I'm trying to show the orthogonality relation for Laguerre Polynomials $L_n(x)$ through their generating function $G(x,t)$.
$$G(x,t)=\frac{1}{1-t}e^{\frac{-xt}{1-t}}=\sum_{n=0}^{\infty} L_n(x) t^n$$
$$\implies \int_{0}^{\infty} e^{-x}L_n(x)L_m(x)\,dx=\delta_{nm}$$
Here is my attempt. Consider the following integral.
$$\int_{0}^{\infty} (G(x,t))^2e^{-x}\,dx$$
I will perform this integral via the explicit generating function-representation, and then via the series-representation. So, on the one hand we have that,
$$\begin{align*}
\int_{0}^{\infty} (G(x,t))^2e^{-x}\,dx&=\int_{0}^{\infty}\frac{1}{(1-t)^2}e^{\frac{-2xt}{1-t}}e^{-x}\,dx\\
&=\frac{1}{(1-t)^2}\int_{0}^{\infty}e^{-\left(1+\frac{2t}{1-t}\right)}\,dx\\
&=\frac{1}{(1-t)^2}\int_{0}^{\infty}e^{-\left(\frac{1+t}{1-t}\right)}\,dx\\
&=\frac{1}{(1-t)^2}\int_{0}^{\infty}e^{-\left(\frac{1-t^2}{(1-t)^2}\right)}\,dx\\
&=\frac{1}{1-t^2}\left[\left.e^{-\left(\frac{1-t^2}{(1-t)^2}\right)}\right|^0_{\infty}\right]\\
&=\frac{1}{1-t^2}\,\,\,\,(|t|<1)\\
&=\sum_{k=0}^{\infty}t^{2k}
\end{align*}$$
On the other hand, we can calculate the integral by the following:
$$\begin{align*}
\int_{0}^{\infty} (G(x,t))^2e^{-x}\,dx&=\int_{0}^{\infty}\left(\sum_{n=0}^{\infty} L_n(x) t^n\right)\left(\sum_{m=0}^{\infty} L_m(x) t^m\right)e^{-x}\,dx\\
&=\int_{0}^{\infty}\sum_{n=0}^{\infty}\left[\sum_{m=0}^n L_m(x) L_{n-m}(x)\right]e^{-x} t^n \,dx\\
&=\sum_{n=0}^{\infty}\left[\sum_{m=0}^n\left(\int_{0}^{\infty}e^{-x}L_m(x)L_{n-m}(x)\,dx\right)\right]t^n
\end{align*}$$
Now that we've done the integral two different ways, we can equate each of their coefficients. This gives us the equality
$$\sum_{m=0}^{2k}\left(\int_{0}^{\infty}e^{-x}L_m(x)L_{2k-m}(x)\,dx\right)=1,\,\,\,\,(k\in\mathbb{N})$$
MY QUESTION: What am I supposed to do from here? I'm trying to show that, in the sum right above, every term is zero, except for when $m=k$. Do I have enough information to show that? Have I done something wrong?
Best Answer
Consider instead \begin{align}\int_0^{\infty}e^{-x}G(x,t)G(x,s)dx&=\frac{1}{(1-t)(1-s)}\int_0^{\infty}e^{-\left(1+\frac{t}{1-t}+\frac{s}{1-s}\right)x}dx=\\ &=\frac{1}{(1-t)(1-s)}\int_0^{\infty}e^{-\frac{1-st}{(1-s)(1-t)}x}dx=\\ &=\frac{1}{1-st}. \end{align} What is the coefficient of $s^mt^n$ in the expansion of the right side when $m\neq n$? and when $m=n$?