Integral of combination of power, exponential, and confluent hypergeometric function I am trying to solve a couple integrals of the form:
\begin{equation}
I(g)=\int_{0}^{\infty} x \, e^{-a(gx-b)^{2}}\,e^{-\beta_{1}x}\, {_{1}}F_{1}(-\alpha_{1};-\alpha_{3};\beta_{3} x) \ \mathrm{d}x
\end{equation}
where, $a>0$, $g\in\mathbb{R}$, $b\in\mathbb{R}$, $0<\beta_{1}<\beta_{3}$, and $0<\alpha_{1}<\alpha_{3}$.
$\alpha_{1}$, and $\alpha_{3}$ are either integers or of the form $\frac{2n+1}{2}$ where $n=0,1,2,\dots$
The shifting of the quadratic in the first exponential term is really giving me issues.  Any thoughts on how to approach this?
Update 08/16/16
I posted a solution to the question for $\alpha_{1}\in \mathbb{Z}^{+}$.  If anyone can find a closed-form solution for the integral when $\alpha_{1}$ is not an integer, I will accept that answer instead.
 A: The integral to be solved is:
\begin{equation}
I(g) = \int_{0}^{\infty}x\,e^{-a(gx-b)^{2}}\,e^{-\beta_{1}x}\, {_{1}}F_{1}(-\alpha_{1};-\alpha_{3};\beta_{3} x) \ \mathrm{d}x
\end{equation}
First, expand the quadratic in the exponential, then simplify to give
\begin{equation}
I(g) = e^{-ab^{2}}\int_{0}^{\infty}x\,e^{-ag^{2}x^{2}-(\beta_{1}-2abg)x}\, {_{1}}F_{1}(-\alpha_{1};-\alpha_{3};\beta_{3} x) \ \mathrm{d}x
\end{equation}
Assuming $\alpha_{1}\in\mathbb{Z}^{+}$, the integrand can be expressed as a sequence using the definition of the confluent hypergeometic function. This gives,
\begin{equation}
I(g) = e^{-ab^{2}}
\sum_{n=0}^{\alpha_{1}}\frac{(-\alpha_{1})_{n}\,\beta_{3}^{n}}{(-\alpha_{3})_{n}\,n!}\int_{0}^{\infty}x^{n+1}\,e^{-ag^{2}x^{2}-(\beta_{1}-2abg)x}\ \mathrm{d}x
\end{equation}
where, $(a)_{n}=\frac{\Gamma(a+n)}{\Gamma(a)}$ is the pochhammer symbol.
From Gradshteyn & Ryzhik (8th ed.) 3.462 #1
\begin{equation}
\int_{0}^{\infty}x^{\nu-1}\, e^{-\beta x^{2}-\gamma x} \ \mathrm{d}x=\frac{\Gamma(\nu)}{(2\beta)^{\nu/2}}\, \exp\left(\frac{\gamma^{2}}{8\beta}\right)\, D_{-\nu}\left(\frac{\gamma}{\sqrt{2\beta}}\right) \quad\quad [\mathrm{Re} \ \beta>0, \ \ \mathrm{Re} \ \nu>0]
\end{equation}
where, $D_{\nu}(z)$ is the parabolic cylinder function.
For this integral, $\nu=n+2$, $\beta=ag^{2}$, and $\gamma=\beta_{1}-2abg$.  Therefore,
\begin{equation}
I(g) = e^{-ab^{2}}
\sum_{n=0}^{\alpha_{1}}\frac{(-\alpha_{1})_{n}\,\beta_{3}^{n}}{(-\alpha_{3})_{n}\,n!}
\frac{\Gamma(n+2)}{(2ag^{2})^{(n+2)/2}}\, \exp\left(\frac{(\beta_{1}-2abg)^{2}}{8ag^{2}}\right)\, D_{-(n+2)}\left(\frac{\beta_{1}-2abg}{\sqrt{2ag^{2}}}\right)
\end{equation}
Simplifying gives,
\begin{equation}
I(g) = \frac{\exp\left(\frac{(\beta_{1}-2abg)^{2}}{8ag^{2}}-ab^{2}\right)}{2ag^{2}}
\sum_{n=0}^{\alpha_{1}}\frac{(-\alpha_{1})_{n}\,(2)_{n}}{(-\alpha_{3})_{n}\,n!}
\left(\frac{\beta_{3}}{\sqrt{2a}\,|g|}\right)^{n}\, D_{-(n+2)}\left(\frac{\beta_{1}-2abg}{\sqrt{2a}\,|g|}\right)
\end{equation}
Note that this solution has a singularity at $g=0$. To get the value of this function at the singularity, plug $g=0$ in the original integral and solve.
