The Gamma Function's Road Through the Fourier Transform

伽马函数的傅里叶变换之路 苏剑林 · 2014-12-08 · Questions complex number Fourier functional equation

The gamma function

$$\Gamma(x)=\int_0^{+\infty}t^{x-1}e^{-t}dt$$

as a generalization of the factorial, tends to confuse many beginners — and I was no exception. The most natural question is: how was such a complicated generalization formula ever arrived at?

In the article "The Magical Gamma Function" on cos.name, there is a fairly detailed account of the history of the gamma function, and reading it carefully taught me a great deal. But one thing I still couldn't find there was a "natural" way of deriving the gamma function. By "natural" I don't mean the simplest, but rather a derivation that starts from some basic properties and definitions and directly solves for the expression of the gamma function. The process and the calculations involved need not be simple, but the underlying idea ought to be direct and clean. Of course, we can't demand that the gamma function was historically born this way, but exploring it after the fact is still worthwhile, since it helps us understand its characteristic properties better. So I tried the following approach, and got some results — but also ran into some puzzles. more

In the real world: it just doesn't work!

As a generalization of the factorial function, the gamma function must naturally satisfy the properties that the factorial function has, such as

$$n!=n\times (n-1)!\tag{1}$$

Let us now define the gamma function $\Gamma(x)$, so that when $x$ is a positive integer we have

$$\Gamma(x)=(x-1)!\tag{2}$$

Then according to $(1)$ we have

$$\Gamma(x+1)=x\Gamma(x)\tag{3}$$

The above equation initially only holds when $x$ is a positive integer, but as a generalization we require that $(3)$ holds for all real numbers (this is precisely the defining property of the gamma function), and in order to obtain a concrete expression for $\Gamma(x)$, we need to solve for $\Gamma(x)$ from $(3)$ — this is a problem of solving a functional equation. Following the idea in "The Art of Operators: Difference, Differentiation, and Bernoulli Numbers", if we write $D\equiv \frac{d}{dx}$, then we can rewrite $(3)$ as

$$e^D\Gamma(x)=x\Gamma(x)\tag{4}$$

We solve this using the Fourier transform. Taking the Fourier transform of both sides, where

$$D\to i\omega,\quad x\to i\frac{d}{d\omega}$$

$(4)$ becomes

$$e^{i\omega}\mathcal{F}[\Gamma(x)]=i\frac{d}{d\omega}\mathcal{F}[\Gamma(x)]$$

This equation is easy to solve, giving

$$\mathcal{F}[\Gamma(x)]=C\exp\left(-e^{i\omega}\right)$$

Finally, applying the inverse transform gives

$$\Gamma(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} C\exp\left(-e^{i\omega}\right)e^{i\omega x}d\omega\tag{5}$$

This seems quite satisfying — but notice that the integral $(5)$ does not yield a well-behaved function (unless $C=0$); for instance, when $x$ is an integer, the integrand is a periodic function! A nonzero periodic function certainly cannot integrate to anything meaningful. Strictly speaking, the integral above does not exist as an ordinary integral — its result can only be regarded as a generalized function (distribution). So this route is a dead end. In fact, this shows that the Fourier transform of $\Gamma(x)$ is not an ordinary function, and hence we cannot study the Fourier transform of $\Gamma(x)$ the way we would study that of an ordinary function — which means there is no way to work backward from its Fourier transform to obtain the gamma function.

In the complex world: light after darkness

There's a phenomenon that recurs often in mathematics: solving some special problem within a limited range can be extraordinarily hard, yet if we generalize the problem and broaden the scope, solving a much larger class of problems all at once turns out to be simpler. The article "Sine Series and Cosine Series", which we discussed not long ago, is one such example. In the present article, we've been treating $\Gamma(x)$ as the extension of the factorial function to general real numbers, but earlier we found no way to solve for $\Gamma(x)$ from equation $(3)$. There is a wonderful trick that lets us extract a meaningful result: simply extend $\Gamma(x)$ into the complex domain, treating it as a complex function $\Gamma(z)$! Doing so lets us solve for $\Gamma(x)$ from equation $(3)$. Having gone through the "mountains blocking every path" of the real-number world, the complex world now offers us the feeling of "light after darkness, a village appears."

Let $x$ be a real variable. Then according to equation $(2)$, for the purely imaginary number $xi$, we have

$$\Gamma\left((x-i)i\right)=\Gamma(x i+1)=x i\Gamma(x i)\tag{6}$$

Likewise, setting $D\equiv\frac{d}{dx}$, and proceeding similarly to $(4)$, we get

$$e^{-i D}\Gamma(x i)=x i\Gamma(x i)\tag{7}$$

Again using the Fourier transform, this time we obtain

$$e^{\omega}\mathcal{F}[\Gamma(x i)]=-\frac{d}{d\omega}\mathcal{F}[\Gamma(x i)]$$

Solving gives

$$\mathcal{F}[\Gamma(x i)]=C\exp\left(-e^{\omega}\right)$$

Taking the inverse transform gives

$$\Gamma(x i)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} C\exp\left(-e^{\omega}\right)e^{i\omega x}d\omega\tag{8}$$

so that

$$\Gamma(z)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} C\exp\left(-e^{\omega}\right)e^{\omega z}d\omega\tag{9}$$

Note that, unlike equation $(5)$, equation $(9)$ is a simple real integral, and it can be further simplified to

$$\begin{aligned}\Gamma(z)=&\frac{1}{2\pi}\int_{-\infty}^{+\infty} C\exp\left(-e^{\omega}\right)e^{\omega z}d\omega\\ =&\frac{C}{2\pi}\int_{-\infty}^{+\infty} \exp\left(-e^{\omega}\right)e^{\omega (z-1)}e^{\omega}d\omega\\ =&\frac{C}{2\pi}\int_{0}^{+\infty} \exp\left(-e^{\omega}\right)e^{\omega (z-1)}de^{\omega}\\ =&\frac{C}{2\pi}\int_{0}^{+\infty} e^{-t} t^{z-1}dt\end{aligned}$$

Finally, using $\Gamma(1)=0!=1$ we can pin down $\frac{C}{2\pi}=1$, giving

$$\Gamma(z)=\int_{0}^{+\infty} e^{-t} t^{z-1}dt\tag{10}$$

Equation $(10)$ is exactly the expression for the gamma function that we set out to find.

This shows that it is only when the gamma function is viewed as a real function of the imaginary part of a complex argument (that is, with the real part of the argument held fixed and the imaginary part treated as the variable) that it possesses a Fourier transform with the properties of an ordinary function's transform.

Summary

This article records the results of my exploratory attempt to work backward to the gamma function by solving the functional equation $(3)$. We've seen the role the Fourier transform plays in this, but we've also seen its shortcomings — some steps require particular tricks before one can push the derivation further. Of course, what matters most is to keep trying and keep exploring; only that way can one truly savor the pleasure of it. The second half of this article — the approach of solving in the complex domain — is something that occurred to me suddenly today, after a long period of fruitless attempts. All in all, keep experimenting, and you'll eventually get somewhere.