On a derivative of Appell's $F_1$ function with respect to a parameter

Question

I'm trying to compute $$\left.\frac{\partial}{\partial\alpha}F_1(2,\alpha,\alpha;3;x,y)\right|_{\alpha=0},$$ where $F_1$ is Appell's hypergeometric function in two variables. What I tried so far is noticing that $$F_1(2,\alpha,\alpha;3;x,y) = \sum_{n=0}^{+\infty}\sum_{m=0}^{+\infty} \frac{(2)_{m+n} (\alpha)_n (\alpha)_m}{n! m! (3)_{n+m}} x^n y^m,$$ and exploiting the fact that $$\frac{\partial}{\partial \alpha}[(\alpha)_n(\alpha)_m] = (\alpha)_n(\alpha)_m (\psi(\alpha+n) + \psi(\alpha+m) - 2\psi(\alpha)),$$ where $\psi$ is the digamma function. Using these expression for the derivative of the Pochhammer symbols, we get $\frac{\partial}{\partial \alpha}[(\alpha)_0(\alpha)_0] = 0$, $\frac{\partial}{\partial \alpha}[(\alpha)_0(\alpha)_1] = 1$, and $\frac{\partial}{\partial \alpha}[(\alpha)_n(\alpha)_m] = \mathcal{O}(\alpha)$ for $n,m \geq 1$. Hence, at $\alpha = 0$, we find that $$\left.\frac{\partial}{\partial \alpha}[(\alpha)_n(\alpha)_m]\right|_{\alpha=0} = \delta_{n,0}\delta_{m,1} + \delta_{n,1}\delta_{m,0},$$ where $\delta_{i,j}$ is the Kronecker delta. This suggests $$\left.\frac{\partial}{\partial\alpha}F_1(2,\alpha,\alpha;3;x,y)\right|_{\alpha=0} = \frac{2}{3}(x+y).$$

This seems about right, but the result is wrong. The result above tells us that the function describes a plane, but a plot of the numerical result with Mathematica yields the figure below, which is not a plane.

What is going wrong? Maybe I'm commuting limits that do not commute? How can I fix this?

Answer 1

You don't have all contributions from the cases (n=0, m>0) and (n>0, m=0). If you just write out the Pochhammer symbols you get

$$ \left.\frac{\partial}{\partial \alpha} \left((\alpha)_n(\alpha)_m \right)\right|_{\alpha=0} = \begin{cases} (m-1)! &\text{, if $n=0$ and $m>0$} \\ (n-1)! &\text{, if $n > 0$ and $m=0$} \\ 0 &\text{, otherwise} \end{cases} $$

and you should then end up with sums you can express as rational functions in $x, y, \log(1-x), \log(1-y)$.

Answer 2

Using integral representations:

$$F_1(2,\alpha,\alpha;3;x,y) =\int_0^1 2 t (1-t x)^{-\alpha } (1-t y)^{-\alpha } \, dt$$ then: $$\left.\frac{\partial}{\partial\alpha}F_1(2,\alpha,\alpha;3;x,y)\right|_{\alpha=0} =\\\underset{\alpha \to 0}{\text{lim}}\frac{\partial }{\partial \alpha }\left(\int_0^1 2 t (1-t x)^{-\alpha } (1-t y)^{-\alpha } \, dt\right)=\\\int_0^1 \left(\underset{\alpha \to 0}{\text{lim}}\frac{\partial \left(2 t (1-t x)^{-\alpha } (1-t y)^{-\alpha }\right)}{\partial \alpha }\right) \, dt=\\\int_0^1 \left(\underset{\alpha \to 0}{\text{lim}}\left(-2 t (1-t x)^{-\alpha } (1-t y)^{-\alpha } \log (1-t x)-2 t (1-t x)^{-\alpha } (1-t y)^{-\alpha } \log (1-t y)\right)\right) \, dt=\\\int_0^1 -2 t (\log (1-t x)+\log (1-t y)) \, dt=\\1+\frac{1}{x}+\frac{1}{y}+\left(-1+\frac{1}{x^2}\right) \log (1-x)+\left(-1+\frac{1}{y^2}\right) \log (1-y)$$