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?

$\begingroup$ 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 $dx^n$. $\endgroup$– Carlo BeenakkerApr 30, 2021 at 13:03

$\begingroup$ Yes you are right sorry $\endgroup$– Adam HammamApr 30, 2021 at 15:00
1 Answer
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) $$\sum_{n=0}^\infty L_n(x)L_n(y)=e^{(x+y)/2}\delta(xy).$$ 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.


$\begingroup$ Is it possible to put $K(t)=\delta(xy) e^{(x+y)/2}e^{(x+y)t}e^{t}$? $\endgroup$ Apr 30, 2021 at 15:29

2$\begingroup$ sure, but it's kind of pointless to allow $K(t)$ to depend on $x$ and $y$ as well, then why introduce this integral? $\endgroup$ Apr 30, 2021 at 15:45