Re: two integrals
- To: mathgroup at smc.vnet.net
- Subject: [mg79157] Re: two integrals
- From: dimitris <dimmechan at yahoo.com>
- Date: Fri, 20 Jul 2007 03:18:24 -0400 (EDT)
- References: <f7kdjq$4bd$1@smc.vnet.net>
On 18 , 09:56, dimitris <dimmec... at yahoo.com> wrote:
> Any ideas for closed form expressions for the
> following integrals?
>
> In[71]:=
> Integrate[ArcTan[(1 - z^2)/z^3]*(1/(z + p)), {z, 0, 1}]
>
> In[73]:=
> Integrate[ArcTan[(1 - z^2)/z^3]*(1/(z - p)), {z, 0, 1}, PrincipalValue
> -> True]
>
> [Parameter p takes values in the range (0,1) so
> the second integral should be interpreted as a Cauchy
> principal value integral]
>
> Thanks, in advance, for your time and effort.
>
> D.A.
(2nd message)
Here is also a workaround for the second integral
In[36]:=
res2 = Integrate[int2, {z, 0, p - e}, Assumptions -> 0 < e < p < 1,
GenerateConditions -> False] +
Integrate[int2, {z, p + e, 1}, Assumptions -> 0 < e < p < 1,
GenerateConditions -> False]
In[37]:=
res3 = Limit[res2, e -> 0]
(*Check*)
In[31]:=
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 -> 3/4, {z, 3/4 +
10^(-7), 1}]
Out[34]=
-3.350006329606476