Let's consider the integral $$\int_0^{\infty}e^{-gx^2-x}dx$$ If I'm not mistaken, $$\int_0^{\infty}e^{-gx^2-x}dx=\frac{\sqrt \pi}{2\sqrt g}e^{1/4g}\left(1-\mathrm{erf}\left(\frac{1}{2\sqrt g}\right) \right)$$ Now I want to try a different method : let's develop the exponential in series : $$\int_0^{\infty}e^{-gx^2-x}dx=\int_0^{\infty}\left(\sum_{n=0}^\infty \frac{(-1)^n g^n}{n!}x^{2n}\right)e^{-x} dx$$ I want to swap the series and integral (which is forbidden here), I get the series : $$\sum_n\frac{(-1)^n g^n}{n!} \int_0^{\infty} x^{2n}e^{-x} dx=\sum_n\frac{(-1)^n g^n}{n!}(2n)!$$ which is clearly divergent for all $g$.
My question is : is there a method to resum (Borel resummation?) this series and get back to the value of the original integral? Are all resummation (which works) methods going to give me the same result?
I'd like to do the same thing with $\int_{-\infty}^{\infty}\exp(-gx^4-x^2)dx=\frac{1}{2\sqrt g}\exp(1/8g)K_{1/4}(1/8g)$ and by developing in powers of $g$ I get the series : $$\sum_n\frac{(-1)^n g^n}{n!}\Gamma\left(2n+\frac{1}{2}\right)$$