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I'm trying to numerically calculate the following Fox-H integral

$$ 1-e^{-100}\sqrt{\frac{x}{\pi}} \left(\frac{1}{2\pi j}\right)^2\\ \oint_{\mathcal{L}_1}\oint_{\mathcal{L}_2} \frac{\Gamma(z_1)\Gamma(z_2)\Gamma(1/2-z_1-z_2)\Gamma(1-z_1-z_2)}{\Gamma(1/2-z_1)\Gamma(3/2-z_1-z_2)}\left(-100 x\right)^{-z_1}x^{-z_2}dz_1 dz_2, $$

for $x$ going from $10^{-7}$ to about $1000$, where the contour $\mathcal{L}_1$ separates the poles of $\Gamma(z_1)$ from the poles of $\Gamma(1/2-z_1-z_2)\Gamma(1-z_1-z_2)$, and the contour $\mathcal{L}_2$ separates the poles of $\Gamma(z_2)$ from the poles of $\Gamma(1/2-z_1-z_2)\Gamma(1-z_1-z_2)$. Three types of contour are possible:

contour1 contour2 contour3

Reference for the images: https://dlmf.nist.gov/16.17

I'm using the NIntegrate function, and I tried all possible contours, but for values of $x$ above some threshold, like $1086.66$, the Mathematica returns the Catastrophic error. I already tried to change the Working Precision but have no better results. From the theory, this integral should be fine, resulting values from $0$ to $1$.

Do you guys have some clues about how can I solve this problem? The code I'm using is:

1 - Re[NIntegrate[
      Exp[-100] \[Sqrt](x/\[Pi]) (1/(2 \[Pi] I))^2
      (
         (Gamma[z1] Gamma[z2] Gamma[1/2 - z1 - z2] Gamma[1 - z1 - z2])/
         (Gamma[1/2 - z1] Gamma[3/2 - z1 - z2])
      ) (-100 x)^-z1 x^-z2,
      {z2, 0.25 - I 20, 0.25 + I 20},
      {z1, -20 - 5 I, 0.125 - 5 I, 0.125 + 5 I, -20 + 5 I}
]]

Edit: There were some typo errors in the Mathematica code, elegantly appointed in the comments. Edit 2: Update the values to achieve the error.

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    $\begingroup$ Why are you using inexact numbers like 0.0025 instead of exact ones like 1/400? $\endgroup$ Jun 30, 2022 at 14:47
  • $\begingroup$ Sorry guys, I tried to clean the code at max to not distract your attention, but I made a mess. The $0.0025x$ term should be $-0.0025x$. @J.M.'sslightlylessbusy, I used these numbers aiming to easy your analysis. Sorry. $\endgroup$ Jun 30, 2022 at 15:23
  • $\begingroup$ I still get only a NIntegrate::slwcon warning no loss of precision error for x = 50, 100, 150, 500, 1000. I'm using V13.0.1, if that makes a difference. $\endgroup$
    – Michael E2
    Jun 30, 2022 at 15:59
  • $\begingroup$ Please, forgive my bad attention. The error occurs when the $0.0025$ is changed to $100$, and for the value of $x=1086.66$. I will update the main question to these new values. $\endgroup$ Jun 30, 2022 at 23:12

1 Answer 1

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I'm not particularly familiar with this integral, but with the factor of Exp[-100], it's pretty small. I changed the 1 -... to 0 -... so we can see the value of the integral:

Block[{x = 108666/100},
 0 - Re[NIntegrate[
    Exp[-100] \[Sqrt](x/\[Pi]) (1/(2 \[Pi] I))^2 ((Gamma[z1] Gamma[
          z2] Gamma[1/2 - z1 - z2] Gamma[1 - z1 - z2])/(Gamma[
          1/2 - z1] Gamma[3/2 - z1 - z2])) (-1/400 x)^-z1 x^-z2,
    {z2, 1/4 - I 20, 1/4 + I 20}, {z1, -20 - 5 I, 1/8 - 5 I, 
     1/8 + 5 I, -20 + 5 I}]]
 ]

NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.

(*  -3.65721*10^-44  *)

Of course the main difference is using exact numbers for the integration path and x. (See @J.M.'s comment.)

The integration takes a long time, maybe more than a minute or minutes, certainly long enough to be worth a warning. I did something else for a while.

At machine precision, 1 minus the integral equals 1. (exactly). Without the Exp[-100] factor, the integral is a little more than -1.

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