The zeta Function, Defining gamma, Continuous Factorials

The gamma Function

The gamma function, developed by Euler (biography), is a continuous extension of the factorials. If you arrived here from a search engine, you might think this an odd place to describe the gama function, under sequences and series. However, γ(s) is related to the zeta function ζ(s), which is itself defined as a series, so I hope it all hangs together.

The function γ(s) is the integral, as x runs from 0 to infinity, of E-xxs-1.

Note that γ(1) = 1, and using integration by parts, γ(s+1) = s×γ(s). Therefore s! may be defined as γ(s+1), for all real s ≥ 0.

For our first noninteger, let s = ½. Substitute x = u2, and integrate 2E-u2, from 0 to ∞. This is the same as integrating E-u2 from -∞ to ∞. This integral is solved by a clever trick, hence γ(½) = sqrt(π). Moving to factorials, ½! = γ(1+½), which is ½γ(½), or sqrt(π)/2. Next time this question comes up at a party - what is ½ factorial - you'll know the answer.

What is γ(0)? Near x = 0, the integrand is bounded below by 1 over 2x, hence the integral is not defined. Thus gamma is not defined at 0. By the same reasoning, gamma is not defined for the negative integers.

Let s lie between 0 and 1. Now γ(s) is dominated by the integral of xs-1, and from 0 to 1, this is well defined. From 1 to infinity, γ(s) is dominated by the integral of e-x, which is also finite. Thus gamma is defined for all s between 0 and 1, including s = 1.

Use integration by parts, as described above, to advance s by 1. Repeat this process, and gamma is defined for all s > 0.

Complex Plane

Assume gamma converges for some s, and consider γ(s+i). (We're moving up in the complex plane.) This multiplies the integrand by xi. Raising a real number to an imaginary power yields a point on the unit circle, having norm 1. For each x, the norm of the integrand has not changed; we're just pushing it around the unit circle. As x runs from 0 to infinity, the real component, and the imaginary component, of γ(s+i), are bounded by the original integrand. Both components are integrable, and γ(s+i) is well defined. This holds for all multiples of i, hence gamma is defined on the entire half plane re(s) > 0.

Note that γ(i+1) is not i×γ(i). Attempt integration by parts, as we did before. We need to evaluate e-xxi at 0 and infinity. Write xi as elog(x)i, and for x near 0, the point spins around the unit circle at break neck speed. There is no limit, and integration by parts fails.

Move to the right, and everything works again. For instance, γ(3+i) is 2+i times γ(2+i). The limit of x2+i, at 0, is the madly spinning xi, squashed down to the origin by x2, so it all works out.

Analytic for s > 1

Fix an s whose real component is greater than 1, and consider a small closed disk about s, greater than 1. The integrand that defines gamma is continuous in s cross x. This is pretty clear, since it is a mathematical formula, but you may want to verify it for x = 0. The curves near s look like xs-1. For example, if s = 3/2, the disk about s defines curves that look like sqrt(x), at least near 0. These all approach 0 uniformly. As s strays into imaginary territory the functions spin around in the complex plane; but they all spiral towards 0 uniformly. We have continuity over the entire 3 dimensional cylinder produced by the closed disk and x ≥ 0. Furthermore, for each x, the integrand becomes analytic in s. With these conditions in place, gamma becomes analytic, and we can interchange symbols to find the first derivative. Taking the derivative with respect to s multiplies xs-1 by log(x). The new integrand is E-xlog(x)xs-1. Integrate with respect to x and find γ(s)′. (No, I don't expect you to perform the integration; it's pretty much a theoretical result.) This can be repeated to characterize higher derivatives.

Analytic for s > 0

In fact, gama is analytic on re(s) > 0, although the above proof is only valid for re(s) > 1. Let's try to leverage what we have already done.

Let c be 1/n, and let gn(s) be the integral of E-xxs-1, as x runs from c to infinity. This is almost the gamma function, except we start integrating at c instead of 0.

Use the reasoning outlined above to show that gn(s) is analytic on the half plane to the right of 0. Thus gn(s) is a sequence of analytic functions that approaches γ(s). If convergence is locally uniform, we're done.

Fix a point p in the complex plane to the right of 0, and choose a radius r so that the closed disk, centered at p, with radius r, lies to the right of 0. The difference between gn(s) and γ(s) is given by an integral from 0 to c. As n increases, c decreases. The integral is largest when the integrand is everywhere positive. This happens when s is real, so slide the disk down until it straddles the x axis. From there, the integral is largest when s is small. The left edge of our disk, s = p-r, is the worst case. Therefore gn approaches γ uniformly over a closed disk, and each gn is analytic, hence γ is analytic.