Gaussian-Like integral

Question

What is the integral of this?

$$\int_0^\infty xe^{-(ax^2+bx)}\,\mathrm{d}x$$

$a$ and $b$ are positive integers.

Answer 1

$$ ax^{2} + bx = a\left[\left(x-\frac{b}{2a}\right)^{2} - \frac{b^{2}}{4a^{2}}\right] $$ make the substitution $$ z = \sqrt{2a}\left(x -\frac{b}{2a}\right) $$ you will end up with an integral like this $$ \frac{\mathrm{e}^{\frac{b^{2}}{4a}}}{2a}\int \left(z-\frac{b}{\sqrt{2a}}\right)\mathrm{e}^{-\frac{z^{2}}{2}}dz $$ or if we split it up $$ \frac{\mathrm{e}^{\frac{b^{2}}{4a}}}{2a}\left[\int^{\infty}_{0} z\mathrm{e}^{-\frac{z^{2}}{2}}dz -\frac{b}{\sqrt{2a}} \int_{0}^{\infty} \mathrm{e}^{-\frac{z^{2}}{2}}dz \right] = \int_{0}^{\infty}x\mathrm{e}^{-\left(ax^{2} + bx\right)}dx $$ at this point you can make use of $$ \int z\mathrm{e}^{-\frac{z^{2}}{2}} = \int -\frac{d}{dz}\mathrm{e}^{-\frac{z^{2}}{2}} = -\mathrm{e}^{-\frac{z^{2}}{2}} $$

the rest you can probably do. Any errors, or need further help comment :).

$\textbf{Edit:}$ $$ \frac{\mathrm{e}^{\frac{b^{2}}{4a}}}{2a}\left[1 - \frac{b}{\sqrt{2a}}\sqrt{\frac{\pi}{2}}\right] $$

you can simplify.

$\textbf{Edit 2:}$ As pointed out by @harrypeter i forgot to choose the appropriate limits for the integral. If one does use the correct limits you will achieve the same result.

Answer 2

The general method is to "complete the square" in the exponent, that is, write $e^{-(ax^2+bx)}$ as $e^{-(\alpha x - \beta)^2}\cdot e^{\gamma}$ for suitable choice of $\alpha, \beta$, and $\gamma$ so as to get the exponential term looking like a Gaussian density function. Then, use the fact that $-xe^{-x^2}$ has antiderivative $e^{-x^2/2}$.