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Math Help - tricky double integral

  1. #1
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    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?).
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  2. #2
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    I think your order of integration is wrong.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    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 e^{b^2x^2} or 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.
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  4. #4
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    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, max(a^2x^2, b^2y^2)= a^2x^2 and in the other 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.
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