\[{\mathsf{i}^{(2)}_{n}}\left(z\right)=\frac{(-1)^{n}}{z^{n+1}}\sum_{k=0}^{n}%
\frac{(2n-2k-1)!!(-\frac{1}{2}z^{2})^{k}}{k!}+\frac{1}{z^{n+1}}\sum_{k=n+1}^{%
\infty}\frac{(\frac{1}{2}z^{2})^{k}}{k!(2k-2n-1)!!}.\]