Can anybody please explain why this limit is not zero?

Question

$$\lim_{x\to0} \Im \left((-1)^x \Gamma(x)\right)=-\pi$$

It seems the imaginary part of $(-1)^x$ should be zero at integer points, including $0$ and the other factor is always real.

Here is a plot (imaginary part is dotted):

Answer 1

The limit is nonzero because $\Gamma$ has a pole at $0$ which annihilates the zero of of the sine there.

The value of the limit depends on how you define $(-1)^x$. The usual ways define it as $\exp (m\pi i x)$ with an odd integer $m$. Most commonly used is $m = 1$. With Euler's formula $\exp (iz) = \cos z + i\sin z$ we then have (for real $x$ not a non-positive integer) $$\Im \bigl((-1)^x\Gamma(x)\bigr) = \Gamma(x)\sin (m\pi x).$$ Using the functional equation of $\Gamma$ it follows that $$\lim_{x \to 0} \Im \bigl((-1)^x\Gamma(x)\bigr) = \lim_{x \to 0} \bigl(x\Gamma(x)\bigr)\frac{\sin (m\pi x)}{x} = \lim_{x \to 0} \Gamma(1+x)\frac{\sin (m\pi x)}{x} = m\pi.$$

A limit of $-\pi$ thus corresponds to the choice $m = -1$.