I studied the integral $F(\lambda)=\int_{-\infty}^\infty \int_{-\infty}^\infty \exp(-(x_1^2+x_2^2+(x_1-x_2)^2)-\lambda(x_1^4+x_2^4))\,dx_1dx_2\;$ as a toy problem related to $\phi^4$ quantum field theory. It can be converted to a power series in $\lambda$
$F(\lambda)=c_0-\lambda c_1+\lambda^2c_2-\lambda^3 c_3+...$, which diverges for all values of $\lambda$. In a move similar to Borel summation, but taking into account that the original integral is Gaussian, I put $b_n = c_n/(\sqrt 2 \,\,4^n\Gamma(2n+1/2))$. The series $G(z)=b_0−zb_1+z^2b_2−z^3b_3+...$ converges for all z. One can formally put
$F(\lambda) = \int_{-\infty}^\infty G(\lambda x^4) \exp(-x^2/2)\,dx \tag{1}$,
because $\int_{-\infty}^\infty x^{4n} \exp(-x^2/2)\,dx = \sqrt 2\,\,4^n\Gamma(2n+1/2)$.
I was able to evaluate $G(z)$ out to $z=145$ by summing the first 200 terms of the power series and using high precision math. When I carried out the integral of equation 1 I was obtained the correct value for $F(\lambda)$ out to $\lambda = 1/2$ accurate to 5 or 6 decimal digits.
QUESTION 1) Why does this summation method work, I based my numerical experiments on intuition, I don't know how to prove it works. Also, would it work to reduce any multidimensional integral of positive definite quadratic term and positive quartic term to a one dimensional integral.
The graph of $G(z)$ is below:
$G(0)=1.44720, G(145)=-0.132149$. In the negative direction $G(z)$ grows exponentially reaching $G(-145)=1.96615\times 10^{32}$
QUESTION 2) How can one derive the asymptotic (large positive $\lambda$) form of $G(z)$, In some cases the large $\lambda$ behaviour of $F(\lambda)$ is known.
SOME DETAILS
$c_n= \frac{2^{n+1}}{\sqrt 3\,n!}\sum_{n_1=0}^n\sum_{n_2=0}^{n-n_1}\frac{n!}{n_1!\,n_2!\,(n-n_1-n_2)!}\Gamma(n+n_1-n_2+1/2)\Gamma(n+n_2-n_1+1/2)6^{(n - n_1 - n_2)}3^{-n-n_1+n_2}$.
$c_0= 3.6276 ,\;\;c_1= 9.6736,\;\; c_2 = 95.6611 $.