Laurent series expansion of the generator function gives,
$g(z,t) = e^{z/2(t - 1/t ) } = \sum J_n (z) t^n. $
The term $t^n$ suggests that this expansion is performed around the origin, so we have,
$J_n(z) = \int \frac{e^{z/2(t - 1/t ) }}{t^{n+1}} dt $
with the integration is to be performed along contour enclosing the origin. My question is how to evaluate this integral when $n=0$. When $n=0$, we have singularity at $z=0$, and the contour integral is being equal to $2 \pi i Res(f(z), z=0)$. Residue at $z=0$ is given as;
$\lim_{t->0} t \frac{ e^{z/2(t - 1/t ) }}{t} $, which gives us zero. There's clearly a mistake in this, but i couldn't find what it is. I'm hoping i can get some help here, thanks.
There is en essential singularity in the numerator, so you simply cannot use the expression for a simple pole to compute the residue. Rather, you need the coefficient of $t^n$ in $e^{(z/2) (t-1/t)}$, which is more involved. The residues in fact form an infinite series (as you'd expect):
$$\frac{1}{n!} (z/2)^n - \frac{1}{(n+1)!} (z/2)^{n+1} \binom{n+1}{1} + \frac{1}{(n+2)!} (z/2)^{n+2} \binom{n+2}{2} - \cdots$$
This holds for $n=0$ as well.