Re[2]: Re: Simplification of definite integral?
- To: mathgroup at smc.vnet.net
- Subject: [mg40739] Re[2]: [mg40701] Re: Simplification of definite integral?
- From: Vladimir Bondarenko <vvb at mail.strace.net>
- Date: Tue, 15 Apr 2003 04:00:34 -0400 (EDT)
- References: <001801c30305$36b29960$53813e44@Dell>
- Reply-to: Vladimir Bondarenko <vvb at mail.strace.net>
- Sender: owner-wri-mathgroup at wolfram.com
"David W. Cantrell" <DWCantrell at sigmaxi.org> wrote on Tuesday, April 15, 2003, 1:12:22 AM:
DWC> any direct way to get Mathematica to give a correct answer for
DWC> this integral. Suggestions anyone?
This bug boils down to a fundamental bug the long-liver I wrote to WRI
about in 2001
N[Integrate[Sin[z]/(z^2 - 1), {z, 2, Infinity}]]
-0.797165 - 1.32178 I
which exists in all versions since Mathematica Windows 387 2.2 (April 9, 1993)
to Mathematica 4.2 for Microsoft Windows (August 23, 2002). A decennial is
coming soon.
However, there is life there is hope ;) I was told that it seems Wolfram
is going to release Mathematica 5 in the near future; maybe they would
fix the mean thing... I am resending it to support at wolfram.com one more
time, just in case.
As for your request, my solution is (please observe Re which means that
it seems there is no straightforward way to solve your task)
int = Integrate[Sin[z - 1]/(z - 1) Sin[z + 1]/(z + 1), {z, 2, a}];
<< Calculus`Limit`
ans = FullSimplify[Re[Limit[int, a -> Infinity]]]
(Cos[2]*(CosIntegral[2] - CosIntegral[6] + Log[3]) + Sin[2]*(Pi + SinIntegral[-2] - SinIntegral[6]))/4
N[ans]
-0.140037
NIntegrate[Sin[z - 1]/(z - 1) Sin[z + 1]/(z + 1), {z, 2, Infinity}, Method -> QuasiMonteCarlo]
Sum[NIntegrate[Sin[z - 1]/(z - 1) Sin[z + 1]/(z + 1), {z, 0 + n, 1 + n}], {n, 2, 99999}]
-0.140004
-0.140035
Best wishes,
Vladimir Bondarenko
Mathematical and Production Director
Symbolic Testing Group
Web : http://www.CAS-testing.org/ GEMM Project (95% ready)
Email: vvb at mail.strace.net
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