Assume $\mu<0$. Let $L^\alpha_n(x)$ be Laguerre polynomials of type $n$ and $f\in L^2(\Bbb C)$.

How to prove that $$\sum^\infty_{k=0}\int_{\Bbb C} f(z)\frac{1}{2(2k+1)-\mu}L^0_k(\lvert z\rvert^2)e^{-\lvert z\rvert^2\over 2}dz= \int_{\Bbb C} \sum^\infty_{k=0}f(z)\frac{1}{2(2k+1)-\mu}L^0_k(\lvert z\rvert^2)e^{-\lvert z\rvert^2\over 2}dz$$ where $dz$ is the usual Lebesgue measure? We recall that $\int^{+\infty}_{0} \lvert L^{\alpha}_{k}(t) \rvert^{2}t^{\alpha}e^{-t}dt=\frac{\Gamma(\alpha+k+1)}{\Gamma(k+1)}$.

We also have $\Gamma(a)\psi\left( a,c,x\right)=\sum^{\infty}_{n=0}\frac{1}{n+a} L^{c-1}_{n}(x)$ where $$\Gamma(a)\psi\left( a,c,x\right)= e^x \int^1_0 \exp(-{x\over 1-t}) t^{a-1}(1-t)^{c-1}\in L^2(\Bbb R) $$. Thanks in advance

Assume $\mu<0$. Let $L^\alpha_n(x)$ be Laguerre polynomials of type $n$ and $f\in L^2(\Bbb C)$.

How to prove that $$\sum^\infty_{k=0}\int_{\Bbb C} f(z)\frac{1}{2(2k+1)-\mu}L^0_k(\lvert z\rvert^2)e^{-\lvert z\rvert^2\over 2}dz= \int_{\Bbb C} \sum^\infty_{k=0}f(z)\frac{1}{2(2k+1)-\mu}L^0_k(\lvert z\rvert^2)e^{-\lvert z\rvert^2\over 2}dz$$ where $dz$ is the usual Lebesgue measure? We recall that $\int^{+\infty}_{0} \lvert L^{\alpha}_{k}(t) \rvert^{2}t^{\alpha}e^{-t}dt=\frac{\Gamma(\alpha+k+1)}{\Gamma(k+1)}$.

We also have $\Gamma(a)\psi\left( a,c,x\right)=\sum^{\infty}_{n=0}\frac{1}{n+a} L^{c-1}_{n}(x)$ where $$\Gamma(a)\psi\left( a,c,x\right)= e^x \int^1_0 \exp(-{x\over 1-t}) t^{a-1}(1-t)^{c-1}\in L^2(\Bbb R) .$$ see https://math.stackexchange.com/questions/1901995/any-reference-for-%24-%5Cgamma%28a%29-u%28a%2Cb%2Cz%29-%3D%5Csum_%7Bj%3D0%7D%5E%7B%2B%5Cinfty%7D-%5Cfrac%7B1%7D%7Bj%2Ba%7D-l%5E%7Bb-1%7D_%7Bj%7D%28z%29-%24 Thanks in advance

Assume $\mu<0$. Let $L^\alpha_n(x)$ be Laguerre polynomials of type $n$ and $f\in L^2(\Bbb C)$.

How to prove that $$\sum^\infty_{k=0}\int_{\Bbb C} f(z)\frac{1}{2(2k+1)-\mu}L^0_k(\lvert z\rvert^2)e^{-\lvert z\rvert^2\over 2}dz= \int_{\Bbb C} \sum^\infty_{k=0}f(z)\frac{1}{2(2k+1)-\mu}L^0_k(\lvert z\rvert^2)e^{-\lvert z\rvert^2\over 2}dz$$ where $dz$ is the usual Lebesgue measure? We recall that $\int^{+\infty}_{0} \lvert L^{\alpha}_{k}(t) \rvert^{2}t^{\alpha}e^{-t}dt=\frac{\Gamma(\alpha+k+1)}{\Gamma(k+1)}$.

Assume $\mu<0$. Let $L^\alpha_n(x)$ be Laguerre polynomials of type $n$ and $f\in L^2(\Bbb C)$.

How to prove that $$\sum^\infty_{k=0}\int_{\Bbb C} f(z)\frac{1}{2(2k+1)-\mu}L^0_k(\lvert z\rvert^2)e^{-\lvert z\rvert^2\over 2}dz= \int_{\Bbb C} \sum^\infty_{k=0}f(z)\frac{1}{2(2k+1)-\mu}L^0_k(\lvert z\rvert^2)e^{-\lvert z\rvert^2\over 2}dz$$ where $dz$ is the usual Lebesgue measure? We recall that $\int^{+\infty}_{0} \lvert L^{\alpha}_{k}(t) \rvert^{2}t^{\alpha}e^{-t}dt=\frac{\Gamma(\alpha+k+1)}{\Gamma(k+1)}$.

We also have $\Gamma(a)\psi\left( a,c,x\right)=\sum^{\infty}_{n=0}\frac{1}{n+a} L^{c-1}_{n}(x)$ where $$\Gamma(a)\psi\left( a,c,x\right)= e^x \int^1_0 \exp(-{x\over 1-t}) t^{a-1}(1-t)^{c-1}\in L^2(\Bbb R) $$. Thanks in advance

Assume $\mu<0$. Let $L^\alpha_n(x)$ be Laguerre polynomials of type $n$ and $f\in L^2(\Bbb C)$

How to prove that $$\sum^\infty_{k=0}\int_{\Bbb C} f(z)\frac{1}{2(2k+1)-\mu}L^0_k(|z|^2)e^{-|z|^2\over 2}dz= \int_{\Bbb C} \sum^\infty_{k=0}f(z)\frac{1}{2(2k+1)-\mu}L^0_k(|z|^2)e^{-|z|^2\over 2}dz$$ where $dz$ is the usual Lebesgue measure. We recall that $\int^{+\infty}_{0} | L^{\alpha}_{k}(t) |^{2}t^{\alpha}e^{-t}dt=\frac{\Gamma(\alpha+k+1)}{\Gamma(k+1)}$,

Thank you in advance.

Assume $\mu<0$. Let $L^\alpha_n(x)$ be Laguerre polynomials of type $n$ and $f\in L^2(\Bbb C)$.

How to prove that $$\sum^\infty_{k=0}\int_{\Bbb C} f(z)\frac{1}{2(2k+1)-\mu}L^0_k(\lvert z\rvert^2)e^{-\lvert z\rvert^2\over 2}dz= \int_{\Bbb C} \sum^\infty_{k=0}f(z)\frac{1}{2(2k+1)-\mu}L^0_k(\lvert z\rvert^2)e^{-\lvert z\rvert^2\over 2}dz$$ where $dz$ is the usual Lebesgue measure? We recall that $\int^{+\infty}_{0} \lvert L^{\alpha}_{k}(t) \rvert^{2}t^{\alpha}e^{-t}dt=\frac{\Gamma(\alpha+k+1)}{\Gamma(k+1)}$.