Dirichlet’s Theorem and Liouville’s Extension of Dirichlet’s Theorem

Topic

Beta & Gamma functions

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 $ .

Brief Theory on Gamma and Beta Functions

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 real number or real part of a complex number. This integral is known as Euler’s Integral. This is of a 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$ .

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 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.

Third definition of 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 the 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:

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


Proof

I haven’t yet learned about the rigorous proof of the Dirichlet’s Theorem. Here is an alternate 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 substuting $ 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)} } $
i.e. 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 $

 

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2 comments
  1. Liouville’s Extension of Dirichlet’s Theorem
    Under the integral, it’s not F(x,y,z)
    It should be F(x+y+z)

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