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numerical integration - Error in integral representation of Appell's series


It was originally asked here.


According to About the confluent versions of Appell Hypergeometric Function and Lauricella Functions the integral


∫10ta−1(1−t)c−a−1(1−xt)−beyt dt


can be expressed as


∫10ta−1(1−t)c−a−1(1−xt)−beyt dt=Γ(a)Γ(c−a)Γ(c)limk→∞F1(a,b,k,c;x,yk).


It look like true, but gives numeric error. Look that, calculated by Mathematica:


F[a_?NumericQ, b_?NumericQ, c_?NumericQ, x_?NumericQ, y_?NumericQ] := 
 NIntegrate[t^(a - 1) (1 - t)^(c - a - 1) (1 - x t)^(-b) Exp[ y t], {t, 0, 1}]


N[F[3/2, 1, 2, .4, .3], 20]
{2.8964403550198865`}

G[a_?NumericQ, b_?NumericQ, c_?NumericQ, x_?NumericQ, y_?NumericQ] := 
 Gamma[a] Gamma[c - a]/Gamma[c] Limit[AppellF1[a, b, k, c, x, y/k],k -> Infinity]

N[G[3/2, 1, 2, .4, .3], 20] 
{2.2854650559595466`}


where I was careful to ensure that |x|<1,|y|<1, and Re(c)>Re(a)>0, which are the condition of F1.Note that the results are different. Furthermore, according to Wolfram Alpha


limk→∞F1[3/2,1,k;2;0.4,any/k]=1.4549


So, is It a math issue or Mathematica issue?



Answer



There is a bug in the calculation of the large k limit of AppellF1. This is easy to illustrate:


Needs["NumericalCalculus`"]
wrongLimit = Limit[AppellF1[3/2, 1, k, 2, 4/10, (3/10)/k], k -> Infinity];
correctLimit = NLimit[AppellF1[3/2, 1, k, 2, 4/10, (3/10)/k], k -> Infinity];
Plot[{AppellF1[3/2, 1, k, 2, 4/10, (3/10)/k], wrongLimit, correctLimit}, {k, .5, 5}
, PlotStyle -> {Blue, Directive[Red, Dashed], Directive[Green, Dashed]}, PlotRange -> {Automatic, {0, 3}}]


enter image description here


Perhaps the bug should be reported. In the meantime, you can use NLimit to get the correct limit. If you use NLimit in your definition of G, you get results that consistent with those of F (up to numerical inaccuracy).


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