Divergence of $\Gamma$ function for complex values

Question

It is said that $\dfrac{1}{\Gamma(ix)}$ (of purely imaginary part) diverges. But why please?

Answer 1

I assume that you want to know why $$\lim_{x \to \infty} \Gamma(ix) = 0$$ Therefore consider the well-known formula $$\Gamma(z)=\lim_{n \to \infty}\frac{n!n^z}{z (z+1)...(z+n)}$$ We obtain:

$$\lim_{x \to \infty} | \Gamma(ix) |=\lim_{x \to \infty} \lim_{n \to \infty} \frac{|n!n^{ix}|}{|ix (ix+1)...(ix+n)|} = \lim_{x \to \infty} \lim_{n \to \infty}\frac{n!}{|x (x-i)...(x-ni)|} = \lim_{x \to \infty} \lim_{n \to \infty}\frac{n!}{\sqrt{x^2 (x^2+1)...(x^2+n^2)}} = \lim_{x \to \infty} \lim_{n \to \infty}\frac{1}{\sqrt{x^2 (x^2+1)((x/2)^2+1)...((x/n)^2+1)}} \le = \lim_{x \to \infty}\frac{1}{\sqrt{x^2}}=0$$ Here $|n^{ix}|=|e^{\log(n)ix}|=1$ follows from the general fact that $|e^z|=e^{Re(z)}$.