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

[benorin]

I found this problem on http://endeavor.macusa.net/mathpropress/ :

Prove that

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

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

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

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

$$\frac{\Gamma (m)\Gamma (n)}{\Gamma (m+n)}\int_{0}^{1} f(z) (1-z)^{m+n-1} \, dz= \int_{0}^{1}\int_{0}^{1} f(z) (1-z)^{m+n-1}t^{m-1}(1-t)^{n-1} \, dz \, dt$$​

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

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

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

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

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 ?

 

[siddharth]

I give up on this.

You should try the calculus forums, you might have a better chance of getting help there.

 

[shmoe]

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

but now what?

Almost there. Change the order of integration to 'protect' the f(w) (the idea being a change of variables on y won't muck up our f(w)):

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

and work on the inner integral. We'd like to get the interval of integration of the inner integral to go from 0 to 1 rather than w to 1, so try a change of variables on y with this in mind. Or you might want to think of trying to match up the terms with "n" in the exponents between this integral and the right hand side (after you substituted the beta integral that is, the one with the "t" and "z" in it).

 

[benorin]

Thanks shmoe!

Almost there.
$$= \int_{0}^{1} f(w)\left(\int_{w}^{1}\left( y-w \right) ^{m-1}(1-y)^{n-1} \, dy\right) \, dw$$

Almost indeed! Recall (or look it up in the CRC table integrals as I did) that

$$\int_{a}^{b} (y-a)^{m-1}(b-y)^{n-1} \, dy = (b-a)^{m+n-1} B(m,n) = (b-a)^{m+n-1}\frac{\Gamma (m)\Gamma (n)}{\Gamma (m+n)}$$

for $$b>a, \Re (m)>0, \Re (n)>0$$

hence

$$\int_{0}^{1} f(w)\left(\int_{w}^{1}\left( y-w \right) ^{m-1}(1-y)^{n-1} \, dy\right) \, dw = \frac{\Gamma (m)\Gamma (n)}{\Gamma (m+n)}\int_{0}^{1} f(w)(1-w)^{m+n-1} \, dw$$

as required.

Thanks shmoe! you kick-ass.

 

[shmoe]

Happy to help, it's a cute looking problem.

(or look it up in the CRC table integrals as I did) that...

No cheating required!

Try the change of variables t=1-(1-y)/(1-w) on the inner "y" integral.

A similar change of variables would likely prove what you found in CRC, something linear to take the interval [a,b] to [0,1]. (I haven't tried the details on this one but I'd wager it will work)