Re: Identity from Erdelyi et al.
- To: mathgroup at smc.vnet.net
- Subject: [mg107896] Re: Identity from Erdelyi et al.
- From: "Dr. C. S. Jog" <jogc at mecheng.iisc.ernet.in>
- Date: Tue, 2 Mar 2010 03:37:15 -0500 (EST)
Dear Dr. McClure:
Since some people suggested that I verify the identity numerically, I am
including the results for some numerical tests that I carried out. They
seem to suggest that my original conjecture about replacing H_{\nu}^2 with
K_{\nu} seems to be correct (although, again, one can't be 100% sure due
to the verification being `numerical')
In: a=2;b=1;lam=1;n=0;
In:
NIntegrate[x*(BesselJ[n,a*x]*BesselY[n,b*x]-BesselJ[n,b*x]*BesselY[n,a*x])/((lam^2+x^2)*(BesselJ[n,b*x]*BesselJ[n,b*x]+BesselY[n,b*x]*BesselY[n,b*x])),{x,0,1000}]
Out: -0.424527
In: -N[(Pi/2)*BesselK[n,lam*a]/BesselK[n,lam*b]]
Out: -0.424926
In: a=2;b=1;lam=1;n=1;
In:
NIntegrate[x*(BesselJ[n,a*x]*BesselY[n,b*x]-BesselJ[n,b*x]*BesselY[n,a*x])/((lam^2+x^2)*(BesselJ[n,b*x]*BesselJ[n,b*x]+BesselY[n,b*x]*BesselY[n,b*x])),{x,0,1000}]
Out: -0.364609
In: -N[(Pi/2)*BesselK[n,lam*a]/BesselK[n,lam*b]]
Out: -0.365008
In: a=5;b=2;lam=2;n=2;
In:
NIntegrate[x*(BesselJ[n,a*x]*BesselY[n,b*x]-BesselJ[n,b*x]*BesselY[n,a*x])/((lam^2+x^2)*(BesselJ[n,b*x]*BesselJ[n,b*x]+BesselY[n,b*x]*BesselY[n,b*x])),{x,0,500}]
NIntegrate::ncvb:
NIntegrate failed to converge to prescribed accuracy after 9
recursive bisections in x near {x} = {32.2188}. NIntegrate obtained
-8
-0.0019889 and 2.02668 10 for the integral and error estimates.
Out: -0.0019889
In: -N[(Pi/2)*BesselK[n,lam*a]/BesselK[n,lam*b]]
Out: -0.00194165
Note that the integrand is highly oscillatory, and NIntegrate can fail to
converge if I give too high a value for `Infinity'
Regards
Jog
On Sun, 28 Feb 2010, Mark McClure wrote:
> On Sun, Feb 28, 2010 at 7:27 PM, Dr. C. S. Jog
> <jogc at mecheng.iisc.ernet.in> wrote:
>
> > I forgot to mention in my previous e-mail the constraints, $0<b<a$, and
> > Re(lam) >0. However, even if I incorporate these constraints using the
> > `Assuming' command, Mathematica still fails to evaluate the said integral.
>
> Then I choose a=2, b=1, and lam=2. The integral still appears to be
> non-zero; less than -0.1, in fact. Of course, I'm still not sure if
> that's what you're after.
>
> Mark
>
>
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.
--
This message has been scanned for viruses and
dangerous content by MailScanner, and is
believed to be clean.