# gaussian integral with changing prefactor

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?

First, you need more than positivity and analyticity to ensure convergence: $A(x)=\frac{1}{1+x^2}$ will make the integral diverge.
BUT, under the assumption that it converges (i.e., $\mathcal{O}(x^{-2})<A(x)$), you should ask what happens if you compare your integral to the Wikipedia solution? In your integral, $d$ and $f$ are zero, which makes the $p=0$ terms undefined, hence this is not applicable in your situation.
Second, since you are concerned about how the integral behaves with respect to $A'$, assuming that $A'(0)>0$ so that the left half of the integral $$\int_{-\infty}^\infty e^{-(A(0)x^2+A'(0)x^3)}dx$$ diverges, perhaps you could explore solutions to $$\int_{0}^\infty e^{-(A(0)x^2+A'(0)x^3)}dx$$ This will converge, and will likely be solvable in terms of $A(0),A'(0)$, since you found an even more general solution, and most importantly give you a feel for how $A'(0)$ affects the integral.
Third, think about how we find the equation you got from Wikipedia: it looks very reminiscent of a Taylor series: remember that $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$. What happens if you try to integrate $$e^{-(A(0)x^2+A'(0)x^3)}=\sum_{n=0}^\infty \frac{\left[-(A(0)x^2+A'(0)x^3)\right]^n}{n!}?$$