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0-x-n-e-ix-z-dx-z-C-

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Question Number 144926 by ArielVyny last updated on 30/Jun/21
∫_0 ^∞ x^n (e^(ix) )^z dx=??? (z∈C)
∫0∞xn(eix)zdx=???(z∈C)
Answered by mathmax by abdo last updated on 30/Jun/21
Ψ=∫_0 ^∞ x^n (e^(ix) )^z dx let z=a+ib ⇒ Ψ=∫_0 ^∞ x^n (e^(ix(a+ib)) )dx =∫_0 ^∞ x^n e^(iax−bx) dx =∫_0 ^∞ x^n e^(−(b−ia)x) dx =_((b−ia)x=t) ∫_0 ^∞ (t^n /((b−ia)^n )) e^(−t) (dt/(b−ia)) =(1/((b−ia)^(n+1) ))∫_0 ^∞ t^n e^(−t) dt =((Γ(n+1))/((b−ia)^(n+1) )) we have −iz=−ia+b ⇒Ψ=((Γ(n+1))/((−iz)^(n+1) ))=((Γ(n+1))/(e^(−i(((n+1)π)/2)) z^(n+1) )) =e^((i(n+1)π)/2) ×((Γ(n+1))/z^(n+1) )
Ψ=∫0∞xn(eix)zdxletz=a+ib⇒Ψ=∫0∞xn(eix(a+ib))dx=∫0∞xneiax−bxdx=∫0∞xne−(b−ia)xdx=(b−ia)x=t∫0∞tn(b−ia)ne−tdtb−ia=1(b−ia)n+1∫0∞tne−tdt=Γ(n+1)(b−ia)n+1wehave−iz=−ia+b⇒Ψ=Γ(n+1)(−iz)n+1=Γ(n+1)e−i(n+1)π2zn+1=ei(n+1)π2×Γ(n+1)zn+1

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