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

$$\int_0^1t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}e^{yt}~dt $$

can be expressed as


I know this is 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]

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] 

Note that $|x|<1,|y|<1$, and $\text{Re}(c)>\text{Re}(a)>0$.

Is this equation true or not?

Edit 1

According to Wolfram Alpha

$$\lim\limits_{k\rightarrow \infty} F_1[3/2,1,k;2;x,y/k] =\frac{2\left(\frac{1}{\sqrt{1-x}}-1\right)}{x}$$

such that $\lim\limits_{k\rightarrow \infty}F_1[3/2,1,k;2;0.4,\text{any/k}]=1.4549$.

So I think

$$\int_0^1t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}e^{yt}~dt \neq \dfrac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\lim\limits_{k\to\infty}F_1\left(a,b,k,c;x,\dfrac{y}{k}\right).$$

What do you say?

Edit 2

It has been solved here.

  • $\begingroup$ @hardmath Sorry about that. I changed the text. The software was Mathematica. $\endgroup$ May 2, 2018 at 14:19
  • $\begingroup$ @hardmath The equation $$\int_0^1t^{a-1}(1-t)^{c-a-1}(1-xt)^{-b}e^{yt}~dt=\dfrac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\lim\limits_{k\to\infty}F_1\left(a,b,k,c;x,\dfrac{y}{k}\right).$$ is true for $|x|<1,|y|<1$, and $\text{Re}(c)>\text{Re}(a)>0$, right? Then why are the numerical results different? This is my question. $\endgroup$ May 2, 2018 at 14:38
  • 1
    $\begingroup$ To distinguish between a math issue and a software issue, I suggest evaluating these expressions with a different CAS. The Mathematica implementation of AppellF1 might, for example, be normalized in a different way than you expect, and so in such cases we cannot explain the error by reasoned mathematical argument. On the other hand it might be an issue related to numerical quadrature or taking limits that does admit a math explanation. $\endgroup$
    – hardmath
    May 2, 2018 at 14:48
  • 1
    $\begingroup$ Nice work finding the discrepancy at Mathematica.SE. Would you still be interested in a comparison of left- and right-hand sides using a different CAS? $\endgroup$
    – hardmath
    May 2, 2018 at 20:31
  • 1
    $\begingroup$ I will post an Answer that shows how to use SageMath's implementation of hypergeometric functions for the comparison. SageMath is free open source software. $\endgroup$
    – hardmath
    May 3, 2018 at 6:00


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