I'm trying to find a closed-form or approximate solution to the following improper integral $$I = \int^\infty_{-\infty}\frac{1}{|x|} \exp \left(- \frac{a}{x^2} - \frac{(x + b)^2}{c}\right)dx$$ for $a\geq0$, $b \neq 0$ and $c > 0$, all known real constants. I know that when $ b=0$, the integral has a solution in terms of the zero-th order modified Bessel function of the second kind, $K_0$. Here's a reference question that treats that case.
Things I've tried so far:
- Since the integral looks very similar to the one on this previous question (it's just missing a $|x|$ on the second term of the exponential), I tried a similar approach of expanding the first term of the exponent with the series definition of an exponential, and then integrating over that. If I skip the formalities of exchanging integral and summation, the problem is reduced to finding the following integral $$I' =\int_{-\infty}^\infty \frac{1}{|x|^{2n + 1}}\exp \left(- \frac{(x+b)^2}{c}\right)dx$$
However, this integral diverges, so this route doesn't seem promising. Note that the original integral does converge, I've verified it via numerical integration.
- Following the same approach as when $b = 0$. Let $x = (ac)^{1/4}u \implies dx = (ac)^{1/4}du $, then $$ I = \exp(-b^2/c) \int_{-\infty}^\infty \frac{1}{|u|}\exp \left(-\sqrt{\frac{a}{c}}(u^{-2} + u^2) -\frac{2b(ac)^{1/4}u}{c}\right) du$$
Now make another change of variables: $u^2 = e^t$, which ultimately leads to $$ I = 2\exp(-b^2/c)\int_0^\infty \exp \left(-2\sqrt{\frac{a}{c}} \cosh t - \frac{2b(ac)^{1/4}e^{t/2}}{c} \right)dt $$
Note again that when $b=0$, the second term of the exponential vanishes and we are left with $$I(b=0) = 2\int_0^\infty \exp \left(-2\sqrt{\frac{a}{c}} \cosh t \right) = 2K_0\left(2\sqrt{\frac{a}{c} }\right)$$
by definition of $K_0$. Now, my question is whether there exists a closed-form solution for the general case when $b \neq 0$. I'd be interested also if there's approximations when $|b|$ is small.