz1−ydy⇒∫011−y−α+121−ydy=Ψ(1−α+12)+γ=Ψ(1−α2)+γ∫011−yα−121−ydy=Ψ(1+α−12)+γ=Ψ(α+12)⇒Φ=12{Ψ(α+12)−Ψ(1−α2)}wehavealsoΨ(1−z)−Ψ(z)=πcotan(πz)⇒Ψ(1+α2)−Ψ(1−α2)=Ψ(1−1−α2)−Ψ(1−α2)=πcotan(π(1−α2))=πcotan(π2−πα2)=πtan(πα2)

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