# How to find $\lim_{x \rightarrow -N} J_{a} = (-1)^N J_{N}$?

So far I have $$\lim_{a \rightarrow -N} J_{a} = \lim_{a \rightarrow -N} \mid \frac{x}{2} \mid^a \sum_{k=0}^{\infty} \frac{(-x^2/4)^k}{k! \; \Gamma(a+k+1)} = \mid \frac{x}{2} \mid^{-N} \sum_{k=0}^{\infty} \frac{(-x^2/4)^k}{k! \; \Gamma(-N+k+1)}$$ I don't see how the $(-1)^n$ gets factored out to get the desired result of $(-1)^n J_{-N}$. Where $J_{a}$ is the Bessel function of the first kind. I am an undergrad student learning Bessel and Legendre ODE's. I know only basic cal and have a reference sheet of gamma and beta function properties. I don't have graduate school background on their meaning etc. so explaining to that detail really won't help me because of my background. I am looking more for math manipulation at this point. Thanks.

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The reflection formula for the Gamma function might help. – marty cohen Oct 7 '12 at 21:28

Use the fact that the reciprocal of the Euler's $\Gamma$-function vanishes at non-positive integers, that is for $k\in \mathbb{Z}_{\leqslant 0}$ $$\lim_{x \to -k} \frac{1}{\Gamma(x)} = 0$$ Thus, all the terms where $0 \leqslant k < N$ vanish in the limit: $$\begin{eqnarray} \lim_{a \to -N} J_a(x) &=& \sum_{k=N}^\infty \left(\frac{x}{2}\right)^{2k-N} \frac{(-1)^k }{k! \cdot \Gamma(1-N +k)} = \sum_{k=0}^\infty \left(\frac{x}{2}\right)^{2k+N} \frac{(-1)^{k+N}}{\Gamma(k+N+1) \cdot k!} \\ &=& (-1)^N J_N(x) \end{eqnarray}$$

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I don't understand how you get the last step in your final result. How does -N become +N ? I initially though you added k+N whenever there was k, but then I noticed that didn't work. Thanks! – renagade629 Oct 7 '12 at 22:27
I made a shift of the summation variable $k=N+m$. Then $k! = \Gamma(k+1) = \Gamma(m+N+1)$, $\Gamma(1-N+k) = \Gamma(1+m) = m!$ and $2k-N = 2(m+N) -N = 2m + N$. Now relabel $m$ as $k$. – user40314 Oct 9 '12 at 21:55