I have a problem solving this gaussian integral:
$$ \int_{-\infty}^{\infty} dx\exp\left(-A(x)\cdot x{}^{2}\right) $$ While A(x)>0, which ensures that the integral doesn't diverge. I'm especialy interested how the integral depends on the first derivative of A. I tried Taylor expansion: $$\begin{eqnarray} \int_{-\infty}^{\infty} dx\exp\left(-A(x)\cdot x{}^{2}\right)&=&\int_{-\infty}^{\infty} dx\exp\left(-A(0)\cdot x{}^{2}-A'(0)\cdot x{}^{3}+...\right) \\&=&diverges \end{eqnarray} $$ Unfortunately, that approach diverges if one includes the x^3 term, since $-A(0)\cdot x{}^{2}-A'(0)\cdot x{}^{3}$ will go to $\infty$ for $x=-\infty$ If one includes the x^4 term, it will converge again. There is a equation for that in wikipedia, without reference: $$\int_{-\infty}^{\infty} e^{a x^4+b x^3+c x^2+d x+f}\,dx =\frac12 e^f \ \sum_{\begin{smallmatrix}n,m,p=0 \\ n+p=0 \mod 2\end{smallmatrix}}^{\infty} \ \frac{b^n}{n!} \frac{c^m}{m!} \frac{d^p}{p!} \frac{\Gamma \left (\frac{3n+2m+p+1}{4} \right)}{(-a)^{\frac{3n+2m+p+1}4}}$$, but this looks way to too complicated to be useful. Is there a better way to solve this integral?