Saturday, October 12, 2019

calculus - Evaluating a parametric integral



I need some help to evaluate the following integral.




0dxxα(x+1) where α(0,1)




I've tried many ways (the best one seems to be developing by Taylor series) but actually I have no solution.




Some ideas?
Thank you.


Answer



I think we will need some complex analysis here.



Take a branch of 1/za defined in C[0,+) and consider the integral
γdzza(z+1)
where γ is a close path composed by an arc of a inner circle of radius $01andtwoparallelsegmentsoverandunderthesegment[r,R].Thenbytheresiduetheoremγdzza(z+1)=2πiRes(1za(z+1),1)=2πieiπa.NowwetakethelimitasR\to+\inftyandr\to 0^+.Itiseasytoseethattheintegralsalongthearcsofthecirclesgoesto0.Hence+0dxxa(x+1)+0dxxae2πia(x+1)=2πieiπawhichimpliesthat+0dxxa(x+1)=2πieiπa1e2iπa=πsin(πa).$


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