How do I find ∫(0,inf)∫(0,inf){y f(x,y)}dxdy, given the following f(x,y)

f(x,y) = exp( - (x^2 + y^2 - 2Rxy) / 2(1-R^2) )

(-1
I've already worked out ∫(-inf,inf) ∫(-inf,inf){f(x,y)} dxdy = 2pi for R=0,
and shown that ∫(-inf,inf){exp(-ax^2)} dx = sqrt(pi/a) for R=0 and a>0.
I'm allowed to use these results in my solution.

What I'm having trouble with now is that I need to find ∫(0,inf) instead of ∫(-inf,inf).

PS. Sorry, this is again related to a question I posted a few days ago. I'm sorry I'm so clueless at calculus. I'd appreciate your help.

Given ∫(y = 0 to ∞) ∫(x = 0 to ∞) y f(x,y) dx dy:
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(i) R = 0 is easy:

We obtain
∫(y = 0 to ∞) ∫(x = 0 to ∞) y exp(-(x^2 + y^2)/2) dx dy
= ∫(y = 0 to ∞) y exp(-y^2/2) dy * ∫(x = 0 to ∞) exp(-x^2/2) dx
= ∫(y = 0 to ∞) y exp(-y^2/2) dy * (1/2) ∫(-∞ to ∞) exp(-x^2/2) dx
= [-exp(-y^2/2) {for y = 0 to ∞}] * (1/2) √(2π)
= 1 * (1/2) √(2π)
= √(π/2).
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(ii) Assume that R is non-zero.

By completing the square, x^2 + y^2 - 2Rxy = (x - Ry)^2 + (1 - R^2) y^2.
So, f(x,y) = exp {[-(x - Ry)^2 - (1 - R^2)y^2] / (2(1 - R^2))}.

Let's make the change of variable u = x - Ry, v = y <==> x = u + Rv, y = v

This has Jacobian ∂(x,y)/∂(u,v) =
|1 R|
|0 1| = 1.

Also, the first quadrant transforms as follows:
x = 0 ==> u = -Ry = -Rv or v = -u/R. [This is why I did R = 0 separately above.]
y = 0 ==> v = 0.

Assuming that R < 0 so that the slope of v = -u/R is positive.
(The case where R > 0 is handled similarly.)

So, the integral transforms to
∫(u = 0 to ∞) ∫(v = 0 to -u/R) v exp {(-u^2 - (1 - R^2) v^2) / (2(1 - R^2))} * 1 dv du
= ∫(u = 0 to ∞) [exp {-u^2/ (2(1 - R^2))} ∫(v = 0 to -u/R) v exp(-v^2/2) dv] du
= ∫(u = 0 to ∞) exp {-u^2/ (2(1 - R^2))} * [-exp(-v^2/2) {for v = 0 to -u/R}] du
= ∫(u = 0 to ∞) exp {-u^2/ (2(1 - R^2))} * [1 - exp(-u^2/(2R^2)] du
= ∫(u = 0 to ∞) [exp {-u^2/ (2(1 - R^2))} - exp {-u^2/(2R^2 (1 - R^2))}] du
= (1/2) ∫(-∞ to ∞) [exp {-u^2/ (2(1 - R^2))} - exp {-u^2/(2R^2 (1 - R^2))}] du
= (1/2) [√(π * 2(1 - R^2)) - √(π * 2R^2 (1 - R^2))]
= √(π/2) (1 - |R|) √(1 - R^2).

Note/Sanity Check: Letting R → 0 yields the first result.
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I hope this helps!

ROFL WHAT

Just draw a sad crying face as your answer.