Finding the Inverse of a Transformation: Solving for u and v

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By some reason I can't access to Latex...

Let be (x, y) = T(u,v) = (uv, v^u)

So x = x(u,v) = uv
and y = y(u,v) = v^u.

I see that T is an injective function, but I can't find u = u(x, y). Can you help me?

Thanks.
 
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This involves a transcendental function. I'd try Mathematica see what it'd come up with, but I am not hopeful.
 
Well let's drop that function, I would thank if you could provide any example of a transformation with its inverse.
 
Suppose x=uv, y=vu^2. Then v=x/u. Substitute into y=vu^2 to get y=xu; solve for u=y/x.
 
OK, so u = u(x,y) = y/x and v = v(x,y) = (x^2)/y.

Thanks, Enuma!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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