 Solution

Put $\Gamma (z)=I(z)$ for $\Re z>0$. Integration by parts gives $\Gamma (z+1)=z\Gamma (z)$, both proving $\Gamma (n)=(n-1)!$ and defining the meromorphic continuation. A keyhole-contour evaluation of ${\int }_0^{\infty }{t}^{z-1}/(1+t)dt$ gives

Forthe reflection formula at $z$ and $z+\frac{1}{2}$ gives $g(z+1)-g(z)=-2\log 2$. Thus $h(z)=g(z)+2z\log 2$ is entire and $1$-periodic; poles and zeros have cancelled in the quotient. Its Fourier expansion and the exponential growth bound make it a finite Laurent polynomial in $e^{2\pi iz}$; the same growth bound for $e^h$ excludes every nonconstant mode. Hence $h=b$. At $z=1/2$, $e^b=2\sqrt{\pi }$, yielding the duplication formula.

Set $z=1/6$ and use reflection at $1/3$:Therefore