8
$\begingroup$

I encountered such expressions in Mathematica

MeijerG[{{ }, {1, c + 1/2}}, {{0, c, c, c}, { }}, 1] + 
MeijerG[{{1}, {c + 1/2}},    {{c, c, c},    {0}}, 1]

which in the notation of Wiki is given by the sum of the following Mellin–Barnes integrals $$G_{2,4}^{4,0}\left(\left.\begin{array}{c} 1, c+\frac12\\ c,c,c,0\end{array} \right| 1\right) = \frac{1}{2\pi i} \int_C\frac{\Gamma\left(-s\right)\Gamma\left(c-s\right)^3}{\Gamma\left(1-s\right)\Gamma\left(c+\frac12-s\right)} ds = - \frac{1}{2\pi i} \int_C \frac{ \Gamma\left(c-s\right)^3}{s\Gamma\left(c+\frac12 -s\right)}ds$$ and $$G_{2,4}^{3,1}\left(\left.\begin{array}{c} 1, c+\frac12\\ c,c,c,0\end{array} \right| 1\right) = \frac{1}{2\pi i} \int_C \frac{\Gamma\left(s\right) \Gamma\left(c-s\right)^3}{\Gamma\left(1+s \right)\Gamma\left(c+\frac12 -s\right)}ds = \frac{1}{2\pi i} \int_C \frac{ \Gamma\left(c-s\right)^3}{s\Gamma\left(c+\frac12 -s\right)}ds$$ and both contours $C$ should be chosen to be the one beginning and ending on $+\infty$. Therefore the two Meijer G functions should be exactly opposite to each other and the sum is identically zero, right?

However, Mathematica yields very different result, by which I mean numerical evaluation of the function with some value of $c$ plugged in. I am wondering what is causing the problem?

$\endgroup$
0

1 Answer 1

14
$\begingroup$

Actually, the contours $C$ in the two integrals are different.

By definition, $C$ goes from $+\infty$ through an clockwise path return to $+\infty$, encircling all poles of $\prod_{j=1}^m\Gamma(b_j-s)$ (each pole exactly once) and none pole of $\prod_{j=1}^n\Gamma(1-a_j-s)$. In this case, the integral-contours of $\textrm{G}^{4, 0}_{2, 4}\Big({1,c+\frac12\atop c,c,c,1}\Big|1\Big)$ encircles poles:

$$c, c+1, c+2,\dotsc, c+k,\dotsc,$$

and poles:

$$0, -1, -2, \dotsc, -k, \dotsc,$$

must be outside the contours. Similar for $\textrm{G}^{3, 1}_{2, 4}\Big({1,c+\frac12\atop c,c,c,1}\Big|1\Big)$, we needs the contour encircles poles:

$$0, 1, 2, \dotsc, k,\dotsc,\quad \mathrm{and}\quad c, c+1,c+2,\dotsc,c+k,\dotsc,$$

As you have found, $1,2,\dotsc,n,\dotsc$ are actually normal points of the integrand, so the contour can "deform" over these points. However, $0$ remains to be a pole, so the contour can not cross over it, which makes the expression obtain a non-zero value.

enter image description here

Thus, for any $c\neq0,-1,-2,\dotsc,-k,\dotsc$ (by definition, they're not allowed), the sum of this two Meijer functions is:

$$-\operatorname{Res}\frac{[\Gamma(c-s)]^3}{s\,\Gamma\big(c+\frac12-s\big)}\Bigg|_{s=0} =-\dfrac{[\Gamma(c)]^3}{\Gamma\big(\frac12+c\big)}.$$

Using Mathematica, we can check our answer:

Table[-MeijerG[{{},{1,c+1/2}},{{0,c,c,c},{}},1,-1]
    -MeijerG[{{1},{c+1/2}},{{c,c,c},{0}},1,-1]
    +Gamma[c]^3/Gamma[1/2+c]//N, 
    {c, SetPrecision[RandomReal[{1,2}, 10]
        +I RandomReal[{1, 2}, 10], 10]}
]//Column

Results Notice that in Mathematica

-MeijerG[{{__}, {__}}, {{__}, {__}}, _, -1]

is corresponding to the definition in Wikipedia(Bateman & Erdelyi).

Edit You may notice that when $c=-k-\frac12$, $s=0$ will also be a normal point. Fortunately, the residue here as we have shown will be zero as we expected.

$\endgroup$
4
  • 3
    $\begingroup$ Just to add to this excellent answer, there is an undocumented function for visualizing the poles of the gamma function ratios that occur in the defining Mellin-Barnes integral: With[{c = 1}, System`MeijerGDump`MeijerGPolePattern[MeijerG[{{}, {1, c + 1/2}}, {{0, c, c, c}, {}}, 1]]]. Also of interest is System`MeijerGDump`MeijerGInfo[MeijerG[{{}, {1, c + 1/2}}, {{0, c, c, c}, {}}, 1]]. $\endgroup$ Jan 17, 2021 at 13:18
  • 1
    $\begingroup$ @J.M., thank you! These functions are really useful. Of course, one also has to be careful about the difference in definitions of the Meijer G function, as pointed out by Chromo Runge. I was going to write such functions myself. Can't believe Mathematica doesn't put such useful functions in the help document. How does one know the existence of such undocumented function? $\endgroup$
    – Jing
    Jan 17, 2021 at 14:05
  • 2
    $\begingroup$ @Jing, just try something like ?? *`*Meijer*. And use GeneralUtilities`PrintDefinitions to get the definitions of symbols. $\endgroup$
    – RungeC
    Jan 17, 2021 at 14:43
  • 2
    $\begingroup$ RawBoxes[ToBoxes[Information["*`*Meijer*"]] /. {ButtonBox[a___] :> ButtonBox[a, ButtonFunction :> (GeneralUtilities`PrintDefinitions[Symbol[#[[2, 2]] <> #[[2, 1]]]] &)]}] may help you, as well. $\endgroup$
    – RungeC
    Jan 17, 2021 at 15:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.