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I would like to solve the following integral:

\begin{equation} \int \prod_i d x_i e^{a x_i^2 + b x_i^4 + c x_i^2 x^2_{i+1}} \end{equation}

This integral can be for sure lead back to a common gaussian integral and solve consequently. I just cannot find how to do so (setting $x=\sqrt{z}$ lead to a $1/\sqrt{z}$ that is hard to treat after the diagonalization).

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  • $\begingroup$ If the last index is $n$, do you take $x_{n+1}=x_1$ in the last term? $\endgroup$
    – FusRoDah
    Commented Dec 11, 2019 at 12:12
  • $\begingroup$ Yes I assume so $\endgroup$ Commented Dec 11, 2019 at 12:15
  • $\begingroup$ Not exactly the same, but not totally different: mathoverflow.net/posts/343912 $\endgroup$ Commented Dec 11, 2019 at 17:25
  • $\begingroup$ I think that post has been deleted since the link does not lead to anything $\endgroup$ Commented Dec 12, 2019 at 12:36

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I think that this integral cannot be reduced to a known Gaussian form. Indeed, let us assume $i=1,2$. Then, you have to evaluate $$ \int dx_1dx_2 e^{ax_1^2+bx_1^4+cx_1^2x_2^2}e^{ax_2^2+bx_2^4+cx_1^2x_2^2} $$ where I used the fact, state by the OP, that $x_{n+1}=x_{n}$ that holds for $n=2$. The first integral can be easily evaluated to give $$ \int_{-\infty}^\infty dx_1 e^{ax_1^2+bx_1^4+2cx_1^2x_2^2}=\frac{1}{2}\sqrt{-\frac{a+2cx_2^2}{b}}e^{-\frac{(a+2cx_2^2)^2}{8b}}K_\frac{1}{4}\left(-\frac{(a+2cx_2^2)^2}{8b}\right) $$ being $K_\frac{1}{4}$ the modified Bessel function of the second kind and it must be $Re(a)\& Re(b)>0$. So, your next integration step is doomed.

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    $\begingroup$ actually you wrongly used the periodic condition, it just gives you that $x_3=x_1$ not that $x_2=x_1$ so the exponent for $n=1$ is $a(x_1^2+x_2^2)+b(x_1^4+x_2^4)+2 c x_1 x_2$ $\endgroup$ Commented Dec 12, 2019 at 15:02
  • $\begingroup$ Of course, I fixed this but does not seem to improve your situation. $\endgroup$
    – Jon
    Commented Dec 12, 2019 at 15:06
  • $\begingroup$ If you put down the integral representation of $\delta (\tau_i - x_i^2)$ you can see that the integral disentagle. However, when performing the integration on $\tau$, I am not sure if a nice form can be found for the term deriving from the integration of the surge term $\endgroup$ Commented Dec 12, 2019 at 17:02
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    $\begingroup$ The point with this integral is that you are managing a 0d quartic field theory and nobody knows how to do it unless with perturbation techniques and possibly resummation. Please, check Zinn-Justin, "Quantum Field Theory and Critical Phenomena", IV ed., Sec. 39.2 starting at page 933 and you will get an idea about. $\endgroup$
    – Jon
    Commented Dec 12, 2019 at 17:46

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