1. ## tricky double integral

Evaluate:

where a and b are positive real numbers. (Here max(s,t) is defined as the larger of the two numbers s and t.)

how do i even begin this? I am not familiar with this max(s,t) function, how does one determine which is greater in this case (depends on x and y, right?).

2. I think your order of integration is wrong.

3. Originally Posted by Prove It
I think your order of integration is wrong.
I'm confused as to how switching the order of integration would help.

Regardless of the order of integration, doesn't $\displaystyle e^{b^2x^2}$ or $\displaystyle e^{a^2y^2}$ have to be integrated, which I know can't be done with elementary functions and I don't know how to do it.

4. a^2x^2> b^2y^2, with all numbers positive, if and only if ax/b> y

You will need to break this into two double integrals will. We have x going from 0 to a and, for each x, y going from 0 to b. Instead, do one integral with y going from 0 to ax/b and another with y going from ax/b to b. In the first integral, $\displaystyle max(a^2x^2, b^2y^2)= a^2x^2$ and in the other $\displaystyle max(a^2x^2, b^2y^2)= b^2y^2$.

Once you have done that, see if changing the order of integration doesn't help.