Computing $ \int_{-\infty}^{\infty} e^{-ax^4+bx^2} dx$ I was wondering if it was possible to compute $ \int_{-\infty}^{\infty} e^{-ax^4+bx^2} dx$  explicitly.
Completing the square doesn't seem to work, since there is no factor of $x$ in order to use the Gaussian integral.
Any hint would be greatly appreciated.
Thanks!
 A: With the change of integration variable $x = (2a)^{ - 1/4} \sqrt t$ and the representation http://dlmf.nist.gov/12.5.E1, we find
\begin{align*}
\int_{ - \infty }^{ + \infty } {e^{ - ax^4  + bx^2 } dx} & = 2\int_0^{ + \infty } {e^{ - ax^4  + bx^2 } dx}  = (2a)^{ - 1/4} \int_0^{ + \infty } {t^{ - 1/2} e^{ - \frac{1}{2}t^2  + \frac{b}{{\sqrt {2a} }}t} dt} 
\\ & = (2a)^{ - 1/4} \sqrt \pi  e^{\frac{{b^2 }}{{8a}}} U\!\left( {0, - \tfrac{b}{{\sqrt {2a} }}} \right).
\end{align*}
Here $U$ is the parabolic cylinder function. By analytic continuation, this formula is valid whenever $\Re a>0$ and $b$ is any complex number. By http://dlmf.nist.gov/12.7.E10, this is also equal to
$$
\frac{1}{2}\sqrt { - \tfrac{b}{a}} e^{\frac{{b^2 }}{{8a}}} K_{1/4}\! \left( {\tfrac{{(-b)^2 }}{{8a}}} \right),
$$
where $K_\nu$ is the modified Bessel function. One has to choose the phase of $-b$ consistently in this form to make the function single-valued (and real valued for real $b$).
A: Your integral is related by a scaling transformation to
$$\int_{-\infty}^{\infty}
\mathrm{e}^{-z(x^4/8-x^2+1)}\mathrm{d}x
= \pi(I_{-1/4}(z)+I_{1/4}(z))\text{,}$$
where $\Re z > 0$ and $I$ denotes a modified Bessel function. To prove this equality,

*

*Write $\mathrm{e}^{zx^2}$ as a Maclaurin series,

*Integrate termwise using Euler's gamma function,

*Use the duplication formula for the gamma function,

*Write the series as a linear combination of two confluent hypergeometric functions, and

*Simplify to the modified Bessel functions.

