I have this integral which I want to solve to forth order in $g$:
$$ Z(g) = \int^\infty_{-\infty} \mathrm{d}x \exp\left(\frac{-1}{2}x^2 + \frac{g}{4!}x^4\right) $$
I can solve this integral quite easily by expanding $e^{\frac{g}{4!}x^4}$ and use this formula: $$\int^\infty_{-\infty} x^{2k} e^{-x^2/2} \mathrm{d}x = (2 \pi)^{1/2} \frac{(2k)!}{2^k k!}.$$
But I want to solve it "using Feynman diagrams" and I'm not sure how to do that. Usually, it's the other way around, I calculate a diagram using integrals. So I'm confused. How does one solve this?
First of all, the one-sided version of the integral identity you mention below is discussed in this MSE post if you happen to be interested in a proof of that fact.
Unfortunately you're not going to find a solution in terms of so-called elementary functions, but we can state the result in terms of the modified Bessel function of the second kind:
$$\int_{-\infty}^\infty \exp\left(\frac{-x^2}{2}+\frac{g}{4!}x^4\right)\mathrm{d}x=\frac{\sqrt{3} e^{-3g/4}K_{1/4}\left(\frac{-3}{4g}\right)}{\sqrt{-g}}$$ For $g<0$.