Calculating the integral $\int_{z}^{\infty} e^{-ax^4+bx^2}dx$

Question

I am looking for a way to calculate the integral $$\int_{z}^{\infty} e^{-ax^4+bx^2}dx$$ with $z>0$. I have only found this so far, which is not quite what I need. I realize that there may not be a simple expression for this integral, in which case can a function of $z$ approximating the integral be found?

Answer 1

First, note that by rescaling $x$, we can eliminate one of $a$ or $b$, so it will suffice to consider an integral of the form

$$ \int\limits_z^\infty dx \ \exp\left(-ax^4+x^2 \right)$$

For $z\rightarrow 0$

Relabeling the constant $a$ to keep future expressions simpler, let

$$ I(z)=\int\limits_z^\infty dx \ \exp\left(-\frac{x^4}{8a}+x^2 \right)$$

Fortunately $I(0)$ is expressible in terms of Bessel functions

$$ I(0)\stackrel{\text{M}}{=} \frac{1 }{2}\pi \sqrt{a} e^a \left(I_{-1/4}(a)+I_{1/4}(a) \right)$$

Where $I_n$ are modified Bessel functions of the first kind, and the symbol $\stackrel{\text{M}}{=}$ denotes Mathematica. Split up the integral

$$ I(z)=I(0)-\int\limits_0^z dx \ \exp\left(-\frac{x^4}{8a}+x^2 \right)$$

For the second term on the right we may expand the exponential around $x=0$

$$ \exp\left(-\frac{x^4}{8a}+x^2 \right)=1+x^2+\left(\frac{1}{2}-\frac{1}{8a} \right)x^4+\cdots$$

Term by term integration yields

$$ I(z)\sim I(0)-\left( z + \frac{z^3}{3}\right) \ \ , \ \ z\rightarrow 0$$

Of course, you may continue the series for higher powers of $z$. Here is plot of the integral for small $z$ and $a=1.2$.

For $z \rightarrow \infty$

Let

$$I(z)= \int\limits_z^\infty dx \ \exp\left(-ax^4+x^2 \right)=\int\limits_z^\infty dx \ \frac{e^{x^2}}{-4ax^3} \frac{d}{dx}\left( e^{-ax^4}\right)$$

We now integrate by parts

$$ I(z)=\frac{e^{-ax^4+x^2}}{-4ax^3} \Bigg\vert_z^\infty-\int\limits_z^\infty dx \ \frac{2x^2 -3}{4ax^4} e^{-ax^4+x^2}$$

For a fixed $a$, we have

$$ I(z) >> \int\limits_z^\infty dx \ \frac{2x^2 -3}{4ax^4} e^{-ax^4+x^2} \ \ , \ \ z \rightarrow \infty$$

Neglecting the small term above, we are left with

$$ I(z) \sim \frac{e^{-az^4+z^2}}{4az^3} \ \ , \ \ z \rightarrow \infty$$

In principle, you could continue to get more terms in the asymptotic series by repeated integration by parts.