Dirichlet’s Theorem and Liouville’s Extension of Dirichlet’s Theorem
The Gamma function is one of the most important special functions in mathematics. It extends the factorial to non-integer values. The Beta function is its close relative, and together they show up everywhere: probability theory, complex analysis, physics, and engineering.
This article covers both functions, their definitions, key properties, and the famous Dirichlet theorem that connects them to multidimensional integrals.
The Gamma Function
You know that \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). But what’s \( 3.5! \)? Or \( (-0.5)! \)? The factorial function only works for non-negative integers.
The Gamma function solves this problem. It’s a smooth curve that passes through all the factorial values and fills in everything in between.
Euler’s Integral Definition
The most common definition uses what’s called Euler’s integral:
$$\Gamma(a) = \int_0^{\infty} e^{-t} t^{a-1} \, dt$$
This integral converges when \( a > 0 \). It’s both an improper integral (upper limit is infinity) and potentially singular at \( t = 0 \) when \( a < 1 \).
The connection to factorials is simple: for positive integers \( n \),
$$\Gamma(n) = (n-1)!$$
Notice the shift. \( \Gamma(5) = 4! = 24 \), not \( 5! \). This off-by-one thing trips people up constantly. Euler defined it this way, and we’re stuck with it.
Euler’s Infinite Product Definition
There’s another way to define the Gamma function that works for all complex numbers (except non-positive integers):
$$\Gamma(a) = \lim_{m \to \infty} \frac{1 \cdot 2 \cdot 3 \cdots m}{a(a+1)(a+2) \cdots (a+m)} m^a$$
This definition extends the Gamma function to negative non-integer values. It shows the function has simple poles at \( 0, -1, -2, -3, \ldots \)
Weierstrass’s Infinite Product Definition
The third definition comes from Weierstrass:
$$\frac{1}{\Gamma(a)} = a e^{a\gamma} \prod_{m=1}^{\infty} \left(1 + \frac{a}{m}\right) e^{-a/m}$$
Here, \( \gamma \) is the Euler-Mascheroni constant:
$$\gamma = \lim_{n \to \infty} \left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} – \ln n\right) \approx 0.5772157$$
This constant shows up everywhere in analysis. Nobody knows if it’s rational or irrational.
All three definitions are equivalent. They give the same values wherever they’re all defined. Use whichever one is most convenient for your problem.
Key Properties of the Gamma Function
The recursion relation is the most useful property:
$$\Gamma(a+1) = a \cdot \Gamma(a)$$
This is what makes it extend the factorial. Just like \( 5! = 5 \times 4! \), we have \( \Gamma(6) = 5 \cdot \Gamma(5) \).
Some specific values worth memorizing:
\( \Gamma(1) = 1 \)
\( \Gamma(1/2) = \sqrt{\pi} \) (this one is famous and useful)
\( \Gamma(n) = (n-1)! \) for positive integers
\( \Gamma(n) = \pm\infty \) for non-positive integers (simple poles)
The reflection formula connects values at \( a \) and \( 1-a \):
$$\Gamma(a)\Gamma(1-a) = \frac{\pi}{\sin(\pi a)}$$
And the duplication formula (Legendre’s formula):
$$\Gamma(a)\Gamma\left(a + \frac{1}{2}\right) = \frac{\sqrt{\pi}}{2^{2a-1}} \Gamma(2a)$$
The Beta Function
The Beta function takes two parameters and is defined as:
$$B(x, y) = \int_0^1 t^{x-1}(1-t)^{y-1} \, dt$$
This integral converges when both \( x > 0 \) and \( y > 0 \).
The Beta function is symmetric: \( B(x, y) = B(y, x) \).
The most important formula connects Beta to Gamma:
$$B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
This relationship is incredibly useful. It lets you convert complicated Beta integrals into Gamma function expressions, which are often easier to evaluate.
Alternative Forms of the Beta Function
The substitution \( t = \sin^2\theta \) gives the trigonometric form:
$$B(x, y) = 2\int_0^{\pi/2} \sin^{2x-1}\theta \cos^{2y-1}\theta \, d\theta$$
The substitution \( t = u/(1+u) \) gives the infinite integral form:
$$B(x, y) = \int_0^{\infty} \frac{u^{x-1}}{(1+u)^{x+y}} \, du$$
Both forms appear regularly in applications.
