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 !.


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 $0$, $$I(\alpha)\approx \frac{1}{4} \sqrt{\pi } (4-3 \alpha )$$

  • $\begingroup$ Thanks a lot. You have helped me several times. $\endgroup$ – Roger209 Jan 3 '15 at 7:56
  • $\begingroup$ Glad to know ! You are very welcome. $\endgroup$ – Claude Leibovici Jan 3 '15 at 7:58

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