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