Friday, 21 July 2017

Improper integral: ,frac1piintinfty0fracsqrtx1+xext,dx



Good evening! How could one evaluate the following integral 1π0x1+xextdx
I have tried the substitution xx2 but still I could not manage to get to a final result. Any ideas would be really appreciated!

Also t>0.


Answer



Yes, sub x=u2 to get



2π0duu21+u2etu2=1πdu(111+u2)etu2=1πt1πduetu21+u2



The latter integral may be evaluated a few different ways. One way is to multiply and divide by et, and differentiate:



1πduetu21+u2=etπduet(1+u2)1+u2




and



ddtduet(1+u2)1+u2=etduetu2=πett



Integrate back with respect to t and get an error function:



duet(1+u2)1+u2=πtdtett=Cπerft



Noting that




du11+u2=π



We finally have the integral taking the value



1πteterfct


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