The Gamma function \Gamma(x)=\int_0^{+\infty}t^{x-1}e^{-t}dt as a generalization of the factorial, often leaves beginners feeling confused, and the author was no exception. A most natural question is: how was such a complex generalization formula obtained?
In the article "The Magical Gamma Function" on cos.name, there is a relatively detailed introduction to the history of the Gamma function, from which the author benefited greatly. However, the slight disappointment was that I still could not find a "natural" way to derive the Gamma function from it. By "natural," I do not mean the simplest way, but rather a method that directly solves for the expression of the Gamma function based on some basic properties and definitions. The process and calculations might not be simple, but the underlying idea should be direct and concise. Of course, we cannot demand that the Gamma function was historically born this way, but it is beneficial as a post-hoc exploration to help us understand its characteristics. Thus, I attempted the following approach and obtained some results, along with some perplexities.
In the Real World: It Doesn’t Work!
As a generalization of the factorial function, the Gamma function must naturally satisfy the properties of the factorial. For instance, the factorial function satisfies n! = n \times (n-1)! \tag{1} Now we define the Gamma function \Gamma(x), which for positive integers x satisfies \Gamma(x) = (x-1)! \tag{2} Thus, according to (1), we have \Gamma(x+1) = x\Gamma(x) \tag{3} Initially, this equation holds only for positive integers x. As a generalization, we require (3) to hold for all real numbers (which is indeed a property of the Gamma function). To find the specific expression for \Gamma(x), we must solve for \Gamma(x) from (3), which is a problem of solving a functional equation. Following the ideas in "The Art of Operators: Differences, Derivatives, and Bernoulli Numbers", let D \equiv \frac{d}{dx}. We can then write (3) as e^D \Gamma(x) = x \Gamma(x) \tag{4} To solve this, we apply the Fourier transform to both sides, where D \to i\omega, \quad x \to i\frac{d}{d\omega} Thus, (4) becomes e^{i\omega} \mathcal{F}[\Gamma(x)] = i\frac{d}{d\omega} \mathcal{F}[\Gamma(x)] This equation is easy to solve, with the solution being \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 perfect, but please note that the integral (5) will not yield a well-behaved function (unless C=0). For example, when x is an integer, the integrand is a periodic function! A non-zero periodic function naturally does not integrate into a meaningful result. Strictly speaking, the aforementioned integral does not exist; its result can only be considered a generalized function (distribution). Therefore, this path failed. In fact, this indicates that the Fourier transform of \Gamma(x) is not an ordinary function, so one cannot study the Fourier transform of \Gamma(x) as one would an ordinary function, and thus there is no way to derive the Gamma function back from its Fourier transform in this manner.
In the Complex World: A Breakthrough
There is often a phenomenon in mathematics: solving a specific problem within a certain range appears extremely difficult, but once the problem is generalized and the range expanded to solve more problems at once, it becomes much more concise. A recent example we discussed is "Sine Series and Cosine Series". In this article, we viewed \Gamma(x) as a generalization of the factorial function to general real numbers, but we were unable to solve for \Gamma(x) from equation (3). There is a wonderful trick that allows us to obtain meaningful results: simply extend \Gamma(x) into the complex domain and treat it as a function \Gamma(z) of a complex variable! Doing so allows us to solve for \Gamma(x) from equation (3). After experiencing a "dead end" in the real world, the complex world gives us a sense of "finding a new path."
Let x be a real variable. Then according to equation (2), for the pure imaginary number xi, we have \Gamma((x-i)i) = \Gamma(xi+1) = xi \Gamma(xi) \tag{6} Similarly, let D \equiv \frac{d}{dx}. Then, analogous to (4), we have e^{-iD} \Gamma(xi) = xi \Gamma(xi) \tag{7} Using the Fourier transform again, we now obtain e^{\omega} \mathcal{F}[\Gamma(xi)] = -\frac{d}{d\omega} \mathcal{F}[\Gamma(xi)] Solving this yields \mathcal{F}[\Gamma(xi)] = C \exp\left(-e^{\omega}\right) Applying the inverse transform gives \Gamma(xi) = \frac{1}{2\pi} \int_{-\infty}^{+\infty} C \exp\left(-e^{\omega}\right) e^{i\omega x} d\omega \tag{8} Then \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, which can be further simplified: \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)} d e^{\omega} \\ &= \frac{C}{2\pi} \int_{0}^{+\infty} e^{-t} t^{z-1} dt \end{aligned} Finally, using \Gamma(1) = 0! = 1, we determine \frac{C}{2\pi} = 1, thus \Gamma(z) = \int_{0^{+\infty}} e^{-t} t^{z-1} dt \tag{10} Equation (10) is exactly the expression for the Gamma function we were seeking.
This indicates that the Gamma function, when viewed as a real function of the imaginary part of a complex variable (that is, fixing the real part of the variable and treating the imaginary part as the variable), possesses a Fourier transform that behaves like an ordinary function.
Summary
This article is the result of the author’s exploratory attempts. By solving the functional equation (3) to derive the Gamma function, we have seen the role of the Fourier transform, but also observed certain limitations where specific tricks are required to proceed. Of course, what is more important is the continuous attempt and exploration to truly appreciate the joy within. The latter part of this article—the idea of solving in the complex domain—was also conceived today after a long period of fruitless attempts. In general, by trying more, there will always be gains.
When reposting, please include the original address: https://kexue.fm/archives/3108
For more detailed reposting matters, please refer to: “Scientific Space FAQ”