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Math Help - Double integrals

  1. #1
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    Double integrals

    Find a solution to this integral I = \int_0^{\infty} e^{-x^2} dx analytically by calculating I^2 = \int_0^{\infty} \int_0^{\infty} e^{-x^2} e^{-y^2} dydx.

    I am not sure about how to go about this question, how did they get I^2 = \int_0^{\infty} \int_0^{\infty} e^{-x^2} e^{-y^2} dydx? Shouldn't I^2 = \int_0^{\infty} e^{-x^2} dx \times \int_0^{\infty} e^{-x^2} dx = \int_0^{\infty}\int_0^{\infty} e^{-x^2} \cdot e^{-x^2} dxdy?

    Thanks very much! (Just a bit confused on this question because I am working ahead of class )
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  2. #2
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    Quote Originally Posted by usagi_killer View Post
    Find a solution to this integral I = \int_0^{\infty} e^{-x^2} dx analytically by calculating I^2 = \int_0^{\infty} \int_0^{\infty} e^{-x^2} e^{-y^2} dydx.

    I am not sure about how to go about this question, how did they get I^2 = \int_0^{\infty} \int_0^{\infty} e^{-x^2} e^{-y^2} dydx? Shouldn't I^2 = \int_0^{\infty} e^{-x^2} dx \times \int_0^{\infty} e^{-x^2} dx = \int_0^{\infty}\int_0^{\infty} e^{-x^2} \cdot e^{-x^2} dxdy?

    Thanks very much! (Just a bit confused on this question because I am working ahead of class )
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  3. #3
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    Quote Originally Posted by usagi_killer View Post
    Find a solution to this integral I = \int_0^{\infty} e^{-x^2} dx analytically by calculating I^2 = \int_0^{\infty} \int_0^{\infty} e^{-x^2} e^{-y^2} dydx.

    I am not sure about how to go about this question, how did they get I^2 = \int_0^{\infty} \int_0^{\infty} e^{-x^2} e^{-y^2} dydx? Shouldn't I^2 = \int_0^{\infty} e^{-x^2} dx \times \int_0^{\infty} e^{-x^2} dx = \int_0^{\infty}\int_0^{\infty} e^{-x^2} \cdot e^{-x^2} dxdy?

    Thanks very much! (Just a bit confused on this question because I am working ahead of class )
    There is a typo in your final line- you surely don't want \int_0^\infty e^{x^2} dx\times \int_0^\infty e^{-x}dx= \int_0^\infty\int_0^\infty e^{-x^2}e^{-x^2} dx dy. Where did that final "dy" come from?

    I suspect you meant \int_0^\infty e^{x^2} dx\times \int_0^\infty e^{-x}dx= \int_0^\infty\int_0^\infty e^{-x^2}e^{-x^2} dx dx but that is also wrong. It is true that I^2= \int_0^\infty e^{-x^2}dx\time\int_0^\infty e^{-x^2} dx but that is NOT the same as \int_0^\infty\int_0^\infty e^{-x^2} e^{-x^2}dx dx. In fact, that last double integral makes no sense. You cannot integrate twice with respect to the same variable.

    Remember that "x" is a dummy variable- the final answer will be a number so that it doesn't matter what you call the variable.
    If I= \int_0^\infty e^{-x^2}dx then it is also true that I= \int_0^\infty e^{-y^2}dy or, for that matter, I= \int_0^\infty e^{-t^2}dt or I= \int_0^\infty e^{-u^2}du.

    We then can say that I^2= \left(\int_0^\infty e^{-x^2}dx\right)\left(\int_0^\infty e^{-y^2} dy\right)= \int_{y= 0}^\infty \int_{x= 0}^\infty e^{-x^2}e^{-y^2} dx dy
    That last part, saying that the product of the two integrals (each integral involving a different variable) is equal to the double integral is a version of "Fubini's theorem".

    Of course, e^{-x^2}e^{-y^2}= e^{-(x^2+ y^2)} so we can write
    I^2= \int_{y=0}^\infty\int_{x=0}^\infty e^{-(x^2+ y^2)} dx dy.

    The whole point of doing that is that now the integral is over the entire first quadrant and we can cast the integral into polar coordinates with \theta going from 0 to \pi/2 and r from 0 to \infty. Of course, in polar coordinates, x^2+ y^2= r^2 and dx dy= r dr d\theta so we have
    I^2= \int_{\theta= 0}^{\pi/2}\int_{r= 0}^\infty e^{-r^2} r drd\theta= \left(\int_{\theta=0}^{\pi/2} d\theta\right)\left(\int_{r=0}^\infty e^{-r^2}r dr\right)
    where I have used Fubini's theorem the other way to separate that double integral into a product of integrals.

    Now that \theta integral is easy and, because of the extra "r" so is the r integral.
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  4. #4
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    Thanks so much both of you, I read the wiki article and HallsofIvy's great explanation, understand it now!
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