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. $\endgroup$ – Dinesh Shankar May 2 '18 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. $\endgroup$ – Dinesh Shankar May 2 '18 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 '18 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 '18 at 20:31
  • $\begingroup$ @hardmath I think it would be cool, but I don't know how to work on other software. $\endgroup$ – Dinesh Shankar May 3 '18 at 2:06

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