Re: Precision available with NIntegrate {Method -> Oscillatory}
- To: mathgroup at smc.vnet.net
- Subject: [mg74072] Re: Precision available with NIntegrate {Method -> Oscillatory}
- From: "Peter" <petersamsimon2 at hotmail.com>
- Date: Thu, 8 Mar 2007 04:41:46 -0500 (EST)
- References: <eslsev$q3h$1@smc.vnet.net>
On Mar 7, 12:17 am, "Peter" <petersamsim... at hotmail.com> wrote:
> Hi. I'm helping someone with a scientific programming problem, and I
> wanted to get a high accuracy result to use as a reference solution.
> I am able to get 15 digits of accuracy for a non-oscillatory integral,
> but it appears that when I specify the option Method -> Oscillatory
> for NIntegrate, it ignores my precision request:
>
> --------------------------------------------------------------------
> a = 0;
> b = 6706902 * 10^\(-7);
> c = 589300 * 10^(-6);
> d = 9802874 * 10^(-7);
> t = -5026548 * 10^(-6);
> alpha = 0;
>
> Iint = NIntegrate[(b^2 - 2 d Cos[t + x] b + d^2 + (a + b c
> x)^2)^(-3/2), {x, alpha, Infinity}, {PrecisionGoal -> 15, WorkingPrecision
> -> 30, MaxRecursion -> 200}]
>
> NIntegrate::slwcon :
> Numerical integration converging too slowly; suspect one of the
> following: singularity, value of the integration being 0, oscillatory
> integrand, or insufficient WorkingPrecision. If your integrand is
> oscillatory try using the option Method->Oscillatory in NIntegrate.
> More...
>
> 1.46869038002327
>
> Cint = NIntegrate[Cos[x]/(b^2 - 2 d Cos[t + x] b + d^2 + (a + b c
> x)^2)^(3/2), {x, alpha, Infinity}, {PrecisionGoal -> 15, WorkingPrecision -
> > 25, MaxRecursion -> 300, Method -> Oscillatory}]
>
> 0.2592156
>
> --------------------------------------------------------------------
>
> I'm not concerned about the slow convergence warning on the first
> integral, Iint. Rather, the second integral, Cint, is only evaluated
> to a precision of 8 digits, despite the explicit request for more in
> the call to Nintegrate. Does anyone have a suggestion to obtain 15
> digits of precision on the second integral?
>
> Thanks,
> Peter
I apologise for some of the symbols appearing wrong in my first post.
I have corrected them in the quoted material above. I'm still looking
for an answer.
Thanks,
Peter