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Can anybody please tell me, how to evaluate a multivariate integral with a gaussian weight function. $$\mathcal{Z_{n}}=\int_{-\infty}^{\infty} dx_{1}dx_{2}dx_{3}dx_{4}...dx_{n}\exp(-\frac{a}{2}\sum_{j}x_{j}^2)\times f(x_{1},x_{2},x_{3},x_{4}....x_{n})$$ where $f(x_{1},x_{2},x_{3},x_{4}....x_{n})=\Pi_{j} \frac{1}{\sqrt{(1+i\,b(x_{j}^2-x_{j+1}^2)^2)}}$, and $i=\sqrt{-1}$. Also we have the condition $x_{n+1}=x_{1}$.

I need a hint to solve this integral. This is how I proceeded,

\begin{eqnarray} \mathcal{Z_{n}}=\int_{-\infty}^{\infty} \Pi_{j=1}^{n} dx_{j} \exp{\Bigg(-\frac{a}{2}\sum_{j=1}^{n}x_{j}^2-\frac{1}{2}\log{\Big(1+i b(x_{j}^2-x_{j+1}^2)^{2}\Big)\Bigg)}} \end{eqnarray}

Now the integral is of the form of the canonical partition function integrated over the configuration space. Hence the integral can be identified as an $n-$particle partition of the canonical ensemble, which is given by \begin{eqnarray} \mathcal{Z}_{n}= \int_{-\infty}^{\infty} \Pi_{j=1}^{n} dx_{j} e^{-\beta \mathcal{H}}, \end{eqnarray} where $$\mathcal{H}=\Bigg(\frac{a}{2}\sum_{j=1}^{n}x_{j}^2+\frac{1}{2}\log{\Big(1+i b(x_{j}^2-x_{j+1}^2)^{2}\Big)\Bigg)}.$$ and $\beta=1$. Then I got stuck !

Sijo Joseph
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    I asked it there, but no response !!!! – Sijo Joseph Mar 06 '14 at 14:42
  • Then I thought that Physicists may do better than Mathematicians ! – Sijo Joseph Mar 06 '14 at 14:42
  • Given the function $f(x_j)$, seems to me that this can only be solved numerically. Where did you find such an integral? – Kyle Kanos Mar 06 '14 at 14:45
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    @SijoJoseph: The integral does not converge for any $n$. Are you sure you typed it correctly? – DumpsterDoofus Mar 06 '14 at 15:03
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    I am pretty sure that the exponential function should read $\exp\left(-\sum_{j} x_j^2\right)$. (More like a Gaussian distribution.) Then the integral converges because of the upper bound given by a Gaussian distribution. I think you are familiar with the integration trick for the Gaussian distribution (changing from Cartesian to Polar coordinates). I am afraid that this does not work here because of the term under the square root. :-( – Tobias Mar 06 '14 at 15:31
  • @Tobias and DumpsterDoofus thank you for pointing out the error. The exponential should contain a minus sign inside the bracket. Sorry for the typo. – Sijo Joseph Mar 06 '14 at 22:07
  • @Kyle Kanos I found this integral in relation to the entanglement dynamics in a bipartite system which is classically chaotic. I am trying to find an analytical expression for the von neumann entropy of entanglement. I am able to formulate my problem in terms of this integral. Now I got stuck since I don't know the tricks to resolve this type of integrals. – Sijo Joseph Mar 06 '14 at 22:08
  • If you have an even number of integral variables you can decompose the integral into a sequence of 2D-integrals (because of $\exp(-x_1^2-x_2^2-x_3^2-x_4^2)=\exp(-x_1^2-x_2^2)\exp(-x_3^2-x_4^2)$ and only 2 vars under the root. The trick for the Gaussian is: $\int_{-\infty}^\infty\int_{-\infty}^\infty \exp(-x^2-y^2)dxdy=\int_{r=0}^\infty \exp(-r^2) 2\pi rdr =[-\pi\exp(-r^2)]_0^\infty = \pi.$ Pityingly the root does not fit in there. – Tobias Mar 06 '14 at 22:19
  • I used a similar reversed procedure to arrive at this integral. This integral corresponds to an entangled state, actually the root term contains entanglement. Hence we cannot split the integral into parts. What I did is, I took the exponential of the logarithm of the square root term. Then I can write it as a partition function with some pseudo-Hamiltonian. – Sijo Joseph Mar 06 '14 at 22:25
  • Yes, my mistake. The pairs are interwoven. So, no splitting. It is a pity. One more thing. Is $i:=\sqrt{-1}$? Then you could/should explicitly state it. Sorry, that I cannot be of more help. – Tobias Mar 06 '14 at 22:33

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