4
$\begingroup$

I'm trying to wrap my mind around computation involving the Meijer $G$ function, as defined here. (Edit: I'm actually using a somewhat mixed notation using the definitions from MathWorld and the Wikipedia page, but you get the idea.) $$G^{m,n}_{p,q}\left(\begin{array}{c|c}\begin{matrix}a_1,\ldots,a_p\\b_1,\ldots,b_q\end{matrix}&z\end{array}\right)=\frac{1}{2\pi i}\int_L \frac{\prod\limits_{j=1}^m \Gamma(b_j-s)\prod\limits_{j=1}^n\Gamma(1-a_j+s)}{\prod\limits_{j=n+1}^p \Gamma(a_j-s)\prod\limits_{j=m+1}^q\Gamma(1-b_j+s)}z^s\,ds$$ For the sake of practice, I'm interested in establishing identity $(5)$: $$G^{1,2}_{2,2}\left(\begin{array}{c|c}\begin{matrix}1,1\\1,1\end{matrix}&z\end{array}\right)=\frac{z}{z+1}$$ From the definition, I'm getting $$\begin{align*}G^{1,2}_{2,2}\left(\begin{array}{c|c}\begin{matrix}1,1\\1,1\end{matrix}&z\end{array}\right)&=\frac{1}{2\pi i}\int_L \frac{\prod\limits_{j=1}^1 \Gamma(b_j-s)\prod\limits_{j=1}^2\Gamma(1-a_j+s)}{\prod\limits_{j=3}^2 \Gamma(a_j-s)\prod\limits_{j=2}^2\Gamma(1-b_j+s)}z^s\,ds\\[2ex] &=\frac{1}{2\pi i}\int_L \frac{\Gamma(1-s)\Gamma(s)}{\prod\limits_{j=3}^2 \Gamma(a_j-s)}z^s\,ds \end{align*}$$ and here's where I'm stuck. How would I approach the product in the denominator?

Edit 2: Since the lower index is larger than the upper index, do I just consider it an empty product and take the term in the denominator to be equal to $1$?

Edit 3: Now that I know how to proceed, here's my attempt at establishing the identity (though I don't use any hypergeometric functions; I'm trying my hand at the contour integral). $$\begin{align*}G^{1,2}_{2,2}\left(\begin{array}{c|c}\begin{matrix}1,1\\1,1\end{matrix}&z\end{array}\right) &=\frac{1}{2\pi i}\int_L\Gamma(1-s)\Gamma(s)z^s\,ds\\[2ex] &=\sum_{k=1}^\infty\underset{s_0=k}{\text{Res}}\,\Gamma(1-s)\Gamma(s)z^s\\[2ex] &=\sum_{k=1}^\infty\lim_{s\to k}(s-k)\Gamma(1-s)\Gamma(s)z^s\\[2ex] &=\sum_{k=1}^\infty\left(\lim_{s\to k}(s-k)\Gamma(1-s)\right)\Gamma(k)z^k\\[2ex] &=\sum_{k=1}^\infty\left(\lim_{s\to k}\frac{s-k}{1-s}\Gamma(2-s)\right)\Gamma(k)z^k\\[2ex] &=\sum_{k=1}^\infty\left(\lim_{s\to k}\frac{s-k}{(1-s)(2-s)}\Gamma(3-s)\right)\Gamma(k)z^k\\[2ex] &\quad\vdots\\[2ex] &=\sum_{k=1}^\infty\left(\lim_{s\to k}\frac{s-k}{\prod\limits_{n=1}^k(n-s)}\right)\Gamma(k)z^k\\[2ex] &=\sum_{k=1}^\infty \left(\lim_{s\to k}\frac{(-1)^{k}}{\prod\limits_{n=1}^{k-1}(s-n)}\right)\Gamma(k)z^k\\[2ex] &=\sum_{k=1}^\infty \left(\lim_{s\to k}\frac{\Gamma(s-(k-1))}{\Gamma(s)}\right)(-z)^k\Gamma(k)\\[2ex] &=\sum_{k=1}^\infty (-z)^k\\[2ex] &=-\frac{z}{1+z} \end{align*}$$ (where $L$ is a contour that's surrounding the poles of $\Gamma(1-s)$, i.e. $s=k$ for integers $k\ge1$.) Unfortunately I seem to be off by a sign.

