This one: $\int_{0}^{\bar{x}}e^{-x^{a}}x^{b}(1-x)^{c}dx,a,b,c\ge0$. When $a=1,c=0,\bar{x}=\infty$ it is the gamma function.
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$\begingroup$ The integral does not even converge for $a>0$. $\endgroup$– LeechLatticeCommented Jun 13, 2019 at 1:41
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$\begingroup$ no name, no closed form evaluation. $\endgroup$– Carlo BeenakkerCommented Jun 13, 2019 at 5:47
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1 Answer
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Maple evaluates $\int e^{-x^a}x^b\;dx$ in terms of the Whittaker M function, which may also be written as ${}_1F_1$ hypergeometrics. $$ \int \!{{\rm e}^{-{x}^{a}}}{x}^{b}\,{\rm d}x= \\ {\frac {{{\rm e}^{-{ x}^{a}/2}}}{ \left( b+1 \right) \left( a+b+1 \right) \left( 2\,a+b+1 \right) } \left( \left( \left( a+b+1 \right) {x}^{b/2+1/2-3/2\,a}+{ x}^{b/2+1/2-a/2}a \right) a{{\rm M}_{-{\frac {a-b-1}{2a}},\,{ \frac {2\,a+b+1}{2a}}}\left({x}^{a}\right)}+{x}^{b/2+1/2-3/2\,a}{ {\rm M}_{{\frac {a+b+1}{2a}},\,{\frac {2\,a+b+1}{2a}}}\left({x }^{a}\right)} \left( a+b+1 \right) ^{2} \right) } $$
answered Jun 13, 2019 at 12:50