Beta function and gamma function

Question

I would like to ask if someone could help me with following equation. \begin{equation} \Gamma(m)\,\Gamma(n) = \int_{0}^{\infty}x^{m-1}e^{-x}\,dx\,\int_{0}^{\infty}y^{n-1}e^{-y}\,dy \end{equation} \begin{equation} \Gamma(m)\,\Gamma(n) = \int_{0}^{\infty}\,\int_{0}^{\infty}x^{m-1}y^{n-1}e^{-(x+y)}\,dx\,dy \end{equation} We use $x = vt$ and $y = v(t-1)$ and in this step I need to explain it more. \begin{equation} \Gamma(m)\,\Gamma(n) = \int_{0}^{1}t^{m-1}(1-t)^{n-1}\,dt\,\int_{0}^{\infty}v^{m+n-1}e^{-v}\,dv \end{equation} Using the definitions of gamma and beta functions, we have \begin{equation} \Gamma(m)\,\Gamma(n) = B(m,n)\,\Gamma(m+n) \end{equation} Thank you in advance.

Answer 1

I would recommend you to work through this paper which introduces the Gamma aswell as the Beta Function quite nicely and also provides a proof of their infamous relation.

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However, the key theorem we are using will working with substitutions within double (or triple, etc.) integrals is a general substitution formula given by

$$\int_{\varphi(U)}f(\mathbf v)\mathrm d\mathbf v~=~\int_Uf(\varphi(\mathbf u))|\det(D\varphi)(\mathbf u)|\mathrm d\mathbf u\tag1$$

Here $\mathbf v$ and $\mathbf u$ are same dimensional vectors used to shorten things up and $\varphi$ is our substitution. What is really intersting about this formula is the differential on the RHS. $|\det(D\varphi)(\mathbf u)|$ refers to the absolute value of the so-called Jacobian Determinant, which is a matrix consisting of partial derivatives of every $\varphi(u_i)$ w.r.t. every $u_i$. Note that for the case that we are dealing with $1$-dimensional functions $(1)$ is just our well-known Integration By Substitution.

Let us get back to the given case. We want to compute the following integral using the subsitutions $x=vt$ and $y=v(1-t)$ (note that there is a little mistake within your post as pointed out by kimchi lover)

$$\int_0^\infty\int_0^\infty x^{m-1}y^{n-1}e^{-(x+y)}\mathrm dx\mathrm dy$$

What we need to compute now are $1.$ the absolute value of the Jacobian Determinant and $2.$ the new borders of integration. The first one is quite simple since we get that

$$|\det(D\varphi(v,t))|=\begin{vmatrix}\frac{\partial}{\partial v}(vt)&\frac{\partial}{\partial t}(vt)\\\frac{\partial}{\partial v}(v(1-t))&\frac{\partial}{\partial t}(v(1-t))\end{vmatrix}=\begin{vmatrix}t&v\\1-t&-v\end{vmatrix}=|-vt-(1-t)v|=|-v|=v$$

Thus, we know that $\mathrm dx\mathrm dy=v~\mathrm dv\mathrm dt$. Since $0<x<\infty$ and $0<y<\infty$ we get by the relation $v=x+y$ and $t=\frac x{x+y}$ that $0<v<\infty$ and $0<t<1$ as our new borders (to be honest this part is tricky sometimes).

Combining what we have accomplished before we finally obtain by $(1)$ that

$$\therefore~\int_0^\infty\int_0^\infty x^{m-1}y^{n-1}e^{-(x+y)}\mathrm dx\mathrm dy=\int_0^1\int_0^\infty (vt)^{m-1}(v(1-t))^{n-1}e^{-v}v~\mathrm dv\mathrm dt$$

From hereon it is rather easy to be finished, i.e. splitting up the double integral, applying the defintions of the Gamma and the Beta Function and you are done.