$\endgroup$
2
+50
$\begingroup$

Regarding the empty product you're right. In case the lower index of a product is greater than the upper index, the product is equal to one (analogously to the empty sum which is zero in such cases).

We can read e.g. in H. Bateman, Higher Transcendental Function Volume 1, section 5.3, p.207

  • Definition of $G$-Function: ... where an empty product is interpreted as $1$ ...

So, the expression reads as \begin{align*}G^{1,2}_{2,2}\left(\begin{array}{c|c}\begin{matrix}1,1\\1,1\end{matrix}&z\end{array}\right) &=\frac{1}{2\pi i}\int_L \Gamma(1-s)\Gamma(s)z^s\,ds \end{align*}

If we use the representation based upon the hypergeometric series

\begin{align*} G^{m,n}_{p,q}\left(\begin{array}{c|c}\begin{matrix}a_1,\ldots,a_p\\b_1,\ldots,b_q\end{matrix}&z\end{array}\right)&= \sum_{h=1}^m \frac{\displaystyle{\prod_{j=1}^{m}}\Gamma(b_j-b_h)^\ast\prod_{j=1}^{n}\Gamma(1+b_h-a_j)}{\displaystyle{\prod_{j=m+1}^{q}}\Gamma(1+b_h-b_j)\prod_{j=n+1}^{p}\Gamma(a_j-b_h)}z^{b_{h}}\\ &\qquad\qquad\times {}_{p}F_{q-1}\left(\begin{array}{c|c} \begin{matrix} 1+b_h-\mathbb{a_p}\\ (1+b_h-\mathbb{b_q})^\ast \end{matrix}&(-1)^{p-m-n}z\end{array}\right) \end{align*} we obtain \begin{align*} G^{1,2}_{2,2}\left(\begin{array}{c|c}\begin{matrix}1,1\\1,1\end{matrix}&z\end{array}\right)&= \sum_{h=1}^1 \frac{\displaystyle{\prod_{j=1}^{1}}\Gamma(b_j-b_1)^\ast\prod_{j=1}^{2}\Gamma(1+b_1-a_j)}{\displaystyle{\prod_{j=2}^{2}}\Gamma(1+b_1-b_j)\prod_{j=3}^{2}\Gamma(a_j-b_1)}z^{b_{1}}\\ &\qquad\qquad\times {}_{2}F_{1}\left(\begin{array}{c|c} \begin{matrix} 1+b_1-\mathbb{a_2}\\ (1+b_1-\mathbb{b_2})^\ast \end{matrix}&-z\end{array}\right)\\ &=z\times{}_{2}F_{1}\left(\begin{array}{c|c} \begin{matrix} 1,1\\ 1 \end{matrix}&-z\end{array}\right)\\ &=\frac{z}{1+z} \end{align*}

$\endgroup$
  • $\begingroup$ Thanks for the answer, now I'm looking into some info about the hypergeometric function. Quick note though: the denominator in the final result should be $1+z$, right? $\endgroup$ – user170231 Aug 17 '15 at 19:58
  • $\begingroup$ @user170231: Typo corrected. Thanks for the hint and for accepting my answer. I suggest to start with $A = B$ by H.S. Wilf. It's fun to read and I'm pretty sure you will like it! :-) $\endgroup$ – Markus Scheuer Aug 17 '15 at 20:15
  • $\begingroup$ @user170231: Thanks for granting the bounty! $\endgroup$ – Markus Scheuer Aug 18 '15 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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