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According to p.244 in "Magnus, W., Oberhettinger, F., Soni, R.: Formulas and Theorems for the Special Functions of Mathematical Physics, Grundlehren der mathematischen Wissenschaften 52, Springer, Berlin et al. (1966)", we have, $$\int_{0}^\infty e^{-zx}\, x^\alpha\, L^{\alpha}_m(x)\, dx = \frac{\Gamma(m+\alpha+1)}{\Gamma(m+1)} z^{-m-\alpha-1}\,(z-1)^{m}; \quad \text{Re}\, z> 0,\, \text{Re}\, \alpha> 0.$$ I am looking for the value of the integral $$I:= \int_{0}^\infty (x+1)^p\, e^{-ax}\, x^\alpha\, L^{\alpha}_m(x)\, dx =?; \quad a> 0,\,p\in\mathbb Z.$$

The Laguerre polynomial of type $\alpha$ and degree $m$ is given by $$L^{\alpha}_m(x)=\sum_{k=0}^{m}(-1)^k \binom{m+\alpha}{m-k} \frac{x^k}{k!}.$$ Then, by this definition, we get
$$I:= \sum_{k=0}^{m} \frac{(-1)^k}{k!} \binom{m+\alpha}{m-k} \int_{0}^\infty (x+1)^p\, e^{-ax}\, x^{\alpha+k}\, dx$$ So to calculate I, it is enough to calculate the following integral: $$J:=\int_{0}^\infty (x+1)^p\, e^{-ax}\, x^{\alpha+k}\, dx=?$$ According to p.277 in "Magnus, W., Oberhettinger, F., Soni, R.: Formulas and Theorems for the Special Functions of Mathematical Physics, Grundlehren der mathematischen Wissenschaften 52, Springer, Berlin et al. (1966)", we have, $$U(\gamma,c,a)=\frac{1}{\Gamma(\gamma)}\int_{0}^\infty (1+x)^{c-\gamma-1}\, e^{-ax}\, x^{\gamma-1}\, dx; \quad \text{Re}\, \gamma> 0,\, \text{Re}\,a> 0.$$ Then for $\gamma=\alpha+k+1$ and $c=\alpha+p+k+2$, we get $$J:=\Gamma(\alpha+k+1) U(\alpha+k+1,\alpha+p+k+2,a).$$ Then, $$I:= \sum_{k=0}^{m} \frac{(-1)^k}{k!} \binom{m+\alpha}{m-k} \Gamma(\alpha+k+1) U(\alpha+k+1,\alpha+p+k+2,a).$$

Z. Alfata
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    What is the context of this problem? Does the $(x+c)^p$ term result from a series expansion? – Kyle Knoepfel Nov 03 '25 at 18:13
  • If I'm not mistaken, the integral $J$ is entry 3.384.3 in Gradshteyn and Ryzhik, Table of Integrals, Series, and Products (8th Edition) – Caesar.tcl Nov 04 '25 at 15:32
  • @ Caesar.tcl, think you for your comment. But in the eqaution 3.384.3 $u>0$? and in our case $u=0$ – Z. Alfata Nov 04 '25 at 15:39
  • hmm that's a good point, but maybe it works regardless. One should check numerically, I've written down the result from GR – Caesar.tcl Nov 04 '25 at 15:41

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The integral $J$ is entry 3.384.3 in Gradshteyn and Ryzhik, Table of Integrals, Series, and Products (8th Edition) and gives

$$\int_0^\infty\mathrm{d}x\,(x+1)^px^{\alpha+k}\mathrm{e}^{-ax}=a^{-\frac{\alpha+k+p}{2}-1}\mathrm{e}^{\frac{a}{2}}\Gamma(\alpha+k+1)W_{\frac{p-\alpha-k}{2},\frac{p+\alpha+k+1}{2}}(a)$$

where $W_{\nu,\mu}(x)$ is the second Whittaker function. I personally don't have any hope the resulting sum has a closed form tho.

Edit: apparently we are not in the valid regime for the parameters, so idk

Caesar.tcl
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