Challenging integral involving exponentials and logarithms

[Steve Zissou]

TL;DR Summary

This is harder than it seems.

Hi friends,

Can anyone offer some insight into this challenging integral?

I can't seem to think my way through this.

Thank you
Stevesie

$$ \int_{0}^{\infty}\frac{1}{x}\exp\left(-\frac{1}{2}\left( \frac{\log\left( x \right)-\mu}{\sigma}\right)^{2} \right)\exp\left(-\frac{1}{2}\left( \frac{ x -\alpha}{\beta}\right)^{2} \right)dx=??? $$

 

[berkeman]

Can you say what this integral is from?

 

[jedishrfu]

Your integral appears to be related to problems involving probability distributions, particularly those involving the product or convolution of different distributions.

Is that what you're working on?

 

[Vanadium 50]

Have you looked it up in a book of integrals?

 

[Steve Zissou]

Can you say what this integral is from?

It's actually related to some statistical Mechanics issues I'm working on, thanks

 

[Steve Zissou]

Have you looked it up in a book of integrals?

You're kidding, right? Ha ha

 

[Steve Zissou]

Your integral appears to be related to problems involving probability distributions, particularly those involving the product or convolution of different distributions.

Is that what you're working on?

Right on, baby

 

[Vanadium 50]

You're kidding, right? Ha ha

No. Why would you think I am kidding?

 

[berkeman]

You're kidding, right? Ha ha

What have you found from your query here? https://www.wolframalpha.com/

 

[Steve Zissou]

What have you found from your query here? https://www.wolframalpha.com/

Looks like Wolfram Alpha can't do it, nor are there any published tables anywhere that have been helpful. That's why I'm enlisting the help of my friends here. Any ideas would be warmly appreciated.

 

[phyzguy]

You may have to do it numerically, or use some approximations. Are the values of alpha, beta, mu and sigma completely arbitrary, or do you have some prior knowledge of what they are? For example, for alpha = 5, beta = 0.5, mu=4.0, sigma=1.0, the second term (the Gaussian) is much narrower than the first term with the log. So you could approximate the first term with a linear function. Then the integral is easy. But this depends on the values of the constant terms. If they are completely arbitrary, it would be pretty easy to build a numerical function that will return the value of the integral given the four constant terms.

 

[Steve Zissou]

Thank you phyzguy! I think you've go the right idea. Still, it seems strange to me that such familiar functions haven't been examined in this way before.

 

[renormalize]

Still, it seems strange to me that such familiar functions haven't been examined in this way before.

I'm not clever enough to evaluate your integral:$$I=I\left(\alpha,\beta,\mu,\sigma\right)\equiv\intop_{0}^{\infty}\exp\left[-\frac{\left(\log x-\mu\right)^{2}}{2\sigma^{2}}\right]\exp\left[-\frac{\left(x-\alpha\right)^{2}}{2\beta^{2}}\right]\frac{dx}{x}\tag{1}$$but I do want to point out that the change-of-variable $x=e^{y-y_{0}+\mu}$ converts it to:$$I=I\left(\varepsilon,y_{0},\sigma\right)=\intop_{-\infty}^{\infty}\exp\left[-\frac{\left(y-y_{0}\right)^{2}}{2\sigma^{2}}\right]\exp\left[-\frac{\left(e^{y}-1\right)^{2}}{2\varepsilon^{2}}\right]dy\tag{2}$$where $\varepsilon\equiv\beta/\alpha$ and $y_{0}\equiv\mu-\log\alpha$. I still can't do the integral in this form either, but it does have the advantage of depending on only 3 parameters rather than 4, and is pretty straightforward to compute numerically for any choice of those parameters.
Also, eq.(2) is reminiscent of integrals that appear in the literature for so-called Normal Log-normal Mixture (NLM) distributions. For example, see: http://repec.org/esAUSM04/up.21034.1077779387.pdf, eq.(4) for a somewhat similar integral. Alas, the author states that the analytical form of it too is "unknown", but does say that the integral can be "readily evaluated...either by simulation or by numerical integration".

 

[Steve Zissou]

Thank you, renormalize! I am familiar with that somewhat cryptic paper. Also I have been mostly unsuccessful in finding more discussion out there of the NLM distribution. Anyways I like your substitution which sets aside one parameter, that's nice. It's the darn double-exponential that seems to be the sticky wicket!