\[\operatorname{Si}\left(a,z\right)=z^{a}\sum_{k=0}^{\infty}\frac{\left(2k+\frac%
{3}{2}\right){\left(1-\frac{1}{2}a\right)_{k}}}{{\left(\frac{1}{2}+\frac{1}{2}%
a\right)_{k+1}}}\mathsf{j}_{2k+1}\left(z\right),\]