# Can you interpret this divergent integral?

In this ArXiv paper by Wilk and Wlodarczyk (published in Physical Review Letters), equation 16 has essentially the following definition of a function: $$\text{f(x)=}\frac{c}{2Dx^2}\exp[\int^x_0 \frac{\mu}{t^2}+\frac{1-2\alpha}{t} dt]$$

They claim that with the normalization condition

$$\int^\infty_0 f(x)\,dx = 1$$

it becomes (with some variable changes)

$$\text{f(x)=}\frac{1}{\Gamma\text{(}\alpha\text{)}}\mu\text{(}\frac{\mu}{x}\text{)}^{\text{(}\alpha-1\text{)}}\exp\text{(-}\frac{\mu}{x}\text{)}.$$

I can't understand the integral in the exponential function, because it appears to be infinite. Can you help me by telling how can I solve this type of integral?

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This question was a repeat of mathoverflow.net/questions/55714. Do not post the same question multiple times - if you think your question should be reopened you can edit it and flag it for moderator attention, or start a thread on tea.mathoverflow.net. –  Zev Chonoles Feb 17 '11 at 16:26
SMH was unable to edit the previous question because two user accounts were created. @SMH: you should register, so your work is less volatile. –  S. Carnahan Feb 17 '11 at 16:35
Can you just email the authors asking what exactly they mean by the function $f$, which---as noted above---is not defined in a traditional sense, at any (!) point? –  Mariano Suárez-Alvarez Feb 17 '11 at 17:33
Why do you expect mathematics published in physics journals to make sense to mathematicians? –  Gerald Edgar Feb 17 '11 at 18:22
@Gerald: Indeed. On the other hand, Talagrand... –  Did Feb 17 '11 at 18:37

The trouble (as was already explained to you) lies in the starting point $t=0$ of the integral in the exponential. Fortunately, W+W are only interested in steady solutions of equation (15) of their arXiv preprint and these can be written as the function in their equation (16) provided one replaces the starting point $t=0$ of the integral by any starting point $t=x_0$ with $x_0>0$. Changing $x_0$ only modifies $f$ by a multiplicative factor so the normalisation condition saves the day.
Assuming for example that $x_0=1$, one gets $$f(x)=\frac{c}{x^2}\exp\left(\int^x_1 \left(\frac{\mu}{t^2}+\frac{1-2\alpha}{t}\right) \mathrm{d}t\right)=\frac{c}{x^2}\exp\left(\mu-\frac{\mu}{x}+(1-2\alpha)\log(x)\right),$$ hence $$f(x)=c\mathrm{e}^{\mu}x^{-1-2\alpha}\mathrm{e}^{-\mu/x}.$$ This is not W+W's formula (18) so either I made a mistake in this post or there is a misprint in W+W's preprint. Note that the function $f_{0}$ written in (18) of W+W's preprint and in your question here is not integrable if $\alpha\le2$ because $f_{0}(x)$ behaves like a multiple of $x^{1-\alpha}$ when $x\to\infty$, hence for such values of $\alpha$, $f_{0}$ cannot be normalized to get a probability density function (even assuming that one got rid of the problem of the starting point $t=0$ as I explained). The function I obtained above is integrable for every positive $\alpha$ and $\mu$.
If $f$ is the density of the distribution of a random variable $X$, the distribution of $Y=\mu/X$ has density $$g(y)=c\mathrm{e}^{\mu}\mu^{-2\alpha}y^{2\alpha-1}\mathrm{e}^{-y},$$ hence $Y$ is a standard Gamma random variable of exponent $2\alpha$ and $$c=\mathrm{e}^{-\mu}\mu^{2\alpha}/\Gamma(2\alpha).$$ (As I said before, this question belongs to math.SE.)
All this might mean that (20) in W+W's preprint should be corrected to $q=1+2\tau D$. Or maybe not. –  Did Feb 17 '11 at 17:55
I want to thank Didier Piau for the answer. It is really helpful. but I think there is a mistake in g(y). g(y) is proportional to $\y^{2\alpha+1}$. I believe that in W+W's article $\2\alpha$ must be just $\alpha$ in obtaining the coefficients of the Fokker-Planck eqn (17). so with the point of evaluationg the integrals that you said, I obtain the distribution like this :$$\text{f(x)=}\frac{1}{\Gamma\text{(}\alpha\text{)}}\mu\text{(}\frac{\mu}{x}\t‌​ext{)}^{\text{(}\alpha+1\text{)}}\exp\text{(-}\frac{\mu}{x}\text{)}.$$ I have $\alpha+1$ not $\alpha-1$. –  SMH Feb 17 '11 at 19:17
@SMH: No, $y^{2\alpha-1}$ is correct. –  Did Feb 17 '11 at 19:22