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.
Function involving the Gamma function
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$\begingroup$
gamma-function
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0*Hint.* $w\Gamma(w) = \Gamma(w+1)$. – 2017-01-11
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0Do you know Euler's reflection formula? – 2017-01-11
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0Do you mean to use the reflection formula with $ z=-z-1/2 $. So that $ \Gamma(-1/2-z)=\frac{\pi}{-\Gamma(z+3/2)cos(\pi z)}=\frac{\pi}{-(z+1/2)\Gamma(z+1/2)cos(\pi z)} $ but this would then give $ k(z+1)=\pi $ I assume I am making a mistake somewhere? – 2017-01-11
1 Answers
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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 $$k(z+1)=k(z)$$