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Unknown variable

  1. Sep 12, 2007 #1
    I am confused about how the y variable came from?

    I = e[tex]^{-x^{2}}[/tex] dx

    I[tex]^{2}[/tex] = e[tex]^{-x^{2}}[/tex] dx e[tex]^{-y^{2}}[/tex] dy

    The equation of I should be like this:

    I = e[tex]^{-x^{2}}[/tex] dx
    I[tex]^{2}[/tex] = (e[tex]^{-x^{2}}[/tex])[tex]^{2}[/tex] d[tex]^{2}[/tex]x
    I[tex]^{2}[/tex] = e[tex]^{-2x^{2}}[/tex] d[tex]^{2}[/tex]x
  2. jcsd
  3. Sep 12, 2007 #2
    You're missing some integral signs, I believe. Remember that things like dx cannot be separated from the integral sign -- and that in the integrand, the variable x is "bound", and is not the same as an x on the outside. So:

    [tex]I = \int_0^{\infty} e^{-x^2}\,dx = \int_0^{\infty} e^{-y^2}\,dy[/tex]
    [tex]I^2 = \int_0^{\infty} e^{-x^2}\,dx \int_0^{\infty} e^{-y^2}\,dy[/tex]

    One more thing: even if you were to be loose with notation and use what I refer to as "physicist shorthand" and treat dx as a variable of sorts, [tex]dx^2 \ne d^2x[/tex]
  4. Sep 12, 2007 #3
    and can you tell me why x = y?
  5. Sep 12, 2007 #4
    It doesn't, but

    [tex] \int_0^{\infty}e^{-x^2}\,dx= \int_0^{\infty}e^{-y^2}\,dy[/tex]
  6. Sep 12, 2007 #5


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    [tex]\int \int f(x) dx dx[/tex]
    would make no sense- once you have integrated with respect to x, there is no "x" left to integrate again!

    It is true, as DeadWolfe said, that
    [tex]\int_{-\infty}^\infty e^{-x^2}dx= \int_{-\infty}^\infty e^{-y^2}dy[/itex]
    because the x and y are "dummy" variables. You would integrate with respect to either, then evaluate at the end points (technically, take the limit) so there is no x or y in the final result- it doesn't matter what you call the variable.

    Of course, it is well known (Fubini's theorem) that the product
    [tex]\left(\int_a^b f(x)dx\right)\left(\int_c^d g(y)dy[/tex]
    is the same as the iterated integral
    [tex]\int_{x=a}^b\int_{y=c}^d f(x)g(y) dydx[/itex]
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