A definite integral - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:29:16Z http://mathoverflow.net/feeds/question/76264 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/76264/a-definite-integral A definite integral Anand 2011-09-24T08:23:48Z 2011-09-29T07:53:19Z <p>Hello,</p> <p>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. </p> <p>Let $$G(t,x)=\frac{e^{-\frac{x^2}{2t}}}{\sqrt{2\pi t}}.$$ The problem is $$\int_0^t \frac{G(s,x)}{\sqrt{t-s}} d s =?$$</p> <p>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: $$\Phi(x) := \int_{-\infty}^x G(1,y) d y.$$</p> <p>Thank you very much for any hints!</p> <p>Wish everyone a nice weekend. :-)</p> <p>Anand</p> http://mathoverflow.net/questions/76264/a-definite-integral/76269#76269 Answer by Jacques Carette for A definite integral Jacques Carette 2011-09-24T11:54:20Z 2011-09-24T11:54:20Z <p>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 $\Phi$.</p> http://mathoverflow.net/questions/76264/a-definite-integral/76271#76271 Answer by Pascal Maillard for A definite integral Pascal Maillard 2011-09-24T12:30:27Z 2011-09-24T12:30:27Z <p>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.</p> <p>Have you checked in books with tables of integrals?</p> <p>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?</p> http://mathoverflow.net/questions/76264/a-definite-integral/76611#76611 Answer by Anand for A definite integral Anand 2011-09-28T08:01:08Z 2011-09-29T07:53:19Z <p>Happy Birthday to Mathoverflow. Wish it flourish and thank many warmhearted people here for their helps! :-)</p> <p>Here is one solution. Let </p> <p>$$G_\sigma(t,x)=\frac{\exp(-\frac{x^2}{2\sigma t})}{\sqrt{2\pi \sigma t}}$$</p> <p>Clearly, </p> <p>$$\int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \int_0^t \frac{e^{-\frac{x^2}{2\nu s}}}{\sqrt{2\pi s (t-s)}} d s\;.$$</p> <p>We assume that $x\ne 0$. Then by change of variable</p> <p>$$s\rightarrow u=\frac{x^2}{2\sigma s}-\frac{x^2}{2\sigma t}, \quad s= \frac{t x^2}{2\sigma t u+x^2},$$</p> <p>the integral becomes</p> <p>$$\frac{|x|e^{-\frac{x^2}{2\sigma t}}}{2\sigma \sqrt{\pi t}}\int_{0}^\infty \frac{e^{-u}}{\sqrt{u}\left(u+\frac{x^2}{2\sigma t}\right)} d u = \sqrt{\frac{\pi}{2\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma t}} \right)\right)\:,$$</p> <p>where we have applied the integral <a href="http://en.wikipedia.org/wiki/Abramowitz_and_Stegun" rel="nofollow">(7.4.9) in P. 302</a></p> <p>$$\int_0^\infty\frac{e^{-at}}{\sqrt{t}(t+z)} d t = \frac{2\pi}{\sqrt{z}}e^{a z} \left(1-\Phi(\sqrt{az})\right),\quad Re(a)>0, z\ne 0, |\arg z|&lt;\pi\:$$</p> <p>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</p> <p>$$\boxed{ \int_0^t \frac{G_\sigma(t-s,x)}{\sqrt{s}} d s = \sqrt{\frac{2\pi}{\sigma}}\left(1-\Phi\left(\frac{|x|}{\sqrt{2\sigma t}} \right)\right)}\:.$$</p> <p>Finally, the case that $x=0$ can be easily verified. This then finishes the proof.</p>