is there a general formula for evaluating Gaussian integrals of the form:
$\int_{\infty}^{\infty}x^{a}e^{-kx^b} dx$. I have checked https://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions but couldn't find a general formula. The reason I am asking is I want to evaluate the above for the cases of (a=4, b=4) and (a=2, b=4).
Kind regards
Assuming $k>0$, $$ \int_{-\infty }^{\infty }\!{x}^{4}{{\rm e}^{-k{x}^{4}}}\,{\rm d}x={\frac {\pi}{4\sqrt{2}{k}^{5/4}\Gamma \left( 3/4 \right) }} \\ \int_{-\infty }^{\infty }\!{x}^{2}{{\rm e}^{-k{x}^{4}}}\,{\rm d}x={\frac {\Gamma \left( 3/4 \right) }{2{k}^{3/4}}} $$ Assuming $k>0$, and $a$ positive even integer, $$ \int_{-\infty }^{\infty }\!{x}^{a}{{\rm e}^{-k{x}^{4}}}\,{\rm d}x={\frac {\Gamma \left( (a+1)/4 \right) }{2{k}^{(a+1)/4}}} $$ (and $a$ odd integer yields $0$ of course). $$ \int_{-\infty }^{\infty }\!{x}^{a}{{\rm e}^{-k{x}^{14}}}\,{\rm d}x={\frac {\Gamma \left( (a+1)/14 \right) }{7{k}^{(a+1)/14}}} $$ To do cases other than even integers, try integrating from $0$, $$ \int_{0}^{\infty }\!{x}^{a}{{\rm e}^{-k{x}^{b}}}\,{\rm d}x={\frac {1}{ b k^{(a+1)/b}}\Gamma \left( {\frac {a+1}{b}} \right)} $$