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Re: Removing singularity at x=0 for integral
Posted:
Nov 28, 2013 3:25 AM


On 27/11/2013 23:20, dspguy2@netscape.net wrote: > On Tuesday, November 26, 2013 3:10:29 PM UTC5, Axel Vogt wrote: >> On 26.11.2013 16:10, dspguy2t wrote: >> >>> Hi I'm trying to numerically integrate the following expression from 0 .. 1 >> >>> f(x)=1/x^1.5 * exp(A/x) , where A is a constant. >> >>> >> >>> As x>0, the exponential goes to zero faster than the power, so f(x)>0, as x> 0. >> >>> I'm evaluating this integral at various values of A  there are cases when A is very small. >> >>> >> >>> I'm trying to find a way to remove the singularity at x=0. I've looked at the usual techniques, like integration by parts, subtracting the singularity out, change of variables. >> >>> >> >>> The change of variables is usually done when the power is less than one e.g.(1/sqrt(x)). exp(A/x) is not analytic so integration by parts seems problematic. >> >>> >> >>> Any suggestions? >> >>> >> >>> Thanks for any assistance. >> >>> David >> >> >> >> Guessing you mean 0 < A the value is Pi^(1/2)*(erf(A^(1/2))1)/A^(1/2) by Maple >> >> >> >> Note that you do not need 'analytic', so x = 1/t^2 is allowed here and will >> >> give what Maple says for Int(2*exp(A*t^2), t = 1 .. infinity) > > Hi Axel, > Thanks for the help. Yes A>0. Given that I need to solve the integral numerically, integrating out to infinity becomes a little problematic (when A is very small). Most texts usually avoid the integration to infinity by making the reverse substitution.
ISTR There are old (forgotten?) numerical methods based on the zeroes of Hermite polynomials that will allow you to do these integrals to infinity with a finite amount of very well targeted effort.
(in the form after the substitution x = 1/t^2)
f(t) = t^3.exp(At^2)
Is easily integrated by parts. But your target problem isn't.
They date from the era of slow computers and WKB approximation for wavefunctions and the like. > > I was able to make some progress by integrating by parts (see post below). The reason I'm integrating numerically is I'm also looking at a similar, but more difficult integral  which definitely can't be done in closed form. > > Thanks, > David
 Regards, Martin Brown



