Closed form expression of $\int_{-\infty}^{+\infty}dx \exp[-\alpha(x^2-a^2)^2]$

Is the following integral $$I(a,\alpha)=\int\limits_{-\infty}^{+\infty}dx \exp[-\alpha(x^2-a^2)^2]$$ analytically solvable i.e., have a closed form expression? Here, $\alpha, a$ are real positive constants. I'm not being able to reduce it to standard improper integrals.

If such a closed form expression for $I(a,\alpha)$ does not exist, what can we say about the limiting values of the integral as $\alpha\rightarrow 0^+$ and $\alpha\rightarrow +\infty$?

• Why do you have the $dx$ before the function? Is it on purpose? – RGS Jan 18 '17 at 14:00
• @RSerrao It's just one of those notation things – Simply Beautiful Art Jan 18 '17 at 14:01
• i think there is no elementary antiderivative – Dr. Sonnhard Graubner Jan 18 '17 at 14:16
• wolframalpha.com/input/… – tired Jan 18 '17 at 14:52
• The choice of $a$ and $\alpha$ as names of the involved parameters makes the integral quite difficult to parse. Additionally, one parameter between $a$ and $\alpha$ is pretty useless, since it can be removed by a suitable substitution. – Jack D'Aurizio Jan 18 '17 at 15:22

If we define $$I(a,b) = \int_{-\infty}^{+\infty}\exp\left[-b(x^2-a^2)^2\right]\,dx$$ for $a,b>0$, by setting $c=ba^4$ and $x=az$ we get: $$I(a,b) = a \int_{-\infty}^{+\infty}\exp\left[-c(z^2-1)^2\right]\,dz = a\int_{0}^{+\infty}\exp\left[-c(z-1)^2\right]\,\frac{dz}{\sqrt{z}}\stackrel{\text{def}}{=} a\,J(c)$$ and: $$J(c) = \int_{-1}^{+\infty}\frac{\exp(-c z^2)}{\sqrt{z+1}}\,dz =\color{blue}{\int_{-1}^{0}\frac{\exp(-cz^2)}{\sqrt{z+1}}\,dz}+\color{red}{\int_{0}^{+\infty}\frac{\exp(-cz^2)}{\sqrt{z+1}}\,dz}$$ where the blue integral can be approximated by expanding the integrand function as a Taylor series and the red integral can be studied by switching to Laplace transforms and getting values of Bessel functions. In any case, the behaviour depends on the magnitude of $\color{green}{ba^4}$.
In terms of modified Bessel functions of the first kind, $$I(a,b) = \frac{\pi a}{2 \exp(ba^4/2)}\left[I_{-1/4}(ba^4/2)+I_{1/4}(ba^4/2)\right].$$ It follows that if $ba^4$ is large we have $$I(a,b) \approx \frac{\pi a }{\sqrt{\pi b a^4}}=\sqrt{\frac{\pi}{b a^2}}$$ while if $ba^4$ is close to zero we have $$I(a,b) \approx \frac{\pi}{2^{3/4}\Gamma(3/4)b^{1/4}}.$$
• @tired: as usual, you are right, but since the OP seemed interested only in asymptotics, I thought it was enough to outline an approach to get them. Essentially, we just have to compute an approximation for the the Laplace transform at $c$ of a function having the form $\frac{1}{z^\alpha (1+z)^\beta}$. – Jack D'Aurizio Jan 18 '17 at 15:56