Disclaimer: I am an undergraduate student about a semester into introductory complex analysis. I am entirely out of my depth here, just curious about something I noticed.

The gamma function $\Gamma(z) := \int_0^\infty\limits z^{z-1}e^{-x} dx$ is a generalization of the standard factorial function $n! = n \times (n-1) \times (n-2) \dots \times 1$ defined over the complex plane (minus nonpositive integers), and with the caveat that for nonnegative integer $n$, $\Gamma(n) = (n-1)!$

As such, on the (positive) reals, $\Gamma$ grows superexponentially. Indeed more generally, for any complex $z$, $\Gamma(z+x) \to \infty$ (in magnitude) as $x \in \mathbb{R}$ tends to infinity. Interestingly though, the opposite is true of the imaginary axis: $\Gamma(z+iy) \to 0$ as $y \in \mathbb{R}$ gets large.

Informally then, the real part of $z$ makes $\Gamma(z)$ large, and the imaginary part makes it small. I was curious: how much stronger is one of these effects than the other?

To make this precise, consider the function $f(\theta) = \lim_{x \to \infty}\limits \left|\Gamma(x e^{i\theta})\right|$; pick an angle, and track the magnitude of $\Gamma$ for inputs of increasing magnitude with that argument in the complex plane. $f(\theta)=\infty, f(\pi/2) = 0$, are there any values for which it's finite and nonzero? Where does it change? Among those values where it tends to infinity, can we say anything about its growth?

I toyed around in Mathematica, specifically with Manipulate[Plot[Abs[Gamma[x*Exp[I*t]]], {x, 0, 50}], {t, 0, Pi/2}]} and found that for a value of $t$ around 1.23 the end behavior seems to change. However when I increase the upper limit for $x$, say Manipulate[Plot[Abs[Gamma[x*Exp[I*t]]], {x, 0, 100}], {t, 0, Pi/2}]}, I need to bring $t$ to about 1.36.

In light of this, I wouldn't be surprised if $f(\theta) = \infty$ for all $\theta < \pi/2$, and looking at the graphs, neither would I be surprised if the growth is always superexponential, even factorial in nature. It's certainly not what I would have expected, I figured there would be some point in $(0, \pi/2)$ where the behavior changes, but it's not entirely out of left field.

Thank you in advance for any possible insight.


1 Answer 1


From Stirling's formula, we see that $$ \left| {\Gamma (x{\rm e}^{{\rm i}\theta } )} \right| \sim \sqrt {\frac{{2\pi }}{x}} {\rm e}^{(\cos \theta) (x\log x - x) - (\theta \sin \theta )x} $$ as $x\to +\infty$ with fixed $|\theta|<\pi$. You can see that as long as $|\theta|<\frac{\pi}{2}$, $\left| {\Gamma (x{\rm e}^{{\rm i}\theta } )} \right| \to +\infty$, since the power of $\mathrm{e}$ becomes positive as soon as $x > {\rm e}^{1 + \theta \tan \theta }$ and then grows faster than $x$. Note that on the imaginary axis, we have an exact formula: $$ \left| {\Gamma ( \pm {\rm i}x)} \right| = \sqrt {\frac{\pi }{{x\sinh (\pi x)}}} \qquad(x>0), $$ giving a more precise description of the exponential decay.

In addition $$ \left| {\Gamma (x + {\rm i}y)} \right| \sim \sqrt {2\pi } \left| y \right|^{x - 1/2} {\rm e}^{ - \pi \left| y \right|/2} $$ for bounded real $x$ as $y\to \pm\infty$.


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