How one can evaluate the following improper integral
∫∞01−e−2x2xex4dx
It seems that one can consider the function
I(α)=∫∞01−e−αx2xex4dx
and try to evaluate I′(α)=∫∞0xe−αx2−x4dx. But this approach did not work for me.
Answer
∫∞01−e−2x2xex4dx=∫∞01−e−2u2ueu2du=∫∞0e−u2du∞∑k=0(−1)k(2u)k(k+1)!=∞∑k=0(−1)k2k−1Γ(k+12)(k+1)!=√π2∞∑n=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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