I recognize that the ∫∞0e−xxndx=Γ(n+1) and ∫+∞−∞e−x2dx=√π. I am having difficulty, however with ∫+∞−∞e−x2xndx. By the substitution u=x2, this can be equivalently expressed as 12∫+∞−∞e−uun−12du. This integral is similar to the first one listed (which equates to the Γ function), except that its domain spans R like the second integral (which equates to √π). Any pointers on how to evaluate this integral would be helpful.
Answer
Let In:=∫+∞−∞e−x2xndx. If n is odd then In=0 and for p≥1:
\begin{align} I_{2p}&=\int_0^{+\infty}e^{-x^2}x^{2p}dx+\int_{-\infty}^0e^{-x^2}x^{2p}dx\\ &=\int_0^{+\infty}e^{-t^2}t^{2p}dt+\int_0^{+\infty}e^{-t^2}(-t)^{2p}dt\quad (\mbox{left: } t=x,\mbox{right: } t=-x)\\ &=2\int_0^{+\infty}e^{-t^2}t^{2p}dt\\ &=2\int_0^{+\infty}e^{-s}s^p\frac 1{2\sqrt s}ds \quad (s=t^2)\\ &=\int_0^{+\infty}e^{-s}s^{p-1/2}ds\\ &=\left[-e^{-s}s^{p-1/2}\right]_0^{+\infty}+\int_0^{+\infty}e^{—s}\left(p-\frac 12\right)s^{p-1-1/2ds}\\ &=\left(p-\frac 12\right)I_{2(p-1)}. \end{align}
Finally we get I_{2p+1}=0 and I_{2p}=\sqrt \pi\prod_{j=1}^p\left(j-\frac 12\right) for all p\geq 0.
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