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