Derivative of the Meijer G-function with respect to one of its parameters 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:
$$\mathcal{D}=\left.\partial_\alpha G_{2,3}^{2,1}\left(1\middle|\begin{array}c1,\alpha\\1,1,0\end{array}\right)\right|_{\alpha=1}$$
 A: 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}
A: 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,
$$\mathcal{D}\approx0.7965995992970531342836758655425240800732066293468318063837458...$$
