I am solving a physics problem and ended up with the following integral:

$\int_{-\infty}^{\infty} exp(-\beta\sum_{i=0}^{N}|\vec{r}_{i+1}-\vec{r}_{i}|^2)dr_{0}dr_{1}...dr_{N+1}$ where $\beta\,$ is just a constant, and I am having some difficulty with evaluating the integral, I tried to do the following substitution $u = \vec{r}_{i+1}-\vec{r}_{i}$ but was not very helpful. any help would be appreciated.

Edit: The physics problem I am solving is:

The Hamiltonian of (𝑁+2) interacting classical particles, that are enclosed in a cube of volume 𝑉 at temperature 𝑇, is given by:

$H = \sum_{i=0}^{N+1} \frac{|\vec{P_{i}}|^2}{2m} +\frac{1}{2}mw^2 \sum_{i=0}^{N}|\vec{r}_{i+1}-\vec{r}_{i}|^2$

Assuming that $\lt{|\vec{r}_{i+1}-\vec{r}_{i}|^2\gt} \,\, \lt\lt \sqrt[3]{V}$ for $0\le i\le N+1,$ Compute the following:

a) The partition function of the sysytem

b) $<|𝑟⃗_{𝑁+1}−𝑟⃗_{0}|^2>$

My Approach:

$Z = \sum exp(-\beta H), where \,\beta = K_{B}T\,$ but in this case since $\lt{|\vec{r}_{i+1}-\vec{r}_{i}|^2\gt} \lt\lt \sqrt[3]{V}$ we can approximate the sum as an integral, So we will have: $\int exp(-\beta\sum_{i=0}^{N+1} \frac{|\vec{P_{i}}|^2}{2m}dP_{i})\int exp(-\beta\sum_{i=0}^{N}|\vec{r}_{i+1}-\vec{r}_{i}|^2 dr_{i})$

  • $\begingroup$ Are the $r_i$ numbers or vectors? If they are vectors, why do you integrate them over the real? $\endgroup$ – klirk Dec 6 '18 at 21:31
  • $\begingroup$ Why did the substitution not work? At first glance there don't seem to be issues. $\endgroup$ – klirk Dec 6 '18 at 21:40
  • $\begingroup$ @klirk because when you carry out the above substitution there will be two consecutive integrals that depend on u, but I want a substitution two make the integrals independent (if it is possible) and I can’t think of one m. $\endgroup$ – fareed Dec 7 '18 at 5:22
  • $\begingroup$ Could you write this down? If I set $u_{i+0}=r_{i+1}-r_i$ and $u_0=r_0-r_N$, then this is a liner change of coordinates, so the Jacobian is just a number. So the integral becomes a product of gaussians. I have to ask again though, what do you mean with integrating your vectors from $-\infty$ to $\infty$ $\endgroup$ – klirk Dec 7 '18 at 8:56
  • $\begingroup$ @klirk I mean that each vector could have any magnitude and could be in the positive or negative direction $\endgroup$ – fareed Dec 7 '18 at 9:08

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