Guidance on reducing this Meijer-G function

According to mathematica, for $\alpha,z>0$ we have \begin{multline} G^{3,1}_{2,3}\left(z\bigg|{0,1\atop 0,0,1-\alpha}\right)=% -\frac{z}{\alpha}\Gamma(1-\alpha){_2F_2}\left({1,1\atop 2,\alpha+1};z\right)\\% \qquad\qquad\qquad-\pi\sin(\pi\alpha)\csc^2(\pi(\alpha-1))\Gamma(1-\alpha)(-z)^\alpha z^{-\alpha}\\% \qquad\qquad\qquad\qquad\qquad\qquad+\pi\sin(\pi\alpha)\csc^2(\pi(\alpha-1))(-z)^\alpha z^{-\alpha}\Gamma(1-\alpha,-z)\\% -\Gamma(1-\alpha)\log z% +\Gamma(1-\alpha)\psi(1-\alpha). \end{multline}

The Wolfram functions website at http://functions.wolfram.com/07.34.03.0987.01 provides a way to write $G^{3,1}_{2,3}()$ as a sum of $_2F_2()$'s but only when the bottom parameters do not differ by integer ammounts. Everywhere else I have looked has also required this restriction. Is there a property of the G function that would allow me to write this G-function such that the bottom parameters do not differ by an integer? How else might I reduce this if not?

It seems like if I were able to get rid of one of the zero's in the bottom argument and assume $\alpha$ is not an integer, then I would be able to apply the relation on the wolfram functions site.

• I believe the answer is essentially the same as here. The integrand of the G-function is a rational combination of gamma functions times $z^y$. The sums of the residues over two sequences of simple poles give two hypergeometric functions. But the double pole at $y=0$ requires taking the derivative of the factor which is regular at zero, producing $\ln z$ and the polygamma function. You cannot write the result in exactly the same form as in the case of simple poles. – Maxim Jun 29 '18 at 19:07
• @Maxim I I tried computing the sum of the residues for $s=1-\alpha+k$ here but am getting zero. What am I missing? The residue of the integrand at $s=0$ seems right – Aaron Hendrickson Jun 29 '18 at 19:29
• Do the computation step by step. You'll see that the residue computation already gives zero before the summation; you need to specify the assumptions on $k$. What you should get is $$\operatorname{Res}_{y=-a-k} \Gamma(a+y) f(y) = \frac {(-1)^k f(-a-k)} {\Gamma(1+k)},$$ where $f$ is the regular factor, and you again have a rational combination of linear gamma functions, the sum of which gives the hypergeometric function. – Maxim Jun 29 '18 at 19:56

As pointed out by @Maxim, this expression is due to the residue series expansion of the $G$ function. Using the residue expansion we write \begin{align} G^{3,1}_{2,3}\left(z\bigg|{0,1\atop 0,0,1-\alpha}\right)% =& \sum_{k=0}^{\infty}\operatorname*{Res}_{s=-k}\left(\frac{\Gamma(s)\Gamma(s+1-\alpha)\Gamma(1-s)}{s}z^{-s}\right)\\% &\quad+\sum_{k=0}^{\infty}\operatorname*{Res}_{s=\alpha-1-k}\left(\frac{\Gamma(s)\Gamma(s+1-\alpha)\Gamma(1-s)}{s}z^{-s}\right). \end{align} As pointed out, the residue at the double pole for $s=0$ yields logarithmic and polygamma terms. The sums of the residues at the poles for $s=-k$ for $k>0$ and $s=\alpha-1-k$ yield hypergeometric and incomplete gamma terms. In total we find
\begin{align} G^{3,1}_{2,3}\left(z\bigg|{0,1\atop 0,0,1-\alpha}\right)=% &-\Gamma(1-\alpha)\biggl(% \log z-\psi(1-\alpha)% +\frac{z}{\alpha}{_2F_2}\left({1,1\atop 1+\alpha,2};z\right)\\% &\quad+\pi\csc(\pi\alpha)(-1)^\alpha\left(1-\frac{\Gamma(1-\alpha,-z)}{\Gamma(1-\alpha)}\right)% \biggr). \end{align} With a bit of algebra one can show that this expression is the same as in the original post above.

• Note that the simplification to $G_{1,2}^{2,1}$ is only formally correct, because, contrary to the definition of the G-function, there isn't an integration contour separating the poles of $\Gamma(-s)$ from the poles of $\Gamma(s)$. – Maxim Jul 1 '18 at 23:30
• Does that mean it is not a valid G function? I was using one of the properties on dlmf to do the reduction. Is this move technically not allowed even though it gave the desired solution? – Aaron Hendrickson Jul 1 '18 at 23:34
• Strictly speaking, it isn't a valid G-function. The computation goes through only because you already know that $y=0$ needs to be included in the sum. Notice that the original $G_{2,3}^{3,1}$ doesn't have that problem, as long as $\alpha \notin \mathbb N \backslash \{1\}$: there is an integration contour such that the poles of $\Gamma(s)^2 \Gamma(1-\alpha+s)$ are to the left and the poles of $\Gamma(1-s)$ are to the right. Then you get the two sums that you've computed. You just don't need the $G_{1,2}^{2,1}$ step. – Maxim Jul 1 '18 at 23:52