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A Evaluation of a horrible Multiple Integral

  1. Apr 6, 2016 #1
    I need to evaluate numerically this integral (second virial coefficient).
    At first, i want to simplify a bit, how could i do it?
    Now, i have a 8-dimensional integral and it's very horrible.
    T, k, tau0 and R are constants.
    r is the position and v is the velocity for two particles 1,2.
    r12 = r1-r2; v12=v1-v2

    $$\int{e^{-\frac{(|\vec{v}_1|^2+|\vec{v}_2|^2)}{2T}}e^{-E(\tau)/T}}d^2r_1d^2r_2d^2v_1d^2v_2$$
    $$E(\tau)=\frac{k}{\tau^2}e^{-\tau/\tau_0}$$
    $$\tau(\vec{r}_{12},\vec{v}_{12})=\frac{-(\vec{r}_{12}\cdot\vec{v}_{12}) -\sqrt{(\vec{r}_{12}\cdot\vec{v}_{12})^2-|\vec{v}_{12}|^2(|\vec{r}_{12}|^2 - (2R)^2)}}{|\vec{v}_{12}|^2}$$
     
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  3. Apr 6, 2016 #2

    stevendaryl

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    Well, you can simplify it a little by changing variables: Let [itex]\vec{r} = \vec{r_{12}}[/itex] and [itex]\vec{v} = \vec{v_{12}}[/itex]. Then you can integrate over [itex]\vec{r_1}[/itex] and [itex]\vec{v_1}[/itex] to leave just an integral over [itex]\vec{r}[/itex] and [itex]\vec{v}[/itex].
     
  4. Apr 7, 2016 #3
    stevendaryl,

    At first, thanks for answering me quickly.

    I'm sorry but i don't understand your method.

    if we do this change then we still have [itex]\vec{v2}[/itex] in the first exponential.
     
  5. Apr 7, 2016 #4

    stevendaryl

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    Okay, so instead, let's let [itex]\vec{v} = \vec{v_2} - \vec{v_1}[/itex] and [itex]\vec{u} = \vec{v_1} + \vec{v_2}[/itex]. Then [itex]|v_1|^2 + |v_2|^2 = \frac{1}{2}(|\vec{u}|^2 + |\vec{v}|^2)[/itex]

    Then you can do the integral over [itex]\vec{u}[/itex].
     
  6. Apr 9, 2016 #5
    Well, i have reached to simplify enough the integral. Thank you.
    $$\int^{\infty}_0\int^{\infty}_0\int^{2\pi}_0 rv\exp\left(-\frac{v^2}{4T}\right)\exp\left(-\frac{E(\tau)}{T}\right)\;\mathrm{d}r\mathrm{d}v\mathrm{d}{\theta}$$
    $$E(\tau)=\frac{k}{\tau^2}\exp\left(-\frac{\tau}{\tau_0}\right)$$
    $$\tau=\frac{-r\cos(\theta)-\sqrt{(2R)^2-r^2\sin^2(\theta)}}{v}$$
     
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