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MathGroup Archive 2007

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Re: Re: Definite Integration in Mathematica 2

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74491] Re: [mg74469] Re: Definite Integration in Mathematica 2
  • From: Daniel Lichtblau <danl at wolfram.com>
  • Date: Fri, 23 Mar 2007 19:03:19 -0500 (EST)
  • References: <etqo3f$10i$1@smc.vnet.net> <200703220615.BAA17037@smc.vnet.net>

dimitris wrote:
> And here is one case that evaluating first the indefinite integral and
> then
> apply the Newton-Leibniz formula is preferable than trust Definite
> Integration
> capabillities of Mathematica

Only if you subscribe to the dictum "two wrongs make a right". More below.


> Integrate[(2 - Sin[x])^(1/4), {x, 1, 4}]
> N[%]
> Integrate[(2 - Sin[x])^(1/4), x]
> Simplify[Limit[%, x -> 4, Direction -> 1] - Limit[%, x -> 1, Direction
> -> -1]]
> Chop[N[%]]
> NIntegrate[(2 - Sin[x])^(1/4), {x, 1, Pi/2, 4}]
> 
> -((1/
> (5*Sqrt[3]*Gamma[7/4]))*(2*I*(5*Sqrt[Pi]*Gamma[5/4]*Hypergeometric2F1[5/4,
> 1/2, 7/4, 1/3] -
>      2*Gamma[7/4]*(AppellF1[5/4, 1/2, 1/2, 9/4, (1/3)*(2 - Sin[1]), 2
> - Sin[1]]*(2 - Sin[1])^(5/4) +
>        AppellF1[5/4, 1/2, 1/2, 9/4, (1/3)*(2 - Sin[4]), 2 - Sin[4]]*(2
> - Sin[4])^(5/4)))))
> 3=2E3356211372370748 - 2.324008709185599*I
> (-(4/5))*AppellF1[5/4, 1/2, 1/2, 9/4, (1/3)*(2 - Sin[x]), 2 -
> Sin[x]]*Sec[x]*Sqrt[1 + (1/3)*(-2 + Sin[x])]*(2 - Sin[x])^(5/4)*
>   Sqrt[-1 + Sin[x]]
> (4*I*(AppellF1[5/4, 1/2, 1/2, 9/4, (1/3)*(2 - Sin[1]), 2 - Sin[1]]*(2
> - Sin[1])^(5/4) +
>     AppellF1[5/4, 1/2, 1/2, 9/4, (1/3)*(2 - Sin[4]), 2 - Sin[4]]*(2 -
> Sin[4])^(5/4)))/(5*Sqrt[3])
> 3=2E3356211372370748
> 3=2E3356211372353624
> 
> Interestingly
> 
> Drop[Expand[Integrate[(2 - Sin[x])^(1/4), {x, 1, 4}]], 1]
> Chop[N[%]]
> 
> (8*I*AppellF1[5/4, 1/2, 1/2, 9/4, (1/3)*(2 - Sin[1]), 2 - Sin[1]]*(2 -
> Sin[1])^(1/4))/(5*Sqrt[3]) -
>   (4*I*AppellF1[5/4, 1/2, 1/2, 9/4, (1/3)*(2 - Sin[1]), 2 - Sin[1]]*(2
> - Sin[1])^(1/4)*Sin[1])/(5*Sqrt[3]) +
>   (8*I*AppellF1[5/4, 1/2, 1/2, 9/4, (1/3)*(2 - Sin[4]), 2 - Sin[4]]*(2
> - Sin[4])^(1/4))/(5*Sqrt[3]) -
>   (4*I*AppellF1[5/4, 1/2, 1/2, 9/4, (1/3)*(2 - Sin[4]), 2 - Sin[4]]*(2
> - Sin[4])^(1/4)*Sin[4])/(5*Sqrt[3])
> 3=2E3356211372370743
> 
> (i.e. ommiting the whole Hypergeometric term)
> 
> Dimitris
> [...]

The antiderivative appears to have two singularities on the integration 
path, one at Pi/2 and one just below 4. I would surmise that latter is 
some transcendental, and Mathematica fails to find it. So the definite 
integral misses a jump and is incorrect.

The blind application of Newton-Leibniz simply misses both, the jumps 
happen to cancel, and the result is "correct". But not the method.


Daniel Lichtblau
Wolfram Research




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