approximate $\int_{u=0}^b e^{-\frac{u^2 + ac}{au}}du.$ I'm trying to find an approximation (or exact solution if possible) for an integral of the form:
$$\int_{u=0}^b e^{-\frac{u^2 + ac}{au}}du.$$
I was thinking of somehow applying a Gauss Hermite Quadriture Expansion, but I'm not sure how I would do this. Does anyone know the best way I should go about this?
 A: I do not know much about numerical integration, but the following trick may help you avoid exponents diverge ad infinitum.
We are dealing with the integral
$$ I = \int_{0}^{b} \exp\left( - \frac{u}{a} - \frac{c}{u} \right) \, du. $$
With the substitution $u = \sqrt{a c} x^{2}$, for $\beta = \left( b / \sqrt{ac} \right)^{1/2}$ we have
\begin{align*}
I
&= 2\sqrt{ac} \int_{0}^{\beta} x \exp\left\{ - \sqrt{\frac{c}{a}} \left( x^{2} + x^{-2} \right) \right\} \, dx \\
&= 2\sqrt{ac} e^{-2\sqrt{c/a}} \int_{0}^{\beta} x \exp\left\{ - \sqrt{\frac{c}{a}} \left( x - x^{-1} \right)^{2} \right\} \, dx.
\end{align*}
With the substitution $y = x - x^{-1}$ and
$$ \alpha := \frac{\beta + \sqrt{\beta^{2} + 4}}{2} = \frac{\sqrt{b} + \sqrt{b + 4\sqrt{ac}}}{2\sqrt[4]{ac}}, \qquad \gamma := \sqrt{\frac{c}{a}}, $$
we find that
\begin{align*}
I &= 2a\gamma e^{-2\gamma} \int_{-\infty}^{\alpha} \frac{\left(\sqrt{y^{2}+4}+y\right)^{2}}{4 \sqrt{y^{2}+4}} e^{-\gamma\,y^{2}} \, dy.
\end{align*}
Now with aid of Mathematica, we can check that 
$$\int_{-\infty}^{0} \frac{\left(\sqrt{y^{2}+4}+y\right)^{2}}{4 \sqrt{y^{2}+4}} e^{-\gamma y^{2}} \, dy = \frac{1}{2} e^{2 \gamma} K_1(2 \gamma)-\frac{1}{4 \gamma}, $$
where $K_{n}(z)$ is the modified Bessel function of the second kind. This allows us to write
\begin{align*}
I &= a\gamma e^{-2\gamma} \left[ e^{2\gamma} K_1\left(2\gamma \right)-\frac{1}{2\gamma} +  2 \int_{0}^{\alpha} \frac{\left(\sqrt{y^{2}+4}+y\right)^{2}}{4 \sqrt{y^{2}+4}} e^{-\gamma\,y^{2}} \, dy \right],
\end{align*}
with each term being much tractable.
