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While solving Legendre's Equation around one of its regular singularity $x = 1$, and investigating the convergence of this solution at another regular singularity $x = -1$, this came up $$P_\nu(-1)=\sum_{n=0}^{\infty} \frac{(-1)^n \Gamma(\nu + n +1)}{(n!)^2 \Gamma(\nu-n+1)} $$ $$= -\frac{\sin(\pi \nu)}{\pi} \ \sum_{n=0}^{\infty}\frac{\Gamma(n + \nu + 1) \Gamma(n-\nu)}{(n!)^2}$$

Where $P_\nu(x) = \sum_{n=0}^{\infty} \frac{\Gamma(n+\nu+1)}{(n!)^2 \Gamma(\nu-n+1)} (\frac{x-1}{2})^n$ is the solution to Legendre Equation solved around its regular singularity $x=1$

Lecturer just said that the summation above can be shown by using Euler Reflection Formula, which is this $$\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}$$ #I am not asking how to proof Euler Reflection Formula :)

Tried several times and still couldn't make it, so I am asking it here

Thanks

MDR_Dope
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1 Answers1

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The Euler reflection formula for the Gamma function for $z\notin \mathbb{Z}$ is given by

$$\Gamma(z)\Gamma(1-z)=\frac\pi{\sin(\pi z)}\tag1$$

Applying $(1)$ with $z=\nu-n+1$ and $1-z=n-\nu$ reveals

$$\begin{align} \Gamma(\nu-n+1)\Gamma(n-\nu)&=\frac\pi{\sin(\pi(\nu-n+1))}\\\\ &=\frac{\pi}{\sin(\pi \nu)\cos(\pi(1-n))+\cos(\pi \nu)\sin(\pi(1-n))}\\\\ &=\frac{(-1)^{n-1}\pi}{\sin(\pi \nu)}\tag2 \end{align}$$

Multiplying both sides of $(2)$ by $\frac{(-1)^n \Gamma(n+\nu+1)}{(n!)^2}$, rearranging, and summing over $n$ yields

$$\sum_{n=0}^\infty\frac{(-1)^n\Gamma(n+\nu+1)}{(n!)^2\Gamma(\nu-n+1)}=-\frac{\sin(\pi\nu)}{\pi}\sum_{n=0}^\infty\frac{\Gamma(n+\nu+1)\Gamma(n-\nu)}{(n!)^2}$$

as was to be shown!

Mark Viola
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