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. :-)


  • 1
    $\begingroup$ I suspect you meant to define $\Phi(x)$ using $G(1,y)$ not $G(t,y)$. $\endgroup$ Commented Sep 24, 2011 at 12:31
  • $\begingroup$ @Brendan McKay, Yes, you are right. :-) $\endgroup$
    – Anand
    Commented Sep 24, 2011 at 13:03

3 Answers 3


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

Here is one solution. Let

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


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

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.


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?

  • $\begingroup$ @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. $\endgroup$
    – Anand
    Commented Sep 24, 2011 at 13:09
  • $\begingroup$ @Anand, I'm glad I could help you. As for the "add comment", it only appears under my answer, maybe I have to earn more reputation in order to comment on others' answers and questions. $\endgroup$ Commented Sep 24, 2011 at 13:27
  • $\begingroup$ @Pascal Maillard, if you can see the comment "@Brendan McKay, Yes, you are right. :-) – Anand 25 mins ago", right below it, you can find where to add a comment. I think your reputation is enough for add comments. :-) $\endgroup$
    – Anand
    Commented Sep 24, 2011 at 13:31
  • $\begingroup$ @Anand, I just checked, it really takes 50 reputation points to leave comments.. :) $\endgroup$ Commented Sep 24, 2011 at 14:40
  • $\begingroup$ @Pascal Maillard, I am glad to see that you have already more than 50 reputation. :-) $\endgroup$
    – Anand
    Commented Sep 28, 2011 at 8:05

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 $\Phi$.

  • $\begingroup$ 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! :-) $\endgroup$
    – Anand
    Commented Sep 24, 2011 at 12:03
  • $\begingroup$ 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. $\endgroup$
    – Anand
    Commented Sep 24, 2011 at 12:12
  • 3
    $\begingroup$ Probably that's because erf and $\Phi$ are not the same. erf is an integral from 0 to something not from $-\infty$ to something. $\endgroup$ Commented Sep 24, 2011 at 12:29
  • $\begingroup$ @Brendan, you are right, after changing $\Phi$ to erf function, numerical calculation suggests that it is the right answer. Thanks! :-) $\endgroup$
    – Anand
    Commented Sep 24, 2011 at 13:29
  • $\begingroup$ @Anand: oops. @Brendan: thanks for noticing! $\endgroup$ Commented Sep 24, 2011 at 18:22

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