in(2)⁡(z)=(-1)nzn+1⁢∑k=0n(2⁢n-2⁢k-1)!!⁢(-12⁢z2)kk!+1zn+1⁢∑k=n+1∞(12⁢z2)kk!⁢(2⁢k-2⁢n-1)!!.