pouzdrotf

Answered

2022-07-02

We are given an integrable function $f:[0,1]\to \mathbb{R}$ and a set $A=\{(x,y)\in {\mathbb{R}}^{2}|x,y\ge 0,x+y\le 1\}$. I need to show that for $\alpha ,\beta >0$ the following holds:

${\int}_{A}f(x+y){x}^{\alpha}{y}^{\beta}d{\lambda}^{2}(x,y)={\int}_{0}^{1}{y}^{\alpha}(1-y{)}^{\beta}d\lambda (y){\int}_{0}^{1}f(x){x}^{\alpha +\beta +1}d\lambda (x).$

We are provided with the hint to use the following map $\varphi :[0,1{]}^{2}\to {\mathbb{R}}^{2},\varphi (x,y)=(xy,x(1-y)).$. I do not see how to use the hint. It is clear that in addition to $f$ all other functions that one finds in the integrals on both sides are integrable on the set $A$ and their product as well. Thus, once one can figure out how to separate the variables, the claim will follow by using the Fubini theorem.

Can somebody provide some insight or a solution proposal? Thanks.

${\int}_{A}f(x+y){x}^{\alpha}{y}^{\beta}d{\lambda}^{2}(x,y)={\int}_{0}^{1}{y}^{\alpha}(1-y{)}^{\beta}d\lambda (y){\int}_{0}^{1}f(x){x}^{\alpha +\beta +1}d\lambda (x).$

We are provided with the hint to use the following map $\varphi :[0,1{]}^{2}\to {\mathbb{R}}^{2},\varphi (x,y)=(xy,x(1-y)).$. I do not see how to use the hint. It is clear that in addition to $f$ all other functions that one finds in the integrals on both sides are integrable on the set $A$ and their product as well. Thus, once one can figure out how to separate the variables, the claim will follow by using the Fubini theorem.

Can somebody provide some insight or a solution proposal? Thanks.

Answer & Explanation

Nirdaciw3

Expert

2022-07-03Added 20 answers

The transformation $(x,y)\mapsto (x+y,\frac{x}{x+y})=(u,v)$ maps $A$ to $(0,1{)}^{2}$ bijectively and has inverse $(u,v)\mapsto (uv,u(1-v))=(x,y)$. Moreover,

$\frac{\mathrm{\partial}(u,v)}{\mathrm{\partial}(x,y)}=-\frac{1}{x+y}=-\frac{1}{u}$

Therefore,

$\begin{array}{rcl}{\iint}_{A}f(x+y){x}^{\alpha}{y}^{\beta}\mathrm{d}A& =& {\iint}_{(0,1{)}^{2}}f(u)(uv{)}^{\alpha}{[u(1-v)]}^{\beta}u\mathrm{d}A\\ & =& {\iint}_{(0,1{)}^{2}}f(u){u}^{\alpha +\beta +1}{v}^{\alpha}(1-v{)}^{\beta}\mathrm{d}A\\ & =& {\int}_{0}^{1}f(x){x}^{\alpha +\beta +1}\mathrm{d}x\cdot {\int}_{0}^{1}{x}^{\alpha}(1-x{)}^{\beta}\mathrm{d}x\end{array}$

$\frac{\mathrm{\partial}(u,v)}{\mathrm{\partial}(x,y)}=-\frac{1}{x+y}=-\frac{1}{u}$

Therefore,

$\begin{array}{rcl}{\iint}_{A}f(x+y){x}^{\alpha}{y}^{\beta}\mathrm{d}A& =& {\iint}_{(0,1{)}^{2}}f(u)(uv{)}^{\alpha}{[u(1-v)]}^{\beta}u\mathrm{d}A\\ & =& {\iint}_{(0,1{)}^{2}}f(u){u}^{\alpha +\beta +1}{v}^{\alpha}(1-v{)}^{\beta}\mathrm{d}A\\ & =& {\int}_{0}^{1}f(x){x}^{\alpha +\beta +1}\mathrm{d}x\cdot {\int}_{0}^{1}{x}^{\alpha}(1-x{)}^{\beta}\mathrm{d}x\end{array}$

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