# Integral, possibly of Bessel or Exponential form.

I'm working with a hierarchical statistical model, whereby the output of a log-normal distribution affects the argument of an exponential distribution. I need to marginalize, obtaining the following integral of interest:

$$I(y;\mu,\sigma,\lambda) \equiv \frac{\lambda}{\sigma\sqrt{2\pi}}\int_{0}^\infty \frac{1}{x^2} \exp\left(-\frac{(\ln x - \mu)^2}{2\sigma^2}\right)\exp\left(-\frac{\lambda}{x}y\right)dx$$ where $\mu, \sigma,\lambda>0$.

I've tried changing variables: $x \rightarrow \exp\theta$, but this doesn't seem to help, as the integral contains both terms like $\ln x$ and powers of $x$.

Q: Can this integration be made in analytic form (series are acceptable, but closed form preferred)?

• I think a change of varibales $x\rightarrow 1/t$ maybe of some help. Furthermore write $\lambda y = \tilde{\lambda}$ – tired Jul 20 '15 at 14:38
• because the resulting integral is strongly related to the Laplace transform of lognormal distribution which is known to have no closed form, i would be very impressed if there is actually one. – tired Jul 20 '15 at 14:46
• Yes, with $x\rightarrow 1/t$, I'm left with something that resembles LT of $\exp\left[-\left\{\ln(1/t)\right\}^2\right]$, and that doesn't (at least) compute on WolframAlpha (not that that is an absolute diagnostic!). – Isambard Kingdom Jul 20 '15 at 15:05
• Given the fact that even the much simpler-looking $\displaystyle\int_0^\infty x^{-2}e^{-\ln^2x-1/x}~dx$ does not seem to possess such a closed form, I would be very skeptical about the existence of such an expression for $I$. – Lucian Jul 20 '15 at 16:39