Morse and Fesbach identity $\sum_{n=0}^{\infty} e^{-(n+\frac{1}{2})|u|} \; P_n(x) =(2\cosh u - 2x)^{-1/2}$

Question

In book called Methods of theoretical physics from Morse and Feshbach, there is identity, which I wanted to prove ($P_n(x)$ are Legendre polynomials): $$\sum_{n=0}^{\infty} e^{-(n+\frac{1}{2})|u|} \; P_n(x) =(2\cosh u - 2x)^{-1/2} \;$$

I did the following: $$\left( 2\cosh u - 2x \right)^{-1/2} = \left( 2\frac{e^{u} + e^{-u}}{2} - 2x \right)^{-1/2} = \left[ e^{u} \left( 1 + e^{-2u} - 2x e^{-u}\right)\right]^{-1/2} =\\ = e^{-u/2} \left( 1 + e^{-2u} - 2x e^{-u}\right)^{-1/2} = \\ = \sum_{n=0}^{\infty} e^{-\left(n + \frac{1}{2}\right)u} P_n(x) \; .$$

My questions are

1. Where does the absolute value $|u|$ come from (where I lost it - mathematically)?
2. In last step I used the generating function of Legendre polynomials $\left(1 + t^2 - 2xt\right)^{-1/2} = \sum_{n=0}^{\infty} t^n P_n(x)$. But the generating function is valid for $|t|<1$ and $t$ in our case is $e^{-u}$ so $e^{-u} < 1$ so $u>0$. Does that mean that our identity is not valid for $u = 0$? Because I was reading some papers and they used it even for $u=0$.

Thanks in advance!

Answer 1

You need $u > 0$ to make your series converge. The absolute value makes the equation the same for $u$ and $-u$, so if it's true for $u > 0$ it's also true for $u < 0$ (it would not be valid for complex $u$, as the left side would not be analytic).

For $u = 0$, the equation says $\sum_{n=0}^\infty P_n(x) = (2 - 2 x)^{-1/2}$, which is your generating function $\sum_{n=0}^\infty t^n P_n(x) = (1+t^2 -2xt)^{-1/2}$ at $t=1$. The question is whether the series converges: Abel's theorem says if $g(t) = \sum_{n=0}^\infty a_n t^n$ has radius of convergence $1$ and $\sum_{n=0}^\infty a_n$ converges, then $\sum_{n=0}^\infty a_n = \lim_{t \to 1-} g(t)$. For $x = \pm 1$ the series doesn't converge, as $P_n(1) = 1$ and $P_n(-1) = (-1)^n$. For $x = 0$ it does, since $$ P_n(0) = \cases{\dfrac{\left(-1\right)^{\frac{n}{2}} \Gamma \! \left(\frac{n}{2}+\frac{1}{2}\right)}{\sqrt{\pi}\, \Gamma \! \left(\frac{n}{2}+1\right)} & for $n$ even\cr 0 & for $n$ odd } $$ and the Alternating Series test can be used to show convergence in this case. For other $x \in (-1, 1)$ I don't know.