Function involving the Gamma function

Question

I Have $$k(z)=\Gamma\left(\frac12+z\right)\Gamma\left(\frac12-z\right)\cos(\pi z)$$ and wish to find a relation linking $k(z+1)$ and $k(z)$. Substituting $z+1$ and using the recurrence relation of the gamma function $$\Gamma(z+1)=z\Gamma(z)$$ and using $$\cos(\pi +z)=-\cos(z)$$ I obtain: $$k(z+1)=-\left(z+\frac12\right)\Gamma\left(z+\frac12\right)\Gamma\left(-\frac12-z\right)\cos(\pi z)$$ However I am stuck with what to do with the $\Gamma\left(-\frac12-z\right)$ term.

Answer 1

Continuing from where you left off with $$k(z+1)$$ We have $$-\left(\frac12+z\right)\Gamma\left(\frac12+z\right)\Gamma\left(-\frac12-z\right)\cos(\pi z)$$ $$=\left(-\frac12-z\right)\Gamma\left(-\frac12-z\right)\Gamma\left(\frac12+z\right)\cos(\pi z)$$ $$=\Gamma\left(-\frac12-z+1\right)\Gamma\left(\frac12+z\right)\cos(\pi z)$$ $$=\Gamma\left(\frac12-z\right)\Gamma\left(\frac12+z\right)\cos(\pi z)$$ Therefore $$k(z+1)=k(z)$$