# A Evaluation of a horrible Multiple Integral

1. Apr 6, 2016

### Korbid

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

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

2. Apr 6, 2016

### stevendaryl

Staff Emeritus
Well, you can simplify it a little by changing variables: Let $\vec{r} = \vec{r_{12}}$ and $\vec{v} = \vec{v_{12}}$. Then you can integrate over $\vec{r_1}$ and $\vec{v_1}$ to leave just an integral over $\vec{r}$ and $\vec{v}$.

3. Apr 7, 2016

### Korbid

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 $\vec{v2}$ in the first exponential.

4. Apr 7, 2016

### stevendaryl

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

Then you can do the integral over $\vec{u}$.

5. Apr 9, 2016

### Korbid

Well, i have reached to simplify enough the integral. Thank you.
$$\int^{\infty}_0\int^{\infty}_0\int^{2\pi}_0 rv\exp\left(-\frac{v^2}{4T}\right)\exp\left(-\frac{E(\tau)}{T}\right)\;\mathrm{d}r\mathrm{d}v\mathrm{d}{\theta}$$
$$E(\tau)=\frac{k}{\tau^2}\exp\left(-\frac{\tau}{\tau_0}\right)$$
$$\tau=\frac{-r\cos(\theta)-\sqrt{(2R)^2-r^2\sin^2(\theta)}}{v}$$