Re: Bug in Mathematica 6 - Integrate - 56 (Log, regression bug, invalid value)
- From: dimitris <dimmechan@xxxxxxxxx>
- Date: Sat, 21 Jul 2007 12:27:31 -0700
On 21 , 16:37, Vladimir Bondarenko <v...@xxxxxxxxxxxxxxx> wrote:
Our little demo continues... Hello again from the VM machine
which is still ignored by CAS manufacturers.
Maple 11/10/9 calculate this integral correctly via dilogarithms
int(ln(z^2+1)/(z^3+z+10), z= 0..infinity);
What about Wolfram Research?
NIntegrate[Log[z^2 + 1]/(z^3 + z + 10), {z, 0, Infinity}]
0.326011
Compare:
N[Integrate[Log[z^2 + 1]/(z^3 + z + 10), {z, 0, Infinity}]]
(* Mathematica 6.0 returns an invalid answer *)
-0.353148
(* Mathematica 5.2/4.2 return a correct answer *)
0.326011
(* Mathematica 3.0 returns the unevaluated integral *)
Integrate[Log[z^2+1]/(z^3+z+10), {z,0,Infinity}]
(**************************************************************)
Let there be a better Computer Algebra,
Vladimir Bondarenko
VM and GEMM architect
Co-founder, CEO, Mathematical Director
http://www.cybertester.com/ Cyber Tester, LLChttp://maple.bug-list.org/ Maple Bugs Encyclopaediahttp://www.CAS-testing.org/ CAS Testing
Assuming that the antiderivative is the same with version 5.2
one can try
(*version 5.2 is used*)
In[1]:=
SessionTime[]
Simplify[Integrate[Log[z^2 + 1]/(z^3 + z + 10), z]];
FullSimplify[Limit[%, z -> Infinity] - Limit[%, z -> 0]]
N[%, 20]
SessionTime[]
Out[1]=
0.40625`7.060338367482902
Out[3]=
(1/104)*(8*ArcCot[2]^2 - 4*ArcTan[3]^2 + Log[2]^2 - 2*Log[5]^2 -
6*ArcTan[3]*Log[10] + Log[10]^2 +
(1/4)*Pi*(-9*Pi + 8*ArcTan[3116/237] + 6*Log[2000000]) + (4 +
6*I)*PolyLog[2, -(1/2) - I/2] +
(4 - 6*I)*PolyLog[2, -(1/2) + I/2] - 8*PolyLog[2, 1/5 - (2*I)/5] -
8*PolyLog[2, 1/5 + (2*I)/5] +
(4 - 6*I)*PolyLog[2, 3/10 - I/10] + (4 + 6*I)*PolyLog[2, 3/10 + I/
10])
Out[4]=
0.32601065867162713006477846934310415002`20.15051499783199 +
0``20.637283198602702*I
Out[5]=
45.984375`9.114155281611692
D.S.A.
.
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