Re: two integrals
- To: mathgroup at smc.vnet.net
- Subject: [mg79152] Re: two integrals
- From: dimitris <dimmechan at yahoo.com>
- Date: Fri, 20 Jul 2007 03:15:48 -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.
(Version 5.2 is used)
Thanks to help from Ajit (I also thank Daniel from WRI for
his response but unfortunately part of his solution needs
the symbolic capabilities of version 6...) I suceeded in getting
a closed form solution for the first integral
In[8]:=
Quit[]
The key is to wrap the integrand to TrigToExp
In[1]:=
int = TrigToExp[ArcTan[(1 - z^2)/z^3]*(1/(z + p))]
Out[1]=
(I*Log[1 - (I*(1 - z^2))/z^3])/(2*(p + z)) - (I*Log[1 + (I*(1 - z^2))/
z^3])/(2*(p + z))
In[2]:=
Timing[res = Integrate[int, {z, 0, 1}, Assumptions -> 0 < p < 1]]
Out[2]=
{64.968*Second, (1/2)*I*(2*Pi^2 + I*Pi*Log[p] - Log[1 - Root[1 -
2*#1^2 + #1^4 + #1^6 & , 1]]*
Log[1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 1])] + Log[-Root[1 -
2*#1^2 + #1^4 + #1^6 & , 1]]*
Log[1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 1])] +
I*Pi*Log[Root[1 - 2*#1^2 + #1^4 + #1^6 & , 2]] +
I*Pi*Log[-(1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 2]))] -
Log[Root[1 - 2*#1^2 + #1^4 + #1^6 & , 2]]*
Log[-(1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 2]))] + Log[1 -
Root[1 - 2*#1^2 + #1^4 + #1^6 & , 2]]*
Log[1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 2])] + Log[1 -
Root[1 - 2*#1^2 + #1^4 + #1^6 & , 3]]*
Log[1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 3])] - Log[-Root[1 -
2*#1^2 + #1^4 + #1^6 & , 3]]*
Log[1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 3])] + I*Pi*Log[-1 +
Root[1 - 2*#1^2 + #1^4 + #1^6 & , 4]] -
I*Pi*Log[Root[1 - 2*#1^2 + #1^4 + #1^6 & , 4]] - Log[-1 + Root[1 -
2*#1^2 + #1^4 + #1^6 & , 4]]*
Log[-(1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 4]))] + Log[Root[1
- 2*#1^2 + #1^4 + #1^6 & , 4]]*
Log[-(1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 4]))] - Log[1 -
Root[1 - 2*#1^2 + #1^4 + #1^6 & , 5]]*
Log[1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 5])] + Log[-Root[1 -
2*#1^2 + #1^4 + #1^6 & , 5]]*
Log[1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 5])] +
I*Pi*Log[Root[1 - 2*#1^2 + #1^4 + #1^6 & , 6]] +
I*Pi*Log[-(1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 6]))] -
Log[Root[1 - 2*#1^2 + #1^4 + #1^6 & , 6]]*
Log[-(1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 6]))] + Log[1 -
Root[1 - 2*#1^2 + #1^4 + #1^6 & , 6]]*
Log[1/(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 6])] - PolyLog[2,
(-1 + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 1])/
(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 1])] + PolyLog[2, Root[1
- 2*#1^2 + #1^4 + #1^6 & , 1]/
(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 1])] + PolyLog[2, (-1 +
Root[1 - 2*#1^2 + #1^4 + #1^6 & , 2])/
(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 2])] - PolyLog[2, Root[1
- 2*#1^2 + #1^4 + #1^6 & , 2]/
(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 2])] + PolyLog[2, (-1 +
Root[1 - 2*#1^2 + #1^4 + #1^6 & , 3])/
(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 3])] - PolyLog[2, Root[1
- 2*#1^2 + #1^4 + #1^6 & , 3]/
(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 3])] - PolyLog[2, (-1 +
Root[1 - 2*#1^2 + #1^4 + #1^6 & , 4])/
(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 4])] + PolyLog[2, Root[1
- 2*#1^2 + #1^4 + #1^6 & , 4]/
(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 4])] - PolyLog[2, (-1 +
Root[1 - 2*#1^2 + #1^4 + #1^6 & , 5])/
(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 5])] + PolyLog[2, Root[1
- 2*#1^2 + #1^4 + #1^6 & , 5]/
(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 5])] + PolyLog[2, (-1 +
Root[1 - 2*#1^2 + #1^4 + #1^6 & , 6])/
(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 6])] - PolyLog[2, Root[1
- 2*#1^2 + #1^4 + #1^6 & , 6]/
(p + Root[1 - 2*#1^2 + #1^4 + #1^6 & , 6])])}
As we see the timing looks quite reasonable.
A chack now
In[3]:=
{(N[#1, 30] & )[res /. p -> 5/7], NIntegrate[ArcTan[(1 - z^2)/z^3]*(1/
(z + p)) /. p -> 5/7, {z, 0, 1}, WorkingPrecision -> 50,
PrecisionGoal -> 30]}
Out[3]=
{1.094102857371793922707221438080436938524900902338`30.15051499783199
+ 0``30.111456845588528*I,
1.09410285737179392270722143808042886494086929837`30.25861428981342}
Now
In[14]:=
ToRadicals[res]
Out[14]=
(*output ommited*)
Can we simplify Output 14 more?
Thanks
Dimitris