How to compute $\int_{0}^{\infty}dx\:\frac{\exp(-ax^2+bx)}{x+1}\:\text{ for }\: a>0, b\in \mathbb{C}$? As the title says I am trying to compute the integral  $I=\displaystyle\int_{0}^{\infty}dx\:\frac{\exp(-ax^2+bx)}{x+1}$ where $a>0$ and $b$ is a complex number. For the special case of $b=-2a$, we have $I=-\displaystyle\frac{1}{2}\exp(a)\:\text{Ei}(-a)$ (where $\text{Ei}$ is the Exponential Integral function). However, for the general complex values of $b$ I am not sure how to compute this integral in terms of some known function.
 A: Using
$${{\rm e}^{bx}}=\sum _{n=0}^{\infty }{\frac { \left( bx \right) ^{n}}{n
!}}
$$
and Maple I am obtaining the following result is terms of  Meijer G function
$$\int _{0}^{\infty }\!{\frac {{{\rm e}^{-a{x}^{2}+bx}}}{x+1}}{dx}=\sum 
_{n=0}^{\infty }{\frac {{b}^{n}
G^{3, 2}_{2, 3}\left(a\, \Big\vert\,^{1/2, 1}_{1, 1/2, n/2+1/2}\right)
}{2n!\,\pi \,{a}^{n/2+1/2}}}
$$
A: Hint:
$I(b)=\int_0^\infty\dfrac{e^{-ax^2+bx}}{x+1}dx$
$\dfrac{dI(b)}{db}=\int_0^\infty\dfrac{xe^{-ax^2+bx}}{x+1}dx$
$\therefore\dfrac{dI(b)}{db}+I(b)=\int_0^\infty e^{-ax^2+bx}~dx$
$=\int_0^\infty e^{-a\left(x^2+\frac{bx}{a}\right)}~dx$
$=\int_0^\infty e^{-a\left(x^2+\frac{bx}{a}+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}\right)}~dx$
$=e^\frac{b^2}{4a}\int_0^\infty e^{-a\left(x+\frac{b}{2a}\right)^2}~dx$
$=e^\frac{b^2}{4a}\int_\frac{b}{2a}^\infty e^{-ax^2}~dx$
$=e^\frac{b^2}{4a}\left(\int_0^\infty e^{-ax^2}~dx-\int_0^\frac{b}{2a}e^{-ax^2}~dx\right)$
$=e^\frac{b^2}{4a}\left(\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}-\int_0^\frac{b}{2\sqrt a}e^{-x^2}~d\left(\dfrac{x}{\sqrt a}\right)\right)$
$=e^\frac{b^2}{4a}\left(\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}-\dfrac{1}{\sqrt a}\int_0^\frac{b}{2\sqrt a}e^{-x^2}~dx\right)$
$=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}e^\frac{b^2}{4a}-\sum\limits_{n=0}^\infty\dfrac{b^{2n+1}n!}{2a^{n+1}(2n+1)!}$ (according to http://en.wikipedia.org/wiki/Dawson_function)
$e^bI(b)=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}\int e^{\frac{b^2}{4a}+b}~db-\int\sum\limits_{n=0}^\infty\dfrac{b^{2n+1}e^bn!}{2a^{n+1}(2n+1)!}db$
$=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}\int e^{\frac{1}{4a}(b^2+4ab)}~db-\int\sum\limits_{n=0}^\infty\dfrac{b^{2n+1}e^bn!}{2a^{n+1}(2n+1)!}db$
$=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}\int e^{\frac{1}{4a}(b^2+4ab+4a^2-4a^2)}~db-\int\sum\limits_{n=0}^\infty\dfrac{b^{2n+1}e^bn!}{2a^{n+1}(2n+1)!}db$
$=\dfrac{1}{2}\sqrt{\dfrac{\pi}{a}}e^{-a}\int e^\frac{(b+2a)^2}{4a}~db-\int\sum\limits_{n=0}^\infty\dfrac{b^{2n+1}e^bn!}{2a^{n+1}(2n+1)!}db$
$=\int\sum\limits_{n=0}^\infty\dfrac{\sqrt\pi e^{-a}(b+2a)^{2n}}{2^{2n+1}a^{n+\frac{1}{2}}n!}db-\int\sum\limits_{n=0}^\infty\dfrac{b^{2n+1}e^bn!}{2a^{n+1}(2n+1)!}db$
$=\sum\limits_{n=0}^\infty\dfrac{\sqrt\pi e^{-a}(b+2a)^{2n+1}}{2^{2n+1}a^{n+\frac{1}{2}}n!(2n+1)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n+1}\dfrac{(-1)^kb^ke^bn!}{2a^{n+1}k!}+C$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
$I(b)=\sum\limits_{n=0}^\infty\dfrac{\sqrt\pi e^{-a-b}(b+2a)^{2n+1}}{2^{2n+1}a^{n+\frac{1}{2}}n!(2n+1)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^{2n+1}\dfrac{(-1)^kb^kn!}{2a^{n+1}k!}+Ce^{-b}$
