Monday, 16 January 2017

calculus - Evaluate intinfty0!frac1e2x2xex4mathrmdx



How one can evaluate the following improper integral
01e2x2xex4dx


It seems that one can consider the function
I(α)=01eαx2xex4dx

and try to evaluate I(α)=0xeαx2x4dx. But this approach did not work for me.


Answer



01e2x2xex4dx=01e2u2ueu2du=0eu2duk=0(1)k(2u)k(k+1)!=k=0(1)k2k1Γ(k+12)(k+1)!=π2n=01(2n+1)n!n=02n(n+1)(2n+1)!!=π21F1(12;32;1)122F2(1,1;32,2;1).




According to WA, the first summand can be represented as:
π4Erfi(1).



Further simplification seems to be not possible.


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