why do these two Meijer G functions not cancel each other?

Question

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?

Answer 1

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.

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

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.