What's the closed-form of the Gaussian-like integral?

Question

I once found that the integral below $$ \,{\rm I}\left(\,\alpha\,\right) =\int_{-\infty}^{\infty}\,{\rm e}^{-\left(\,x^{2}\,\, +\ \alpha\,x^{4}\,\right)} \,\,\,{\rm d}x\,,\qquad \left(\,\alpha > 0\,\right) $$ has a closed-form, but I don't know the exact result now. So please provide me with some references or carry on the calculation here. Thank you !.

Answer 1

According to a CAS, provided $\Re(\alpha )>0$, $$I(\alpha)=\int_{-\infty}^{\infty}e^{-(x^2+\alpha x^4)}dx=\frac{e^{\frac{1}{8 \alpha }}}{2 \sqrt{\alpha }}K_{\frac{1}{4}}\left(\frac{1}{8 \alpha }\right)$$ where appears the modified Bessel function of the second kind.

For values of $\alpha$ close to $0$, $$I(\alpha)\approx \frac{1}{4} \sqrt{\pi } (4-3 \alpha )$$