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Integral question

  1. Sep 15, 2007 #1
    Given that [tex] \int_{0}^{\infty} \exp{-x^2}dx =\frac{\sqrt{\pi}}{2} \\ \ \mbox{ show that } \ \int_0^{\infty} x^4 \exp{-x^2} dx= \frac{3\sqrt{\pi}}{8} \\ [/tex]. From the kinetic Theory the root mean square velocity of the molecules [tex] (\overline{v^2})^{\frac{1}{2}} [/tex] is the square root of the integral [tex] 4\pi \frac{m}{2\pi kT} \int_0 ^{\infty} \exp{\frac{-mv^2}{2kT}}v^4 dv, where k is Boltzmann constant. T is the absolute temperature, m the mass of each molecule and v the speed of any molecule. Using the substitution [tex] x^2=\frac{mv^2}{2kT} \\ \ \mbox{ show that (\overline{v^2})^\frac{1}{2} = (\frac{3kT}{m})^\frac{1}{2} \\ [/tex] Using integration by parts I get the following:
    [tex] \int x^4\exp{-x^2}dx =\lim_{x\rightarrow \infty} x^4|\frac{\sqrt{\pi}}{2} - \lim_{X\rightarrow \infty}\int 4x^3 dx \\ \ \mbox{ which does not give} \ 3\frac{\sqrt{\pi}}{8} [/tex]. Any help would be welcome.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Sep 15, 2007
  2. jcsd
  3. Sep 15, 2007 #2
    The missing integral is: [tex] 4 \pi \frac{m}{2 \pi kT} \int_{0}^{\infty} \exp{\frac{-mv^2}{2kT}}v^4 dv [/tex]
     
  4. Sep 15, 2007 #3

    arildno

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    NO, NO, NO!!

    You are given that the definite integral:
    [tex]\int_{0}^{\infty}e^{-x^{2}}=\frac{\sqrt{\pi}}{2}[/tex]

    When you do integration by parts, you need to find AN ANTI-DERIVATIVE of the integrand that you are to evaluate between the two limits.

    The anti-derivative (a function!!)[tex]\int{e}^{-x^{2}}dx[/tex] is NOT equal to the value of the definite integral you were given!

    Thus, you must proceed differently:
    Let us rewrite:
    [tex]\int_{0}^{\infty}x^{4}e^{-x^{2}}dx=\frac{1}{2}\int_{0}^{\infty}x^{3}(2xe^{-x^{2}})dx[/tex]
    Now, set:
    [tex]u(x)=x^{3}, \frac{dv}{dx}=2xe^{-x^{2}}[/tex]

    See if you can use this hint properly..
     
  5. Sep 17, 2007 #4
    Thanks for the reply.
    [tex] \mbox{ Let } \ v=\int2x\exp{-x^2}dx \ \mbox{ therefore } \ v=-\int\exp{u}du = -\exp{u}du \ \mbox{ but } u=-x^2 \ \mbox{therefore } \ v=-\exp{-x^2} \\ [/tex]
    For the second part [tex]x=v(\frac{m}{2kT})^{\frac{1}{2}} \ \mbox{ therefore } \ dv=dx(\frac{2kT}{m})^{\frac{1}{2}} \ \mbox{ and also therefore } \ v^3=x^3(\frac{2kT}{m})^{\frac{3}{2}} \\ \mbox{ and } v=\sqrt{\frac{2kT}{m}} [/tex].The result I get then is [tex]\frac{1}{2}(\frac{1}{2})^{\frac{3}{2}}\frac{3kT}{m}\\[/tex] which is out by the constant factor [tex] \frac{1}{2}(\frac{1}{2})^{\frac{3}{2}}\\ [/tex].Thanks for the help.
     
  6. Sep 18, 2007 #5
    Actually the missing integral is [tex] 4\pi(\frac{m}{2\pi kT})^{\frac{3}{2}}\int \mbox{ e.t.c.} [/tex]
     
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