Home » Math » Dirichlet’s Theorem and Liouville’s Extension of Dirichlet’s Theorem

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

Topic

Beta & Gamma functions

Statement

\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 .

Proof

I haven’t yet learnt about the rigorous proof of the Dirichlet’s Theorem. Here is an alternate proof.  We have to evalute 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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