Re: two integrals
- To: mathgroup at smc.vnet.net
- Subject: [mg79204] Re: two integrals
- From: chuck009 <dmilioto at comcast.com>
- Date: Sat, 21 Jul 2007 04:21:33 -0400 (EDT)
How about just using the built-in CauchyPrinipalValue routine for the numerical check:
In[68]:=
<< NumericalMath`CauchyPrincipalValue`
In[68]:=
CauchyPrincipalValue[int2 /. p -> 3/4, {z, 0, {3/4}, 1},
WorkingPrecision -> 50, PrecisionGoal -> 30]
Out[68]=
-3.350007071973017439190457717672333631156356729013`31.4358463\
52995448
> On 18 , 09:56, dimitris <dimmec... at yahoo.com>
> wrote:
> N[res3 /. p -> 3/4]
>
> Out[31]=
> -3.3500070719730175 + 0.*I
>
> In[34]:=
> NIntegrate[ArcTan[(1 - z^2)/z^3]*(1/(z - p)) /. p ->
> 3/4, {z, 0, 3/4 -
> 10^(-7)}] +
> NIntegrate[ArcTan[(1 - z^2)/z^3]*(1/(z - p)) /. p
> p -> 3/4, {z, 3/4 +
> 10^(-7), 1}]
>
> Out[34]=
> -3.350006329606476
>
>