I am interested in calculating the integral $$ I(a)=\int_0^\infty e^{-x^2-ax^4}dx, $$
where $a>0$. After expanding $e^{-ax^4}=\sum_{n=0}^\infty \frac{(-a)^n}{n!}x^{4n}$, we find $$ I(a)= \sum_{n=0}^\infty \frac{(-a)^n}{n!}\int_0^\infty x^{4n}e^{-x^2}dx. $$ We can solve the integral $$ \int_0^\infty x^{4n}e^{-x^2}dx=\frac{d^{2n}}{dk^{2n}}\left.\int_0^\infty e^{-kx^2}dx\right|_{k=1}=\left.\frac12\frac{d^{2n}}{dk^{2n}}\sqrt{\frac\pi k}\;\right|_{k=1}=\frac{\sqrt{\pi}}2\frac{(4n-1)!!}{2^{2n}} $$ Hence, we find $$ I(a)= \frac{\sqrt{\pi}}2\sum_{n=0}^\infty \frac{(-a)^n}{n!}\frac{(4n-1)!!}{2^{2n}}, $$ where $(-1)!!=1$. In the above, I assumed I can change the order between the summation and the integration. I think that is fine, since the expansion of $e^{-ax^4}$ converges uniformly. I have some old notes in which I stated, based on a Mathematica integration, that $$ I(a)=\frac1{4\sqrt{a}}e^{\frac1{8a}}K_{1/4}\left(\frac1{8a}\right), $$ where $K_{1/4}$ is the modified Bessel function of the second kind. This is a black box to me and I am not sure if it will make the numerical expansion easier to obtain.
- Is my demonstration correct?
- Is there a simpler alternative (closed) form?
- How fast does this sum converge? Can I split the alternating summation into a positive and a negative part? I believe it converges absolutely.
- Is the Mathematica expression equivalent to the summation I found?
a similar question has been asked before: mathexchange question 4011739
Remark: I would like to add that a version of this integral is in Gradshteyn-Ryzhik, formula 3.323.3. I did not see a proof in the reference given there: A. Erdelyi "Tables of Integral Transformations", vol. 1 page 147. I found three "proofs" in Viktor Moll's Special Integrals of Gradshteyn and Ryzhik: the Proofs - Volume II, pages 109-116. All these demonstrations seem very complicated to me though.