Dirichlet’s Theorem
Dirichlet’s theorem extends the Beta-Gamma relationship to triple integrals. It’s essential for evaluating certain volume integrals.
Statement
If \( V \) is the region where \( x \geq 0 \), \( y \geq 0 \), \( z \geq 0 \), and \( x + y + z \leq 1 \), then:
$$\iiint_V x^{l-1} y^{m-1} z^{n-1} \, dx \, dy \, dz = \frac{\Gamma(l)\Gamma(m)\Gamma(n)}{\Gamma(l+m+n+1)}$$
The region \( V \) is a tetrahedron with vertices at the origin and the three points \( (1,0,0) \), \( (0,1,0) \), and \( (0,0,1) \).
Proof
We evaluate the integral by setting up the limits carefully. The region is bounded by the coordinate planes and the plane \( x + y + z = 1 \).
$$\int_0^1 \int_0^{1-x} \int_0^{1-x-y} x^{l-1} y^{m-1} z^{n-1} \, dz \, dy \, dx$$
Integrating with respect to \( z \) first:
$$= \int_0^1 \int_0^{1-x} x^{l-1} y^{m-1} \left[\frac{z^n}{n}\right]_0^{1-x-y} dy \, dx$$
$$= \frac{1}{n} \int_0^1 \int_0^{1-x} x^{l-1} y^{m-1} (1-x-y)^n \, dy \, dx$$
Now substitute \( y = (1-x)t \), so \( dy = (1-x) \, dt \). When \( y = 0 \), \( t = 0 \). When \( y = 1-x \), \( t = 1 \).
$$= \frac{1}{n} \int_0^1 \int_0^1 x^{l-1} (1-x)^{m-1} t^{m-1} (1-x)^n (1-t)^n (1-x) \, dt \, dx$$
$$= \frac{1}{n} \int_0^1 \int_0^1 x^{l-1} (1-x)^{m+n} t^{m-1} (1-t)^n \, dt \, dx$$
The integrals separate:
$$= \frac{1}{n} \int_0^1 x^{l-1} (1-x)^{m+n} \, dx \times \int_0^1 t^{m-1} (1-t)^n \, dt$$
These are Beta functions:
$$= \frac{1}{n} \cdot B(l, m+n+1) \cdot B(m, n+1)$$
Converting to Gamma functions using \( B(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} \):
$$= \frac{1}{n} \cdot \frac{\Gamma(l)\Gamma(m+n+1)}{\Gamma(l+m+n+1)} \cdot \frac{\Gamma(m)\Gamma(n+1)}{\Gamma(m+n+1)}$$
Since \( \Gamma(n+1) = n \cdot \Gamma(n) \), we have:
$$= \frac{1}{n} \cdot \frac{\Gamma(l)\Gamma(m+n+1)}{\Gamma(l+m+n+1)} \cdot \frac{\Gamma(m) \cdot n \cdot \Gamma(n)}{\Gamma(m+n+1)}$$
The \( n \) cancels, and \( \Gamma(m+n+1) \) cancels:
$$= \frac{\Gamma(l)\Gamma(m)\Gamma(n)}{\Gamma(l+m+n+1)}$$
Which completes the proof. ∎
Liouville’s Extension of Dirichlet’s Theorem
Liouville generalized Dirichlet’s theorem to handle functions of \( x + y + z \) and different boundary conditions.
If \( x, y, z \) are all positive and \( h_1 < x + y + z < h_2 \), then:
$$\iiint_V x^{l-1} y^{m-1} z^{n-1} F(x+y+z) \, dx \, dy \, dz = \frac{\Gamma(l)\Gamma(m)\Gamma(n)}{\Gamma(l+m+n)} \int_{h_1}^{h_2} F(h) h^{l+m+n-1} \, dh$$
This is powerful. It reduces a triple integral involving \( F(x+y+z) \) to a single integral in \( h \). The geometric complexity gets absorbed into the Gamma function ratio.
Special Case
When \( F = 1 \), \( h_1 = 0 \), and \( h_2 = 1 \), Liouville’s extension reduces to Dirichlet’s original theorem (with a slight adjustment in the Gamma function argument).
