TITLE: Challenge: Non-Gaussian quartic integral and path integral in Quantum field theory
QUESTION [15 upvotes]: (1) It is well-known that we can get a Gaussian integral of this type, where $x$ is in $\mathbb{R}$: 
$$
\int_{-\infty}^{\infty} dx e^{-ax^2}=\sqrt{(2\pi)/a}. \tag{i}
$$
We can generalize this integral from a number $a$ to a matrix value $A$, from a number $x$ to a vector $X=(x_1,x_2,...,x_N)$.
So we generalize
$$
\int_{-\infty}^{\infty} Dx  \;
e^{\frac{i}{2}X^T \cdot A \cdot X+ i JX }
=
\int_{-\infty}^{\infty} dx_1
\int_{-\infty}^{\infty} dx_2
\dots
\int_{-\infty}^{\infty} dx_N
e^{\frac{i}{2}X^T \cdot A \cdot X+ i JX }=\left(\frac{(2\pi i)^N}{\det(A)}\right)^{1/2}
e^{-\frac{i}{2} J^T \cdot A^{-1} \cdot J}.
\tag{ii}
$$
(2) We can generalize again the integral to a functional integral or the so-called path integral in Quantum Field Theory.
So for a real scalar field $\phi$, can be regarded as the earlier Eq(i) real valued $x$ generalized to the field
$$
x \to \varphi,
$$
and the matrix value $A$ is  generalized to the matrix operator $M$, so we can evaluate the path integral in the $N$-dimensional spacetime manifold $\Sigma_N$:
$$
\int_{-\infty}^{\infty} D \varphi e^{i \int_{\Sigma_N} (d^N x)\big( -\frac{1}{2}\varphi  \cdot M  \cdot \varphi + J  \cdot \varphi \big)} . \tag{iii}
$$
where the Green's function $G(x_i-x_f)$ is a solution of the delta function $\delta^N(x_i-x_f)$ on right hand side (r.h.s):
$$
-M G(x_i-x_f)=\delta^N(x_i-x_f).
$$
Therefore, the $A$ matrix in Eq.(ii) is related to $M$, while also related to $G$ via the inverse generalization
$$
A \to M,
$$
$$
A^{-1}_{if}=G(x_i-x_f).
$$
Thus we obtain:

$$
\int_{-\infty}^{\infty} D \varphi e^{i \int_{\Sigma_N} (d^N x)\big( -\frac{1}{2}\varphi  \cdot M  \cdot \varphi + J  \cdot \varphi \big)}=Constant\cdot
e^{\frac{-i}{2} \int d^N x_i  \int d^N x_f
J(x_i)
G(x_i-x_f)
J(x_f)
}. \tag{iii}
$$
  involving the Green's function $G$.
  up to a not crucial constant $Constant$ which I do not care.

(3) Question:
Suppose I have a quartic integral from (i) to (i')

$$
\int_{-\infty}^{\infty} dx  \; e^{h x - a x^2-b x^4}=?. \tag{i'}
$$
  I suppose that we can still easily get an answer.

How about the matrix value integral from (ii) to (ii'):

$$
\int_{-\infty}^{\infty} Dx  \;
e^{\frac{i}{2}X^T \cdot A \cdot X+ i J \cdot X +(X^T   \cdot X)^2 }
=?.
\tag{ii'}
$$
  Do we also have a QFT path integral analogy:
  $$
\int_{-\infty}^{\infty} D \varphi e^{i \int_{\Sigma_N} (d^N x)\big( -\frac{1}{2}\varphi  \cdot M  \cdot \varphi + J  \cdot \varphi +  \varphi^4 \big)} =?. \tag{iii'}
$$
My question concerns the neat evaluation of (i'), (ii') and (iii'). Thank you!

Here we can focus on the real valued function field $\varphi(x) \in \mathbb{R}$

REPLY [2 votes]: For your integral (i'), at least the $h=0$, $a>0$, $b>0$ case can be done analytically (or using Mathematica or Gradshteyn-Ryzhik):
$$
\int_{-\infty}^\infty e^{-ax^2-bx^4}{\rm d}x = \frac{1}{2} \sqrt{\frac{a}{b}} e^{\frac{a^2}{8 b}}
    K_{\frac{1}{4}}\left(\frac{a^2}{8 b}\right).
$$
The $h\not=0$ case seems not to permit a closed-form solution, as far as I can tell (but of course the integral exists).
For symmetric positive-definite $A$, the multidimensional case can be expressed in terms of the one-dimensional integrals in an eigenbasis of $A$. I don't think it will have a closed-form solution unless $J=0$ and $A$ is a multiple of the identity.
The path integral case is very different, because it only makes sense upon renormalization (the parameters of the action will depend on the regulator used to define and evaluate the integral). It is generally believed that the integral exists in a meaningful sense only for $\lambda=0$ (i.e. "$\phi^4$ theory is trivial").