Analytic solution to integral with exponentiated logarithm squared I came across this integral in Vaughan (2005) (eq. 21) and I was wondering if there could be an analytical solution to it.
$$I = \int_0^\infty e^{-a \ln^2(x) - b x} \,\mathrm{d}x$$
with $a,b > 0$.   
So far it has proven impenetrable to my usual arsenal of integration tricks. 
Motivation: I've approximated this integral numerically in python using scipy, but I typically have to do this for $>10^4$ different values of $a$ and $b$ which tends to take a tremendously long time.  I'm hoping that a analytical solution will be faster to compute.

What I have tried:
Making the substitution: $u = \ln(x)$, the integral can be rewritten as
$$I = \int_{-\infty}^\infty \exp\left(-a u^2 + u - b e^u \right) \,\mathrm{d}u = \int_{-\infty}^\infty f'(u) g(u) \,\mathrm{d}u$$
Using $f(x) = -\frac{1}{b} e^{-be^x}$ and $g(x) = e^{-ax^2}$
 with $f'(x) = e^{-be^x + x}$ and $g'(x) = -2ax e^{-ax^2} $,
I tried integration by parts. However this just leads to another similar integral which is no easier to solve.
$$ I = -\frac{2a}{b}\int_{-\infty}^\infty u \exp(-a u^2 -b e^u)\,\mathrm{d}u$$
Any leads would be appreciated!
 A: Hint:
Approach $1$:
$\int_0^\infty e^{-a\ln^2x-bx}~dx$
$=\int_{-\infty}^\infty e^{-au^2-be^u}~d(e^u)$
$=-\dfrac{1}{b}\int_{-\infty}^\infty e^{-au^2}~d(e^{-be^u})$
$=-\left[\dfrac{e^{-au^2-be^u}}{b}\right]_{-\infty}^\infty+\dfrac{1}{b}\int_{-\infty}^\infty e^{-be^u}~d(e^{-au^2})$
$=-\dfrac{2a}{b}\int_{-\infty}^\infty ue^{-au^2}e^{-be^u}~du$
$=-\dfrac{2a}{b}\int_{-\infty}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nb^nue^{nu-au^2}}{n!}~du$
$=\int_{-\infty}^\infty\sum\limits_{n=0}^\infty\dfrac{2(-1)^{n-1}ab^{n-1}ue^{-a\left(u^2-\frac{nu}{a}\right)}}{n!}~du$
$=\int_{-\infty}^\infty\sum\limits_{n=0}^\infty\dfrac{2(-1)^{n-1}ab^{n-1}ue^{-a\left(u^2-\frac{nu}{a}+\frac{n^2}{4a^2}-\frac{n^2}{4a^2}\right)}}{n!}~du$
$=\int_{-\infty}^\infty\sum\limits_{n=0}^\infty\dfrac{2(-1)^{n-1}ab^{n-1}e^\frac{n^2}{4a}ue^{-a\left(u-\frac{n}{2a}\right)^2}}{n!}~du$
$=\int_{-\infty}^\infty\sum\limits_{n=0}^\infty\dfrac{2(-1)^{n-1}ab^{n-1}e^\frac{n^2}{4a}\left(v+\dfrac{n}{2a}\right)e^{-av^2}}{n!}~dv$
$=\int_{-\infty}^\infty\sum\limits_{n=0}^\infty\dfrac{2(-1)^{n-1}ab^{n-1}e^\frac{n^2}{4a}ve^{-av^2}}{n!}~dv+\int_{-\infty}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^{n-1}b^{n-1}ne^\frac{n^2}{4a}e^{-av^2}}{n!}~dv$
$=\sum\limits_{n=1}^\infty\dfrac{(-1)^{n-1}b^{n-1}e^\frac{n^2}{4a}\sqrt\pi}{(n-1)!\sqrt a}$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^nb^ne^\frac{(n+1)^2}{4a}\sqrt\pi}{n!\sqrt a}$
But this approach fails as the infinite series diverges
