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Proof that integral = root pi

  1. Mar 25, 2005 #1
    A problem in my textbook guides you through this proof using a multiple integral.

    I follow the whole thing except for one step. It requires that you show that (sorry don't know latex, I(a,b) will denote integral from a to b, e the exponential)


    where R is the rectangle such that u and v lie between plus and minus x and the second integral is a multiple integral over this region.

    How can you prove this? I tried working with the Riemann definitions of integrals but couldn't get anywhere. Thank you in advance for any help.

    by they way, could anyone post the above equation in Latex so I can see how it would be done? Thanks again.
  2. jcsd
  3. Mar 25, 2005 #2


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    U can prove this

    [tex]\left[\int_{-x}^{x} e^{-u^{2}} \ du \right]^{2}=\int_{-x}^{x}\ du \int_{-x}^{x} dv \ e^{-u^{2}-v^{2}} [/tex]

    using FUBINI's theorem.So how about opening your book at this theorem...?

  4. Mar 25, 2005 #3


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    Notice that [tex]\left[\int_{-x}^x e^{-u^2}du\right]^2[/tex]
    is the same as [tex]\left[\int_{-x}^x e^{-u^2}du\right]\left[\int_{-x}^x e^{-v^2}dv\right][/tex]
    Last edited: Mar 25, 2005
  5. Mar 25, 2005 #4
    Unfortunately my book doesn't have Fubini's theorem, or more likely it just isn't calling it that. HallsofIvy, I know that it can be rewritten in that way, but from there how do you know that you can multiply the two functions? I searched for Fubini's theorem online and the versions I found only said that you can switch the order of the iterated integral. How can this be used to prove the desired result?
  6. Mar 25, 2005 #5
    In an iterated integral, if the integrand can be factored into functions of only one of the variables to be integrated, you can always split it up into several multiplied integrals and visa-versa. Let [itex]F(x), \ G(y)[/itex] be antiderivatives of [itex]f(x)[/itex] and [itex]g(y)[/itex], respectively. Then

    [tex] \int_a^b \int_c^d f(x)g(y) \ dx \ dy = \int_a^b \left[ g(y) \int_c^d f(x) dx \right] \ dy = \int_a^b g(y) \left( F(d) - F(c) \right) \ dy = \left(F(d) - F(c)\right) \int_a^b g(y) \ dy [/tex]

    [tex]=\left(F(d) - F(c)\right)\left(G(b) - G(a)\right) = \int_c^d f(x) \ dx \int_a^b g(y) \ dy[/tex]
  7. Mar 27, 2005 #6


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    Out of curiosity, I googled on "Fubini's Theorem" and the very first hit was
    which includes the fact that the doubled integral is equal to the product of two integrals (PROVIDED F(x,y)= f(x)g(y) and the region of integration is a rectangle).

    In fact, I just notice that this theorem is an EXERCISE on page 968 of Salas, Helle, and Etgen's "Calculus", edition 9.

    The basic idea is this:
    [tex]\int_{x=a}^b\int_{y= c}^d f(x)g(y) dydx = \int_{x=a}^bf(x)\left[\int_{y=c}^d g(y)dy\right]dx[/tex]
    because f(x) is a "constant" does not involve y and so can be treated like a constant (that is, taken out the integral) when integrating with respect to y.

    But [tex]\int_{y=c}^d g(y)dy[/tex] is a constant, a number. We can take that out of the x-integration:
    [tex]\int_{x=a}^b\int_{y= c}^d f(x)g(y) dydx= \left[\int_{x=a}^b f(x)dx\right]\left[\int_{y=c}^d g(y)dy\right][/tex]
    Last edited: Mar 28, 2005
  8. Mar 28, 2005 #7
    Thank you. Totally makes sense now, I was making the problem too complicated the way I was thinking about it.
  9. Mar 28, 2005 #8


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    You must realize that the integral u're computing is solvable only for an infinite domain,like the real axis,or the positive semiaxis.

  10. Mar 29, 2005 #9


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    I don't know, maybe I'm just mising stuf. Anyway, looks like to me, need to show:

    [tex]\int_{-\infty}^{\infty} e^{-x^2}dx=\sqrt{\pi}[/tex]

    So, let:

    [tex]A=\int_{-\infty}^{\infty} e^{-x^2}dx[/tex]


    [tex]A^2=[\int_{-\infty}^{\infty} e^{-x^2}dx]^2[/tex]

    As per Hall:

    [tex]A^2=\left[\int_{-\infty}^{\infty} e^{-u^2}du\right]\left[\int_{-\infty}^{\infty} e^{-v^2}dv\right][/tex]

    Converting to polar coordinates:

    [tex]A^2=\int_{0}^{2\pi}\int_0^{\infty} e^{-r^2}rdrd\theta[/tex]

    That one's doable so:




    Don't quite understand how Fubini's theorm enters the equation though since really don't need to switch the order of integration as I see it.
  11. Mar 29, 2005 #10


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    It does.U define the square of the initial integral as

    [tex] \iint_{R^{2}} e^{-u^{2}-v^{2}} \ du \ dv [/tex]

    Fubini's theorem assures that u can u can identify the square of the initial integral with the double integral on R^{2}...

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