Timeline for Estimating an integral with a singularity at the interval's endpoint
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 12, 2011 at 19:01 | vote | accept | Mikhail Kagalenko | ||
| May 12, 2011 at 19:01 | comment | added | Mikhail Kagalenko | OK, I missed that this integral evaluates exactly. Thank you for your answer | |
| May 12, 2011 at 17:18 | comment | added | Michael Renardy | Mathematica gives the value $${\pi\over 2}\sqrt{y_0}e^{-y_0^2/2}(I_{-1/4}(y_0^2/2)+I_{1/4}(y_0^2/2)),$$ | |
| May 12, 2011 at 13:15 | comment | added | Mikhail Kagalenko | Put another way, there's still problem of analytic estimate for $\int_0^\infty\frac{1}{\sqrt{y}}\exp(-(y-y_0)^2)dy$ and the saddle-point approximation doesn't work for $у_0\in[0,1]$. Sure, it's easy to get the value $2\ \Gamma\left(5/4\right)$ for this integral when $y_0=0$, but that wasn't my question. | |
| May 12, 2011 at 10:03 | comment | added | Mikhail Kagalenko | I am interested in this integral as a function of $x_0$ for fixed large, but finite $a$. The interval of values $x_0\in[0, \sqrt{a}]$ is of particular interest. | |
| May 12, 2011 at 2:16 | comment | added | Michael Renardy | If $x_0/\sqrt{a}$ is bounded, the calculation above works. If $x_0/\sqrt{a}$ is large, the saddle point method works. | |
| May 12, 2011 at 2:12 | comment | added | Robert Israel | For $x_0 = 0$, i.e. $y_0 = 0$, Maple says $\int_0^\infty \frac{1}{\sqrt{y}} {\rm exp}(-y^2)\, dy = \frac{\pi \sqrt{2}}{2 \Gamma(3/4)}$ so your integral is asymptotic to $\frac{\pi \sqrt{2}}{2 \Gamma(3/4)} a^{-1/4}$. | |
| May 12, 2011 at 1:43 | comment | added | Robert Israel | If $x_0$ does not depend on $a$, then either $x_0 = 0$ (which does require a different analysis) or $x_0 > 0$, in which case $a x_0^2 \to \infty$ as $a \to \infty$. If you want something for $x_0 \sim 1/\sqrt{a}$, that says $x_0$ does depend on $a$. | |
| May 11, 2011 at 23:51 | comment | added | Mikhail Kagalenko | No, $x_0$ doe not depend on $a$. I mean that I need the value of the integral for all $x_0\in [0,1]$, including the region $x_0\sim 1/\sqrt{a}$. Sorry if my comment was unclear. | |
| May 11, 2011 at 23:36 | history | answered | Michael Renardy | CC BY-SA 3.0 |