Does the integral $\int_0^{\infty}e^{cx^2+dx}dx/(a+bx)$ have a closed form?

Question

The integral is $$\DeclareMathOperator{\dm}{d\!} \int_0^{\infty}\frac{e^{-cx^2+dx}}{a+bx}\dm x. $$ Here I assume that $a,b,c,d$ are chosen to make this integral convergent. Rewritting the rational function as a integral over exponential, I get $$ I=\int_{x,y\ge 0}e^{-cx^2-bxy+dx-ay}\dm x\!\dm y. $$ The exponential $\exp(-cx^2-byx)$ can be expressed as the generating function $$ e^{-cx^2-byx}=\sum_n\frac{(\sqrt{c}x)^n}{n!}H_n\left(-by/2\right), $$ where $H_n(x)$ is the Hermite polynomials. Substituting into integrand, I can integrate over $x$ and get $$ I=\sum_n\frac{c^{n/2}}{n!(-d)^{n+1}}\Gamma(1+n)\int_0^{\infty}H_n(-by/2)e^{-ay}\ \dm y. $$ This integral over $y$ is nothing but the Laplace-transform of Hermite polynomials. Seeding it to MMA then I end up with a pretty awful sum of series to $n$ like this $$ I\sim\frac{\Gamma(n+1)}{n!}\frac{c^{n/2}}{(-d)^{n+1}}\frac{2^{n-1}e^{-in\pi/2-a^2/b^2}\left(-2a/b\right)^{n+1}\Gamma(-n/2)\Gamma(1+n/2,-a^2/b^2)}{\Gamma(-n)}. $$ Is it possible to complete this summation over $n$?

Answer 1

This is not an answer but an alternate approach for another infinite summation.

For sure, I assume that $a$, $b$ and $c$ are positive.

$$\DeclareMathOperator{\dm}{d\!} \frac{e^{-cx^2+dx}}{a+bx}=\frac{e^{-cx^2}}{a+bx}+\sum_{n=1}^\infty\frac 1{n!} \frac{e^{-c x^2} x^n}{(a+b x)}\,d^n$$

As already said in comments (replacing $\text{Shi}$ and $\text{Chi}$ by more "trivial" functions) $$\int_0^\infty\frac{e^{-cx^2}}{a+bx}\,\dm x=\frac 1{2b}\,e^{-\frac{a^2 c}{b^2}} \left(\pi \, \text{erfi}\left(\frac{a \sqrt{c}}{b}\right)-\text{Ei}\left(\frac{a^2 c}{b^2}\right)\right)$$

$$J_n=\int_0^\infty\frac{e^{-cx^2}}{a+bx}x^n\,\dm x=\frac {i^n}{4b}\left(\frac{a}{b}\right)^n\,\,e^{-\frac{a^2 c}{b^2}}\, A_n$$ where $$A_n=n\, \Gamma \left(\frac{n}{2}\right)\, \Gamma \left(-\frac{n}{2},-\frac{a^2 c}{b^2}\right)+2\, i\, \Gamma \left(\frac{n+1}{2}\right)\, \Gamma \left(\frac{1-n}{2},-\frac{a^2 c}{b^2}\right)$$

Computing the partial sums for the case of $a=b=c=d=1$

$$\left( \begin{array}{cc} 1 & 0.886227 \\ 2 & 0.995680 \\ 3 & 1.033048 \\ 4 & 1.044539 \\ 5 & 1.047780 \\ 6 & 1.048629 \\ 7 & 1.048837 \\ 8 & 1.048886 \\ 9 & 1.048896 \\ 10 & 1.048899 \\ \end{array} \right)$$