## 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 2011 at 12:31
@Brendan McKay, Yes, you are right. :-) – Anand Sep 24 2011 at 13:03

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 2011 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 2011 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 2011 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 2011 at 13:29
@Anand: oops. @Brendan: thanks for noticing! – Jacques Carette Sep 24 2011 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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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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