# A definite integral

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

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I suspect you meant to define $\Phi(x)$ using $G(1,y)$ not $G(t,y)$. –  Brendan McKay Sep 24 '11 at 12:31
@Brendan McKay, Yes, you are right. :-) –  Anand Sep 24 '11 at 13:03

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.

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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 $\Phi$.

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Thanks Jacques Carette. I only use Mathematica and Matlab. It seems to me that Mathematica is most powerful symbolic calculation engine. I will have a look of Maple. Thanks a lot! :-) –  Anand Sep 24 '11 at 12:03
Hello, Jacques Carette, I plot the ratio function between the numerical integral of my integral and your solution. But it doesn't give a constant 1 function. –  Anand Sep 24 '11 at 12:12
Probably that's because erf and $\Phi$ are not the same. erf is an integral from 0 to something not from $-\infty$ to something. –  Brendan McKay Sep 24 '11 at 12:29
@Brendan, you are right, after changing $\Phi$ to erf function, numerical calculation suggests that it is the right answer. Thanks! :-) –  Anand Sep 24 '11 at 13:29
@Anand: oops. @Brendan: thanks for noticing! –  Jacques Carette Sep 24 '11 at 18:22

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?

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@Pascal Maillard, your comment is very valuable. Why not just put it as an (partial) answer. The problem that I am working on has been dealt by others using local times of Brownian bridges. Since I am not very familiar with those matters, I am trying to solve the problem analytically. Thank you very much. By the way, to add comment, just search "add comment" in this page, the first result is what you are looking for. –  Anand Sep 24 '11 at 13:09
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