A definite integral - MathOverflow

A definite integral

2011-09-24T08:23:48Z

Hello,

I am trying to find an explicit form of the following definite integral. I have tried Mathematica and it failed to give an answer. I am wondering whether anyone knows this integral. It might relate to certain special functions.

Let $$ G(t,x)=\frac{e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}. $$ The problem is $$ \int_0^t \frac{G(s,x)}{\sqrt{t-s}} d s =? $$

One integral, that might be useful, is $$ \int_0^t G(s,x) d s = |x|\left(\Phi\left(\frac{|x|}{\sqrt{2t}}\right)-1\right) + 2t G(t,x) $$ where $\Phi(x)$ is the distribution function of the standard normal random variable: $$ \Phi(x) := \int_{-\infty}^x G(1,y) d y. $$

Thank you very much for any hints!

Wish everyone a nice weekend. :-)

Anand

Answer by Jacques Carette for A definite integral

2011-09-24T11:54:20Z

Maple very quickly gives me $$\frac{\sqrt{2\pi}}{2}\left(1-\mathit{signum}(x)\Phi(\frac{x}{\sqrt{2t}})\right)$$ (assuming $t>0$ and $x$ real), where it uses 'erf' for your $\Phi$.

Answer by Pascal Maillard for A definite integral

2011-09-24T12:30:27Z

I don't know whether this helps, but a probabilistic interpretation of your integral is the following: When multiplied by $\sqrt{t} e^{x^2/2t}$, it is the expectation of the local time at $x$ (or at $0$) of a Brownian bridge from $0$ to $x$ of length $t$. So basically, if one knows the law of the hitting time of $x$ of this process, one should be able to calculate this integral. You might search for that.

Have you checked in books with tables of integrals?

I wanted to post this as a comment, but could not find out how (I'm new to MO). Can you help me on that, please?

Answer by Anand for A definite integral

2011-09-28T08:01:08Z

Happy Birthday to Mathoverflow. Wish it flourish and thank many warmhearted people here for their helps! :-)

Here is one solution. Let

$$ G_\sigma(t,x)=\frac{\exp(-\frac{x^2}{2\sigma t})}{\sqrt{2\pi \sigma t}} $$

Clearly,

$$ \int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \int_0^t \frac{e^{-\frac{x^2}{2\nu s}}}{\sqrt{2\pi s (t-s)}} d s\;. $$

We assume that $x\ne 0$. Then by change of variable

$$ s\rightarrow u=\frac{x^2}{2\sigma s}-\frac{x^2}{2\sigma t}, \quad s= \frac{t x^2}{2\sigma t u+x^2}, $$

the integral becomes

$$ \frac{|x|e^{-\frac{x^2}{2\sigma t}}}{2\sigma \sqrt{\pi t}}\int_{0}^\infty \frac{e^{-u}}{\sqrt{u}\left(u+\frac{x^2}{2\sigma t}\right)} d u = \sqrt{\frac{\pi}{2\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma t}} \right)\right)\:, $$

where we have applied the integral (7.4.9) in P. 302

$$ \int_0^\infty\frac{e^{-at}}{\sqrt{t}(t+z)} d t = \frac{2\pi}{\sqrt{z}}e^{a z} \left(1-\Phi(\sqrt{az})\right),\quad Re(a)>0, z\ne 0, |\arg z|<\pi\: $$

with $a=1$ and $z=\frac{x^2}{2\sigma t}$, where we have used the fact that $\text{Erfc}(x)=2(1-\Phi(x))$. This then proves

$$ \boxed{ \int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \sqrt{\frac{2\pi}{\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma t}} \right)\right)}\:. $$

Finally, the case that $x=0$ can be easily verified. This then finishes the proof.