Applications
These functions aren’t just mathematical curiosities. They appear constantly in real problems.
The Beta distribution in statistics uses the Beta function. It models probabilities of probabilities (like conversion rates, click-through rates, or any proportion with uncertainty).
The Gamma distribution models waiting times. It’s everywhere in queuing theory, reliability engineering, and Bayesian statistics.
In physics, the Gamma function appears in the volume of n-dimensional spheres, partition functions in statistical mechanics, and Feynman integrals in quantum field theory.
String theory uses the Beta function extensively. The Veneziano amplitude (one of the first string theory results) is literally a Beta function.
Suggested Reading
Calculus Made Easy
Silvanus P. Thompson (Author), Martin Gardner (Author)
Hardcover : 336 pages
Frequently Asked Questions
What is the Gamma function?
The Gamma function extends the factorial to all complex numbers except non-positive integers. It’s defined by the integral Γ(a) = ∫₀^∞ e^(-t) t^(a-1) dt. For positive integers, Γ(n) = (n-1)!, so Γ(5) = 24 = 4!. The off-by-one relationship to factorials is a common source of confusion.
What is the Beta function?
The Beta function B(x,y) is defined as ∫₀¹ t^(x-1)(1-t)^(y-1) dt. It’s symmetric (B(x,y) = B(y,x)) and related to the Gamma function by B(x,y) = Γ(x)Γ(y)/Γ(x+y). This relationship makes it useful for evaluating complicated integrals.
Why is Γ(1/2) equal to √π?
Substituting a = 1/2 into the Gamma integral and using u = √t gives Γ(1/2) = 2∫₀^∞ e^(-u²) du. This is the Gaussian integral, which equals √π. This result connects the Gamma function to probability theory and the normal distribution.
What is the Euler-Mascheroni constant?
The Euler-Mascheroni constant γ ≈ 0.5772 is defined as the limit of (1 + 1/2 + 1/3 + … + 1/n – ln n) as n approaches infinity. It appears in the Weierstrass definition of the Gamma function and throughout analysis. Whether it’s rational or irrational remains an open problem.
What is Dirichlet’s theorem for integrals?
Dirichlet’s theorem states that the triple integral of x^(l-1)y^(m-1)z^(n-1) over the region where x,y,z ≥ 0 and x+y+z ≤ 1 equals Γ(l)Γ(m)Γ(n)/Γ(l+m+n+1). It extends the Beta-Gamma relationship to three dimensions and is useful for evaluating volume integrals.
How are Beta and Gamma functions related?
The fundamental relationship is B(x,y) = Γ(x)Γ(y)/Γ(x+y). This lets you convert Beta function integrals into Gamma function expressions. Since Gamma function values are well-tabulated and have known properties, this conversion often simplifies calculations significantly.
What is Liouville’s extension of Dirichlet’s theorem?
Liouville extended Dirichlet’s result to integrals involving a function F(x+y+z). The triple integral becomes [Γ(l)Γ(m)Γ(n)/Γ(l+m+n)] times a single integral of F(h)h^(l+m+n-1). This reduces three-dimensional integrals to one-dimensional ones when the integrand depends on the sum of variables.
Where is the Gamma function used in physics?
The Gamma function appears in the volume formula for n-dimensional spheres, partition functions in statistical mechanics, Feynman integrals in quantum field theory, and solutions to differential equations in quantum mechanics. The Beta function famously appears in the Veneziano amplitude in string theory.
What is the recursion formula for Gamma?
The recursion formula is Γ(a+1) = a·Γ(a). This mirrors the factorial property n! = n·(n-1)!. Combined with Γ(1) = 1, it lets you compute Γ(n) for any positive integer as (n-1)!. The formula also extends Gamma to negative non-integers by rearranging: Γ(a) = Γ(a+1)/a.
Why does Γ(n) = (n-1)! instead of n!?
Euler originally defined it this way, and the convention stuck. Some authors use Π(n) = n! = Γ(n+1) to avoid the shift, but Γ remains standard. The off-by-one relationship means Γ(5) = 4! = 24, not 120. Always double-check which convention a source uses.
