I'm interested in the following integral: \begin{equation} P_\beta(x) = \int_{-\infty}^{\infty}\exp\left(-\frac{\beta}{4}\left(y^2+x^2-1\right)^2\right)dy, \end{equation} where $\beta>0$.
For context, this integral comes from the probability density function of the displacement (or velocity) of a Van der Pol-Rayleigh oscillator, as shown by Talmadge 1991 (eqn. 9).
This integral is very similar in form to this question which has already been answered on here, however Did's method was only correct for $a>0$, $b >0$. For my specific case, $b<0$ ($x^2-1<0$) forms an important solution case. Through some trial and error with Mathematica/MATLAB, I was able to find the general solution for my case:
\begin{equation} P_\beta(x) = \sqrt{|x^2-1|}\exp\left(-\frac{\beta}{4}\left(x^2-1\right)^2\right) F(x^2-1,\beta), \end{equation} where \begin{equation} F(x^2-1,\beta) = \sqrt{2}{K}_{1/4}\left(\frac{\beta}{4}\left(x^2-1\right)^2\right)+2 H\left(-(x^2-1)\right){I}_{1/4}\left(\frac{\beta}{4}\left(x^2-1\right)^2\right), \end{equation}
and where $I_{\nu}(z)$, $K_{\nu}(z)$ are the modified Bessel functions of the first and second kind, respectively, and where $H(x)$ is the Heaviside function. When $x^2-1>0$, I can see how the solution is formed, but I'm struggling with the $x^2-1<0$ case. Any insights or hints would be greatly appreciated.
Assume $\beta>0$. Note that by $(12.5.1)$ \begin{align*} P_\beta (x) &= 2\exp \left( { - \tfrac{\beta }{2}(x^2 - 1)^2 } \right)\int_0^{ + \infty } {\exp \left( { - \tfrac{\beta }{2}y^4 - \beta y^2 (x^2 - 1)} \right){\rm d}y} \\ & = \exp \left( { - \tfrac{\beta }{2}(x^2 - 1)^2 } \right)\frac{1}{{\beta ^{1/4} }}\int_0^{ + \infty } {t^{ - 1/2} \exp \left( { - \tfrac{1}{2}t^4 - \sqrt \beta (x^2 - 1)t} \right){\rm d}t} \\ & = \exp \left( { - \tfrac{\beta }{4}(x^2 - 1)^2 } \right)\frac{{\sqrt \pi }}{{\beta ^{1/4} }}U( 0,\sqrt \beta (x^2 - 1)), \end{align*} where $U$ is one of the parabolic cylinder functions. If $x^2>1$, then by $(12.7.10)$ this further simpliefies to $$ \exp \left( { - \tfrac{\beta }{4}(x^2 - 1)^2 } \right)\sqrt {\frac{{x^2 - 1}}{2}} K_{1/4} \!\left( {\tfrac{\beta}{4} (x^2 - 1)^2 } \right). $$ If $x^2<1$, we may write it as $$ \exp \left( { - \tfrac{\beta }{4}(x^2 - 1)^2 } \right)\frac{\pi }{{\beta ^{1/4} }}V(0,\sqrt \beta (1 - x^2 )) $$ by $(12.2.15)$.