I have the following contour integral with the integrand
1/(2 Pi I)Gamma[-z]^(2)Gamma[1+z]^(2)
where the contour is an infinite semi circle running parallel to the imaginary line from -0.6-Infinity to -0.6+Infinity and semi-circular part in towards the positive real z-axis. On numerical integration using Mathematica, I obtain the result to be 1.0( code is given below)
NIntegrate[(1/(2Pi I))(Gamma[-z])^(2)(Gamma[1+z])^(2),{z,-0.6-20I,-0.6+20I}]
but when I apply Cauchy residue theorem I get zero because the residue for all poles at z=0,1,2,... and at z=-1,-2,-3... are all zero. Why is there a contradiction in the result?
This is my first question on Math StackExchange. Pardon me if my question is not well presented.
From the reflection formula
$$ \Gamma(1+z) \Gamma(-z) = - \frac{\pi}{\sin(\pi z)} $$
Its square is then equal to
$$ \pi^2 \csc^2(\pi z) $$
whose primitive we actually know to be $-\pi \cot(\pi z)$. So for any $a \in \mathbb{R} \setminus \mathbb{Z}$
$$ \frac{1}{2\pi i} \int_{a- \infty i }^{a + \infty i} \left[ \Gamma(1+z) \Gamma(-z)\right]^2 ~dz = \lim_{s \to \infty} -\frac1{2i} [\cot(\pi(a + is)) - \cot(\pi(a-is))] = 1$$
so the evaluation by Mathematica is not wrong.
The above result also indicates that the integral along the semi-circular arc is not negligible as you have assumed.