# Laplace transform and Laguerre Polynomials

What is the kernel $$K(t)$$ of the following Laplace transform equation: $$\int_{0}^{+\infty}e^{-(x+y)t} K(t) dt= \sum_{n=0}^{\infty}\varphi_{n}^{\alpha}(x)\varphi_{n}^{\alpha}(y),$$ where $$\varphi_{n}^{\alpha}(x)=(\frac{n!}{\Gamma(\alpha+n+1)})^{1/2} \frac{e^{x}}{n!} \frac{d^{n}}{x^{n}}(e^{-x}x^{n})$$ is the normalized Laguerre polynomial. If the result is not known, where can I find this type of equation?

• I presume your expression for $\varphi^\alpha_n(x)$ is missing a factor $x^{-\alpha}$ before the derivative, and a factor $x^\alpha$ after the derivative, and that the denominator $x^n$ should be $dx^n$. Apr 30, 2021 at 13:03
• Yes you are right sorry Apr 30, 2021 at 15:00

Let me check this for $$\alpha=0$$, when $$\varphi_n^\alpha(x)=L_n(x)$$ is the Laguerre polynomial. The completeness relation is (see eq. 1.10 of Li and Wong - Integral and series representations of the Dirac delta function) $$\sum_{n=0}^\infty L_n(x)L_n(y)=e^{(x+y)/2}\delta(x-y).$$ This depends on both the sum and difference of $$x$$ and $$y$$, so the representation given in the OP, which depends only on $$x+y$$, does not exist.
• Is it possible to put $K(t)=\delta(x-y) e^{(x+y)/2}e^{(x+y)t}e^{-t}$? Apr 30, 2021 at 15:29
• sure, but it's kind of pointless to allow $K(t)$ to depend on $x$ and $y$ as well, then why introduce this integral? Apr 30, 2021 at 15:45