# Proving an identity involving the derivative of the Laguerre polynomials with respect to $n$

I've recently come across the following equality in a paper: suppose one defines an analytic function $L(n,x)$ which is equal to the $n$th Laguerre polynomial for $n\in\{0,1,\ldots\}$, and let* $L^{(1,0)}(n,x) = \frac{\partial}{\partial n}L(n,x)$. Then apparently

$$L^{(1,0)}(-1,-x) = -[\gamma_E + \Gamma(0,-x) + \log(-x)]e^{-x}$$

where $\gamma_E$ is the Euler-Mascheroni constant and $\Gamma$ is the incomplete gamma function.

Numerically, this checks out, but I would be interested in a symbolic proof that the equality holds. I've been playing around with the generating function for the Laguerre polynomials (and some other representations), but I haven't figured out a way to make the connection. Can anyone help me out?

*$L^{(1,0)}$ is called the "multivariate Laguerre polynomial" in the paper; I couldn't find a source to tell me whether this is a widely recognized function or not.

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$\gamma_E$ is Euler-Mascheroni, I take it. –  Ｊ. Ｍ. Aug 7 '12 at 4:30
In any event, this seems to be the applicable formula. You have $$\lim_{\nu\to-1}(\pi\cot(\nu\pi)+\psi(1+\nu))=\lim_{\nu\to-1}\psi(-\nu)=-\gamma‌​$$ and the series will then evaluate in terms of the incomplete gamma function (or exponential integral, whichever you prefer). –  Ｊ. Ｍ. Aug 7 '12 at 4:36
@J.M. true, I forgot to specify that. Thanks for the tip. –  David Z Aug 7 '12 at 5:25

Laguerre function is defined in terms of the hypergeometric ${}_1F_1$ as follows: $$L_\nu(x) = {}_1F_1\left(-\nu; 1; x\right)$$ Thus: $$\left.\frac{\partial}{\partial \nu} L_\nu(z)\right|_{\nu = -1} = \sum_{n=0}^\infty \frac{(1)_n}{n!} \frac{z^n}{n!} \left( \psi(n+1) - \psi(1) \right) = \sum_{n=1}^\infty \frac{z^n}{n!} H_n = \sum_{n=1}^\infty \frac{z^n}{n!} \sum_{m=1}^n \frac{1}{m} \\= \sum_{m=1}^\infty \frac{1}{m} \sum_{n=m}^\infty \frac{z^n}{n!} = \sum_{m=1}^\infty \frac{1}{m} \mathrm{e}^z \frac{\gamma(m,z)}{\Gamma(m)} = \mathrm{e}^z \int_0^z \sum_{m=1}^\infty \frac{t^{m-1}}{m!} \mathrm{e}^{-t} \mathrm{d} t = \mathrm{e}^z \int_0^z \frac{1- \exp(-t)}{t} \mathrm{d} t = \mathrm{e}^z \left( \gamma + \log(z) + \Gamma(0,z) \right)$$
Thanks, that works! I missed the connection between $L$ and the hypergeometric function. –  David Z Aug 8 '12 at 3:09