I have the generating function $\hat{\beta}(z)=\sum_{k=0}^\infty C_k z^k$ where $C_k$ are Catalan numbers.
Now, for $|z|<1/4$ we've
$$\hat{\beta}(z)=\sum_{k=0}^{\infty} z^k \int x^{2k}\sigma(x) dx = \int \frac{1}{1-zx^2}\sigma(x) dx = \int \frac{1}{1-\sqrt{z}x}\sigma(x)dx,$$
where $\sigma(x) := \frac{1}{2\pi}\sqrt{4-x^2}1_{|x|\le 2} $.
It's written that the third equality above (in the chain of equalities of the second $\hat{\beta}(z)$) follows from the fact that the support of $\sigma$ is the interval $[-2,2]$. I don't see why does that imply the third equality?
Can anyone please explain it to me?
Thanks in advance.
By expanding the term $\frac{1}{1-\sqrt{z}x}$ as a power series, you see that the odd powers (combined with the density function $\sigma(x)$), gives rise to an odd function, hence its integration over a symmetric interval around the origin is equal to zero.