close form solution needed for following integration I wan to solve integration of following form $$\int  xe^{-ax^m-bx^2}dx$$where $a>0,b>0,m>2$. I have looked into Gradeshteyn book but i have not found anything relevant to this. Actually, I need some expression for the integral when the integral limits are from $0$ to $\infty$. Your help will be highly appreciated. Thanks in advance.
 A: A closed-form antiderivative does not seem to exist in general.  
For the integral from $0$ to $\infty$, start with the change of variables
$x = t^{1/m}$, which makes it
$$ \dfrac{1}{m} \int_0^\infty t^{-1+2/m} e^{-at} e^{-b t^{2/m}}\; dt$$
Expand out $\exp(-b t^{2/m})$ as a power series and integrate term-by-term.
I get 
$$ \sum_{k=0}^\infty {\frac { \left( -1 \right) ^{k}{b}^{k}}{k!\; m}
\Gamma \left( 2\,{\frac {k+1}{m}} \right) {a}^{-2\,{\frac {k+1}{m}}}}
$$
For each integer $m > 2$, it seems this can be written in terms of hypergeometric functions.
A: Concerning the antiderivative, I do not think you any expression would be found as soon as $m>2$ (even for integer values of $m$.
If we consider the definite integral $$I_m=\int_0^\infty  x\,e^{-ax^m-bx^2}\,dx$$ a CAS seems to produce some nice monsters $$I_3=\frac{9 b^2 \, _2F_2\left(1,\frac{3}{2};\frac{4}{3},\frac{5}{3};-\frac{4 b^3}{27
   a^2}\right)-4 \sqrt[3]{3} \pi  a^{2/3} e^{-\frac{2 b^3}{27 a^2}}
   \left(\sqrt[3]{3} b \text{Bi}\left(\frac{b^2}{3 \sqrt[3]{3} a^{4/3}}\right)-3
   a^{2/3} \text{Bi}'\left(\frac{b^2}{3 \sqrt[3]{3} a^{4/3}}\right)\right)}{54 a^2}$$ where appear hypergeometric function, Airy function and derivative.$$I_4=\frac{\sqrt{\pi } e^{\frac{b^2}{4 a}} \text{erfc}\left(\frac{b}{2
   \sqrt{a}}\right)}{4 \sqrt{a}}$$ For $m=5$ and above, appear awful linear combinations of hypergeometric functions. For $m=6$, a "simple" one which seems to be $$I_6=\frac{\frac{3 b^3 \, _1F_2\left(1;\frac{4}{3},\frac{5}{3};-\frac{b^3}{27
   a}\right)}{a}+4\ 3^{2/3} \pi  \left(\frac{b^{3/2}}{\sqrt{a}}\right)^{2/3}
   \text{Bi}\left(-\frac{\left(\frac{b^{3/2}}{\sqrt{a}}\right)^{2/3}}{\sqrt[3]{3}}\right)}{36 b}$$
So, I do not think that there is any hope for closed form and infinte series would nedd to be considered.
I stop here since, while I was typing, came Robert Israel's solution.
