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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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    @J.M. true, I forgot to specify that. Thanks for the tip.2012-08-07

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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) $

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    Thanks, that works! I missed the connection between $L$ and the hypergeometric function.2012-08-08