close form solution needed for following integration

Question

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.

Answer 1

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.

Answer 2

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.