Evaluating a multidimensional integral












1














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



$int_{-infty}^{infty} exp(-betasum_{i=0}^{N}|vec{r}_{i+1}-vec{r}_{i}|^2)dr_{0}dr_{1}...dr_{N+1}$ 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 $lt{|vec{r}_{i+1}-vec{r}_{i}|^2gt} ,, ltlt sqrt[3]{V}$ for $0le ile N+1,$ 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 $lt{|vec{r}_{i+1}-vec{r}_{i}|^2gt} ltlt sqrt[3]{V}$ we can approximate the sum as an integral, So we will have:
$int exp(-betasum_{i=0}^{N+1} frac{|vec{P_{i}}|^2}{2m}dP_{i})int exp(-betasum_{i=0}^{N}|vec{r}_{i+1}-vec{r}_{i}|^2 dr_{i})$






integration definite-integrals improper-integrals







share|cite|improve this question
























  • Are the $r_i$ numbers or vectors? If they are vectors, why do you integrate them over the real?
    – klirk
    Dec 6 at 21:31










  • Why did the substitution not work? At first glance there don't seem to be issues.
    – klirk
    Dec 6 at 21:40










  • @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.
    – fareed
    Dec 7 at 5:22










  • 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 $infty$
    – klirk
    Dec 7 at 8:56










  • @klirk I mean that each vector could have any magnitude and could be in the positive or negative direction
    – fareed
    Dec 7 at 9:08
















1














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



$int_{-infty}^{infty} exp(-betasum_{i=0}^{N}|vec{r}_{i+1}-vec{r}_{i}|^2)dr_{0}dr_{1}...dr_{N+1}$ 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 $lt{|vec{r}_{i+1}-vec{r}_{i}|^2gt} ,, ltlt sqrt[3]{V}$ for $0le ile N+1,$ 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 $lt{|vec{r}_{i+1}-vec{r}_{i}|^2gt} ltlt sqrt[3]{V}$ we can approximate the sum as an integral, So we will have:
$int exp(-betasum_{i=0}^{N+1} frac{|vec{P_{i}}|^2}{2m}dP_{i})int exp(-betasum_{i=0}^{N}|vec{r}_{i+1}-vec{r}_{i}|^2 dr_{i})$






integration definite-integrals improper-integrals







share|cite|improve this question
























  • Are the $r_i$ numbers or vectors? If they are vectors, why do you integrate them over the real?
    – klirk
    Dec 6 at 21:31










  • Why did the substitution not work? At first glance there don't seem to be issues.
    – klirk
    Dec 6 at 21:40










  • @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.
    – fareed
    Dec 7 at 5:22










  • 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 $infty$
    – klirk
    Dec 7 at 8:56










  • @klirk I mean that each vector could have any magnitude and could be in the positive or negative direction
    – fareed
    Dec 7 at 9:08














1












1








1







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



$int_{-infty}^{infty} exp(-betasum_{i=0}^{N}|vec{r}_{i+1}-vec{r}_{i}|^2)dr_{0}dr_{1}...dr_{N+1}$ 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 $lt{|vec{r}_{i+1}-vec{r}_{i}|^2gt} ,, ltlt sqrt[3]{V}$ for $0le ile N+1,$ 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 $lt{|vec{r}_{i+1}-vec{r}_{i}|^2gt} ltlt sqrt[3]{V}$ we can approximate the sum as an integral, So we will have:
$int exp(-betasum_{i=0}^{N+1} frac{|vec{P_{i}}|^2}{2m}dP_{i})int exp(-betasum_{i=0}^{N}|vec{r}_{i+1}-vec{r}_{i}|^2 dr_{i})$






integration definite-integrals improper-integrals







share|cite|improve this question















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



$int_{-infty}^{infty} exp(-betasum_{i=0}^{N}|vec{r}_{i+1}-vec{r}_{i}|^2)dr_{0}dr_{1}...dr_{N+1}$ 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 $lt{|vec{r}_{i+1}-vec{r}_{i}|^2gt} ,, ltlt sqrt[3]{V}$ for $0le ile N+1,$ 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 $lt{|vec{r}_{i+1}-vec{r}_{i}|^2gt} ltlt sqrt[3]{V}$ we can approximate the sum as an integral, So we will have:
$int exp(-betasum_{i=0}^{N+1} frac{|vec{P_{i}}|^2}{2m}dP_{i})int exp(-betasum_{i=0}^{N}|vec{r}_{i+1}-vec{r}_{i}|^2 dr_{i})$






integration definite-integrals improper-integrals




integration definite-integrals improper-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 at 10:13

























asked Dec 6 at 16:30









fareed

193




193












  • Are the $r_i$ numbers or vectors? If they are vectors, why do you integrate them over the real?
    – klirk
    Dec 6 at 21:31










  • Why did the substitution not work? At first glance there don't seem to be issues.
    – klirk
    Dec 6 at 21:40










  • @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.
    – fareed
    Dec 7 at 5:22










  • 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 $infty$
    – klirk
    Dec 7 at 8:56










  • @klirk I mean that each vector could have any magnitude and could be in the positive or negative direction
    – fareed
    Dec 7 at 9:08


















  • Are the $r_i$ numbers or vectors? If they are vectors, why do you integrate them over the real?
    – klirk
    Dec 6 at 21:31










  • Why did the substitution not work? At first glance there don't seem to be issues.
    – klirk
    Dec 6 at 21:40










  • @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.
    – fareed
    Dec 7 at 5:22










  • 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 $infty$
    – klirk
    Dec 7 at 8:56










  • @klirk I mean that each vector could have any magnitude and could be in the positive or negative direction
    – fareed
    Dec 7 at 9:08
















Are the $r_i$ numbers or vectors? If they are vectors, why do you integrate them over the real?
– klirk
Dec 6 at 21:31




Are the $r_i$ numbers or vectors? If they are vectors, why do you integrate them over the real?
– klirk
Dec 6 at 21:31












Why did the substitution not work? At first glance there don't seem to be issues.
– klirk
Dec 6 at 21:40




Why did the substitution not work? At first glance there don't seem to be issues.
– klirk
Dec 6 at 21:40












@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.
– fareed
Dec 7 at 5:22




@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.
– fareed
Dec 7 at 5:22












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 $infty$
– klirk
Dec 7 at 8:56




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 $infty$
– klirk
Dec 7 at 8:56












@klirk I mean that each vector could have any magnitude and could be in the positive or negative direction
– fareed
Dec 7 at 9:08




@klirk I mean that each vector could have any magnitude and could be in the positive or negative direction
– fareed
Dec 7 at 9:08















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