I am trying to solve the following integral $$ \int_0^{\infty} \frac{1}{\sqrt{a / \theta (1-e^{-2\theta t})} } \exp \left\{ - \frac{[x-\omega t+x_0 (1-e^{-\theta t})]^2 }{a/\theta(1-e^{-2\theta t})} \right\} dt $$ It is a Guassian with a mean and variance which are time dependent, and $\theta$, $\omega$ ,$a$ are positive constants, and in principle also $x>0$ and can be used a constant in the integration.

I been looking in many places in internet, in integral tables, and I couldn't find anything similar, of course I am not pretending to find this exact integral, but I was hopping to find something similar, anyway, that didn't happen. Also numerical integration of this expression converges to finite values for different sets of parameters.

I think that trying to solve this integral without a change of variables probably is not possible, or at least to me, so I used the following change of variables that seem to simplify a little bit the original integral. $$ s = e^{-\theta t} $$ and so $t=-\log{s}/\theta$ and $dt = -\frac{1}{\theta s} ds$ and the new limit of integration is $s \in (1,0)$

so the new integral is

$$ \int_1^{0} \frac{1}{\sqrt{a / \theta (1-s^2)} } \exp \left\{ - \frac{[x+\omega/\theta \log(s)+x_0 (1-s) ]^2 }{a/\theta(1-s^2)} \right\} (-\frac{1}{\theta s})ds $$ Both limits $\lim_{t \to 0}$ and $\lim_{t \to \infty}$ of the original integrand are $0$, and also $\lim_{s \to 0}$ and $\lim_{s \to 1}$ are $0$.

I think that the advantage of this change of variables is that this makes me remind the kind of integrals that can be solved using the Residues theorem, but I am stuck here cause I have some doubts. These are my questions/doubts.

i) do you think that there is a better change of variables or another way to solve the original integral without using the residues theorem. Or do you suggest another changes of variables in which the expression becomes even simpler?

ii) In case that you agree that I can use this change of variables and the residues theorem, which is the right close curve that I should use in order to apply the theorem. I used residues theorem to solve improper integrals but I don't know if I can use it for this particular case. In my integration domain there are two simple poles at $s=0$ and at $s=1$.

well, that's all, any piece of help will be really appreciated.

  • $\begingroup$ Assuming you did you calculation correctly (and that I did as well) by redefining some of your constants and absorbing them into each other we can write this as $$n\int_0^{1} \frac{1}{s\sqrt{1-s^2} } \exp \left\{ - \frac{[a(1-s)+c+b\log(s)]^2 }{1-s^2} \right\} ds$$ At this point I would recommend making any substitutions you can to simplify the exponential, even at the cost of making the rest of the integrand more complicated $\endgroup$ Aug 24 '17 at 16:12
  • $\begingroup$ @BrevanEllefsen Thanks for the replay!, If I manage to find a substitution, which is the advantage of simplifying the exponential? Do you have any suggestion to follow after that? $\endgroup$
    – Iván
    Aug 24 '17 at 17:12
  • $\begingroup$ admittedly, it was just an idea... I would have to tinker with it to figure out where to go after that, or even if the idea is viable. It might even be that this isn't expressible in closed form. As a general rule though, it is easier to algebraically manipulate things when everything is outside of an exponential. Mathematica still chokes on my rewrite. $\endgroup$ Aug 24 '17 at 17:19
  • $\begingroup$ @BrevanEllefsen I tried to use this expression in mathematica as well but I din't have any answer back. I will try your suggestion and try to simplify it a little bit and then I will see if I have more luck. Thanks again. $\endgroup$
    – Iván
    Aug 24 '17 at 17:26
  • $\begingroup$ Perhaps of interest is that, if we let $s^2 \mapsto u$, we get $$\frac{1}{2}\int _0^1\frac{1}{u\sqrt{1-u}}\exp \left(\frac{-\left(a\left(1-\sqrt{u}\right)+c+\frac{b}{2}\ln \left(u\right)\right)^2}{1-u}\right)du$$ $\endgroup$ Aug 24 '17 at 17:33

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