Re: Numerical integration
- To: mathgroup at smc.vnet.net
- Subject: [mg78822] Re: Numerical integration
- From: dh <dh at metrohm.ch>
- Date: Tue, 10 Jul 2007 06:35:57 -0400 (EDT)
- References: <f6sh65$7ob$1@smc.vnet.net>
Hi Jose,
your integral can be looked at as a contour integral on a circle around
-1 with radius 1 of the following function:
Exp[I t](Exp[I t] - 1)Sec[Exp[I t] - 1] dt == z Sec[z] /I dz
by resdiue theory the integral is 2 Pi I time the residue at the pole at
z=-Pi/2 what gives Pi^2 = 9.8696.
If I compute:
N[Integrate[h[t], {t, 0, 2Pi}]] in version 5.1 I get the wrong result of
-3.58, on the other hand:
N[Integrate[h[t], {t, 0, 2Pi}]] in version6 I get the correct result.
Therefore, you seesm to have found an bug that has been corrected in
version 6.
hope this helps, Daniel
José Carlos Santos wrote:
> Hi all:
>
> I detected a problem concerning numerical integration under Mathematica
> 5.1. Consider this function:
>
> h[t_] := Exp[I t](Exp[I t] - 1)Sec[Exp[I t] - 1]
>
> If I compute:
>
> N[Integrate[h[t], {t, 0, 2Pi}]]
>
> I get -3.58642. But if I compute
>
> N[Integrate[Re[h[t]], {t, 0, 2Pi}] + Integrate[Im[h[t]], {t, 0, 2Pi}]I]
>
> instead of getting the same answer, I get -9.869604401106042 +
> 2.2377264147293624^(-16)I. Of course, the imaginary part doesn't bother
> me, but why are the results so different? BTW the difference is equal
> to 2*pi.
>
> Best regards,
>
> Jose Carlos Santos
>