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MathGroup Archive 2007

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Definite Integration vs Newton-Leibniz formula

  • To: mathgroup at smc.vnet.net
  • Subject: [mg74314] Definite Integration vs Newton-Leibniz formula
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Sun, 18 Mar 2007 00:48:32 -0500 (EST)

Consider the following function

f[x_] = Sin[x^3]Log[x]

Then

Integrate[f[x], {x, 0, 1}]
{% // N, NIntegrate[f[x], {x, 0, 1}]}

(-(1/16))*HypergeometricPFQ[{2/3, 2/3}, {3/2, 5/3, 5/3}, -(1/4)]
{-0.060865478966467726, -0.06086547896286495}

Let evaluate first the indefinite integral and then apply the Newton-
Leibniz formula.

F[x_] = Integrate[Sin[x^3]Log[x], x]

(1/(6*x^2))*(-3*I*x^3*HypergeometricPFQ[{1/3, 1/3}, {4/3, 4/3}, (-
I)*x^3] +
   3*I*x^3*HypergeometricPFQ[{1/3, 1/3}, {4/3, 4/3}, I*x^3] +
   (3*(((-I)*x^3)^(2/3) + (I*x^3)^(2/3))*Gamma[4/3] - ((-
I)*x^3)^(2/3)*Gamma[1/3, (-I)*x^3] - (I*x^3)^(2/3)*Gamma[1/3,
I*x^3])*Log[x])

Limit[F[x], x -> 1, Direction -> 1] - Limit[F[x], x -> 0, Direction ->
-1]
(1/2)*I*(-HypergeometricPFQ[{1/3, 1/3}, {4/3, 4/3}, -I] +
HypergeometricPFQ[{1/3, 1/3}, {4/3, 4/3}, I])

Of course

%//N//Chop
-0.060865478966467726

as it should be.

But why the definite integral return a different value?

I believe that integrals like the above are evaluated by first
determination of the antiderivative and then evaluation
at endpoints.

Using a code for Limit adopted from a post Daniel Lichtblau, in order
to see what limits are evaluated
by Integrate I got more confused.

Unprotect[Limit];
Limit[a___] := Null /; (Print[InputForm[limit[a]]]; False)

Integrate[f[x],{x,0,1}]

limit[0, x -> 0, Direction -> -1, Assumptions -> True]
limit[(-1 + x)^2*(3*Cos[1] - Sin[1]/2) + (-1 + x)*Sin[1], x -> 1,
Direction -> 1,
    Assumptions -> True]
(-(1/16))*HypergeometricPFQ[{2/3, 2/3}, {3/2, 5/3, 5/3}, -(1/4)]

Can somebody give me insight what is going?

Thanks a lot
Dimitris



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