After separating variables in a Laplace equation in polar coordinates, I have to solve the resulting Bessel equation for the $R$-variables (the $\Theta$ variable I do not consider in this post as it yields a regular exponential function):
$$
\begin{equation}
R_{rr}+\frac{1}{r}R_r+(1-\frac{\lambda}{r^2})R=0
\end{equation}
$$
I used the Frobenius method, and got as follows:
I made the ansatz $R(r)=\sum_{n=0}^\infty a_nr^{n+q}$, and get thus
$$
\begin{split}
R'(r) & =\sum_{n=0}^\infty(n+q)a_nr^{n+q-1},\\
&\qquad\Downarrow\\
R''(r) &=\sum_{n=0}^\infty(n+q)(n+q-1)a_nr^{n+q-2}.
\end{split}
$$
Inserting in the PDE:
$$
\sum_{n=0}^\infty(n+q)(n+q-1)a_nr^{n+q-2}+\frac{1}{r}\sum_{n=0}^\infty(n+q)a_nr^{n+q-1}+\Big(1-\frac{\lambda}{r^2}\Big)\sum_{n=0}^\infty a_nr^{n+q}=0.
$$
Simplified:
$$
\sum_{n=0}^\infty a_nr^{q+n-2}\big[(q^2+n^2)+2qn-1)\big]+\sum_{n=0}^\infty a_nr^{n+q}-\lambda\sum_{n=0}^\infty a_nr^{n+q-2}=0
$$
Now, I set $n=0$ and get:
$$
a_0r^{q}\big[(q^2-1-\lambda)\big]+a_0r^{q}=0.
$$
Solving the quadratic equation $q^2-1-\lambda+1=0\rightarrow q=\pm\sqrt{\lambda}$.
Then, solving the recursion equation $[(n+q)^2-\lambda]a_n=a_{n-2}$ $$ a_{n+2}=\frac{a_{n}}{[(n+2+q)^2-\lambda]}. $$ But then, how does one use gamma function and beta function here? For instance, $\Gamma(n)=(n-1)!$ and the relationship between $\Gamma$ and $\beta$ function is $$ \begin{equation} \beta(p,q)=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)} \end{equation} $$ But re-writing $$ a_{n+2}=\frac{a_{n}}{[(n+2+q)^2-\lambda]}, $$ with the use of $\Gamma$ and $\beta$ is not clear to me. Does anyone have any suggestions?
Thanks