Dirichlet's Theorem and Liouville's Extension of Dirichlet's Theorem

Topic: Beta & Gamma functions

The Gamma function and Beta functions belong to the category of special transcendental functions and are defined in terms of improper definite integrals. Definitions of Beta and Gamma functions are given below.

But before that, let's quote the statement of Dirichlet's theorem so that we can work around Liouville's extension of Dirichlet's theorem.

Statement of Dirichlet's Theorem

$ \int \int \int_{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)} } $ ,
where V is the region given by $ x \ge 0 y \ge 0 z \ge 0 x+y+z \le 1 $ .

Gamma and Beta Functions

As I told earlier, the Gamma and Beta functions are special transcendental functions defined in terms of improper definite integrals.

Gamma Function

If we consider the integral $ I =\displaystyle{\int_0^{\infty}} e^{-t} t^{a-1} \mathrm dt$ , it is once seen to be an infinite and improper integral.

This integral is infinite because the upper limit of integration is infinite, and it is improper because $ t=0$ is a point of infinite discontinuity of the integrand, if $ a<1$ , where $ a$ is either a real number or real part of a complex number.

This integral is known as Euler's Integral.

Euler's Integral is of great importance in mathematical analysis and calculus.

The result, i.e., integral, is defined as a new function of real number $ a$ , as $ \Gamma (a) =\displaystyle{\int_0^{\infty}} e^{-t} t^{a-1} \mathrm dt$ .

Three Definitions for Gamma Function

Let $ a$ be any positive real number, then we can integrate the Eulerian Integral $ I =\displaystyle{\int_0^{\infty}} e^{-t} t^{a-1} \mathrm dt$ by assuming $ t^{a-1}$ as first function and $ e^{-t}$ as second function, integrating it by parts.

After a little work, one might get
$ I =\displaystyle{\int_0^{\infty}} e^{-t} t^{a-1} \mathrm dt=(a-1)(a-2) \ldots 2\cdot 1$ .

This is defined as gamma function of $ a$ (i.e., $ \Gamma (a)$ ) and $ \Gamma (a) =\displaystyle{\int_0^{\infty}} e^{-t} t^{a-1} \mathrm dt =(a-1)(a-2) \ldots 2 \cdot 1$ .


If $ a$ is a positive integer, then we can write $ \Gamma a=(a-1)!$ .

This definition is not defined for Gamma Function for negative numbers and zero. The second definition of Gamma Function is given in terms of Euler's infinite limit
$ \Gamma (a)=\displaystyle{\lim_{m \to \infty}} \dfrac{1\cdot 2 \cdot 3 \cdots m}{a(a+1)(a+2) \ldots (a+m)} m^a$ , where $ a$ be either real or complex number.


The third definition of the Gamma function is given in terms of Weierstrass's infinite product, as $ \Gamma (a)$ for any number $ a$ is,
$ \dfrac{1}{\Gamma a} =a e^{a \gamma} \displaystyle{\prod_{m=1}^{\infty}} \left({1+\frac{m}{a}}\right) e^{-a/m}$ ; where $ \gamma =\lim_{n \to {\infty}} \left({1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n}-\log n}\right)$ is a constant, called Euler-Mascheroni Constant and its value is approximately $ 0.5772157$ .


All three definitions defined above are equivalent to each other.

Important Properties of Gamma Functions

  1. If $ n$ is a positive integer:
    $ \Gamma (n)=(n-1)!= \\ (n-1)(n-2)(n-3)\ldots 2\cdot 1$
  2. If $ n$ is a negative integer or zero: $ \Gamma (n)=\infty$
  3. If $ n$ is a non-integer number, then $ \Gamma (n)$ exists and has a finite value.

Beta Function

The beta function of two values $x$ and $y$ is defined as an integral:

$$ \displaystyle B \left({x, y}\right) := \int_{\mathop 0}^{\mathop 1} t^{x - 1} \left({1 - t}\right)^{y - 1} \ \mathrm d t $$

In terms of Gamma Functions it is defined by:

$$ B \left({x, y}\right) := \dfrac {\Gamma \left({x}\right) \Gamma \left({y}\right)} {\Gamma \left({x + y}\right)} $$

Suggested Reading

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Calculus Made Easy

Silvanus P. Thompson (Author), Martin Gardner (Author)

Hardcover ‏ : ‎ 336 pages

Proof of Dirichlet's Theorem

The rigorous proof of Dirichlet's Theorem is a tough thing to handle. So, here is an alternate and easy proof.

We have to evaluate the given triple integral over the volume enclosed by the three coordinate planes and the plane $ x+y+z=1$.

Hence we may write the given triple integral as

$ \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} x^{l-1} y^{m-1} z^{n-1} dx dy dz $

=$ \int_{0}^{1} ! \int_{0}^{1-x} ! x^{l-1} y^{m-1} {[z^{n}/n]}_{0}^{1-x-y} dx dy$

=$ \frac {1}{n} \int_{0}^{1} \int_{0}^{1-x} x^{l-1} y^{m-1} {(1-x-y)}^{n} dx dy$

Now substituting $ y=(1-x)t$ or $ dy=(1-x) dt$ we get

$= \frac {1}{n} \int_{0}^{1} \int_{0}^{1} x^{l-1} {(1-x)}^{m-1} t^{m-1} {[1-x-(1-x)t]}^{n} (1-x) \\ dx dt$

$= \frac {1}{n} \int_{0}^{1} \int_{0}^{1} x^{l-1} {(1-x)}^{m-1} t^{m-1} {(1-x)}^{n} {(1-t)}^{n} (1-x) \\ dx dt$

$= \frac {1}{n} \int_{0}^{1} \int_{0}^{1} x^{l-1} {(1-x)}^{m+n} t^{m-1} {(1-t)}^{n} dx dt$

$= \frac {1}{n} \times \int_{0}^{1} x^{l-1} {(1-x)}^{m+n} dx \times \int_{0}^{1} t^{m-1} {(1-t)}^n dt $

=$ \frac {B(l m+n+1) \times B(m n+1)}{n} $

=$ \frac {1} {n} \times \frac { \Gamma {(l)} \Gamma {(m+n+1)} } { \Gamma {(l+m+n+1)} } $ $ \times $ $ \frac { \Gamma{(m)} \Gamma {(n+1)} } { \Gamma {(m+n+1)} } $

=$ \frac { \Gamma {(l)} \Gamma {(m)} \Gamma {(n)} }{ \Gamma{(l+m+n+1)} } $
which is the statement.

Liouville's Extension of Dirichlet's Theorem

If $ x, y, z$ are all positive such that $ h_1 < (x+y+z) < h_2 $ then

$\int \int \int_{V} x^{l-1} y^{m-1} z^{n-1} F (x,y,z) dx dy ,dz$

$=\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 the Liouville's Extension of Dirichlet's Theorem.