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Math Help - Is this function integrable?

  1. #1
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    Is this function integrable?

    Is the function f(x,y) = \exp(-xy) integrable over the region

    \{(x,y) : 0 < x < y < x+x^2\}?

    I can show that f is not integrable over the whole positive quadrant, but this seems a lot tricker...I'm not even sure what the answer "should" be (I'm assuming it is integrable though). I guess we could use comparison though or some other technique or trick.
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  2. #2
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    So integrate over y first:

     \int_{ x}^{x+x^2 } e^{-xy} =

    \frac{e^{-x^2 (x+1)} \left(-1+e^{x^3}\right)}{x}

    Now show that this goes to zero faster than something you know is integrable from zero to infinity, like E^-(x+1)

    Limit x->Infinity of  \frac{ \left ( \frac{e^{-x^2 (x+1)} \left(-1+e^{x^3}\right)}{x}  \right )  }{e^{-(x+1)} }

    Do L'Hoptial on the numerator once, then bring down the negative powers and you get an E^(x^3) on top and an E^(x^4) on the bottom. Then this will converge to zero.

    Then the integral is integrable.

    I suppose there are also problems at zero, but take the limit of the function at zero and you get zero so all is well.
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  3. #3
    Member Mentia's Avatar
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    I made a mistake, you dont get E^(x^4), thats obvious. It will still converge though due to the other terms. There might be something better to compare it to though besides E^-(x+1).
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  4. #4
    Member Mentia's Avatar
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    It is even more clear if you just write it as:

    \frac{e^{-x^2 (x+1)} \left(-1+e^{x^3}\right)}{x} = - \frac{1 }{ xe^{x^{2}(1+x)}} +  \frac{ 1}{ xe^{x^{2}}}

    and then just compare it to 1/(x^2) or something similar.
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  5. #5
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    Comparison to 1/x^2 will only show integrability over [1,\infty). I think the real difficulty is showing integrability over (0,1). Indeed I'm not even sure if it integrable over this interval.

    In fact all someone needs to be able to do is to prove whether or not 1/(x\exp(x^2)) is integrable over (0,1) and then we'll be done.
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  6. #6
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    I think you can show that it isn't integrable by making the substitution x^2 = t and then using the fact that 1/(t\exp(t)) is not integrable.
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  7. #7
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    Quote Originally Posted by Mentia View Post
    I suppose there are also problems at zero, but take the limit of the function at zero and you get zero so all is well.
    Quote Originally Posted by HenryB View Post
    Comparison to 1/x^2 will only show integrability over [1,\infty). I think the real difficulty is showing integrability over (0,1). Indeed I'm not even sure if it integrable over this interval.
    Mentia is right, there should be no problem at 0: \int_0^1 \int_x^{x+x^2} e^{-xy} dy dx amounts to integrating a continuous function on a compact subset of \mathbb{R}^2, so it is finite, without computation. It only remains to show that \int_1^\infty \int_x^{x+x^2} e^{-xy} dy dx is finite, and a comparison with \frac{1}{x^2} works fine. In order to simplify the computation, you can even say: for x\geq 1,

    0\leq \int_x^{x+x^2} e^{-xy} dy\leq  \int_x^\infty e^{-xy} dy = \frac{e^{-x^2}}{x}\leq \frac{C}{x^2},

    where C is some constant. So that the integral on x\in[1,+\infty) converges.


    Just for the sake of not letting a question answered: the function  \frac{e^{-x^2}}{x} is clearly not integrable on (0,1]: the numerator converges to 1 as x\to0^+, hence the function is greater than \frac{1/2}{x} for small enough x (or we can say we have the asymptotic equivalence \frac{e^{-x^2}}{x}\sim_{x\to 0^+} \frac{1}{x}), and \frac{1}{x} is not integrable on (0,1].
    So how come I proved the function is integrable ? Just because there is another term than \frac{e^{-x^2}}{x}, and their difference tends to 0 when x\to 0: no divergence.
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