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Prove that

[tex]\int_{0}^{1}\int_{0}^{1} f(xy)(1-x)^{m-1}y^{m}(1-y)^{n-1} \, dx \, dy = \frac{\Gamma (m)\Gamma (n)}{\Gamma (m+n)}\int_{0}^{1} f(z)(1-z)^{m+n-1} \, dz[/tex]

where f(xy) is an arbitrary function of xy.

My work: The beta integral gives [tex] B(m,n) =\frac{\Gamma (m)\Gamma (n)}{\Gamma (m+n)}=\int_{0}^{1}t^{m-1}(1-t)^{n-1} \, dt[/tex]

hence the the righthand-side of the given problem becomes:

[tex]\frac{\Gamma (m)\Gamma (n)}{\Gamma (m+n)}\int_{0}^{1} f(z) (1-z)^{m+n-1} \, dz= \int_{0}^{1}\int_{0}^{1} f(z) (1-z)^{m+n-1}t^{m-1}(1-t)^{n-1} \, dz \, dt [/tex]

also, as for the lefthand-side of the given problem I have considered applying the change of variables [tex]w=xy\Rightarrow dw = y dx[/tex] so that [tex] 0\leq x\leq 1\Rightarrow 0\leq w\leq y[/tex] and the integral becomes

[tex]\int_{0}^{1}\int_{0}^{1} f(xy)(1-x)^{m-1}y^{m}(1-y)^{n-1} \, dx \, dy = \int_{0}^{1}\int_{0}^{y} f(w)\left( 1-\frac{w}{y} \right) ^{m-1}y^{m}(1-y)^{n-1} \, \frac{dw}{y} \, dy [/tex]

[tex]= \int_{0}^{1}\int_{0}^{y} f(w)\left( y-w \right) ^{m-1}(1-y)^{n-1} \, dw \, dy[/tex]

but now what? the alternative change of variables [tex]w=xy\Rightarrow dw = x dy[/tex] so that [tex] 0\leq y\leq 1\Rightarrow 0\leq w\leq x[/tex] yields instead this

[tex]\int_{0}^{1}\int_{0}^{1} f(xy)(1-x)^{m-1}y^{m}(1-y)^{n-1} \, dx \, dy = \int_{0}^{1}\int_{0}^{x} f(w)(1-x)^{m-1} \left( \frac{w}{x} \right) ^{m}\left( 1-\frac{w}{x} \right) ^{n-1} \, \frac{dw}{x} \, dx [/tex]

which also seems to be somewhat lacking. Should I try expanding by the binomial theorem (I do not assume that m and n are integers, but rather that their real parts are > 0) ?

Any thoughts? A transformation of variables of the form T:{u=u(x,y), v=v(u,v)} perhaps ?

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# Double int needs a good T-form, got one?

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