Derivative of the Meijer G-function with respect to one of its parameters

Question

Are there any approaches that allow to find a derivative of the Meijer G-function with respect to one of its parameters in a closed form (or at least numerically with a high precision and in reasonable time, with all found digits provably correct)? I am particularly interested in this case: $$\mathcal{D}=\left.\partial_\alpha G_{2,3}^{2,1}\left(1\middle|\begin{array}c1,\alpha\\1,1,0\end{array}\right)\right|_{\alpha=1}$$

Answer 1

Yes, it is possible in some cases. For example, $$\begin{align}\mathcal{D}&={_2F_2}\left(\begin{array}c1,1\\2,2\end{array}\middle|-1\right)\\&=\gamma-\operatorname{Ei}(-1),\end{align}$$ where ${_pF_q}$ is the generalized hypergeometric function, $\gamma$ is the Euler–Mascheroni constant, and $\operatorname{Ei}(z)$ is the exponential integral. In case you need a numeric value, $$\mathcal{D}\approx0.7965995992970531342836758655425240800732066293468318063837458...$$

Answer 2

To prove the result stated in @Cleo answer, we can use the definition of the Meijer function: \begin{equation} G_{2,3}^{2,1}\left(1\middle|\begin{array}c1,\alpha\\1,1,0\end{array}\right)=\frac{1}{2i\pi}\int_\mathcal L\frac{[\Gamma(1+s)]^2\Gamma(-s)}{\Gamma(\alpha+s)\Gamma(1-s)}\,ds \end{equation} Here (case (i) of the above definition), $\mathcal L$ can be a straight line $(\gamma-i\infty,\gamma+i\infty)$ with $-1<\gamma<0$.

The above expression can be simplified by the Gamma functional relation to express \begin{equation} G_{2,3}^{2,1}\left(1\middle|\begin{array}c1,\alpha\\1,1,0\end{array}\right)=-\frac{1}{2i\pi}\int_\mathcal L\frac{[\Gamma(1+s)]^2}{\Gamma(\alpha+s)}\,\frac{ds}{s} \end{equation} By differentiation (assuming the validity), \begin{equation} \left.\partial_\alpha G_{2,3}^{2,1}\left(1\middle|\begin{array}c1,\alpha\\1,1,0\end{array}\right)\right|_{\alpha=1}=\frac{1}{2i\pi}\int_\mathcal L\Psi(s+1)\Gamma(s+1)\,\frac{ds}{s} \end{equation} The poles are situated at $s=0$ and $s=-n-1$ with $n=0,1,2,\cdots$. The residue at the pole $s=-n-1$ is $\frac{(-1)^n}{(n+1)\Gamma(n+2)}$. To evaluate the integral, we close the contour on the left side (the contribution of the left half-circle being asymptotically vanishing) to express \begin{align} \left.\partial_\alpha G_{2,3}^{2,1}\left(1\middle|\begin{array}c1,\alpha\\1,1,0\end{array}\right)\right|_{\alpha=1}&=\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)\Gamma(n+2)}\\ &=-\sum_{k=1}^\infty\frac{(-1)^k}{k\,k!} \end{align} By comparison with the series for the exponential integral \begin{align} \left.\partial_\alpha G_{2,3}^{2,1}\left(1\middle|\begin{array}c1,\alpha\\1,1,0\end{array}\right)\right|_{\alpha=1}&=\gamma+E_1(1)\\ &=\gamma-\operatorname{Ei}(-1) \end{align}