# 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, 2017 at 14:00
• @RSerrao It's just one of those notation things Jan 18, 2017 at 14:01
• i think there is no elementary antiderivative Jan 18, 2017 at 14:16
• wolframalpha.com/input/… Jan 18, 2017 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. Jan 18, 2017 at 15:22

## 1 Answer

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 $$I(a,b) \approx \frac{\pi}{2^{3/4}\Gamma(3/4)b^{1/4}}.$$

• i think that the complete integral should be expressible in a sum of modified besselfunctions of the first kind, so no approximation is needed here (but i'm too lazy for doing all the algebra at the moment) Jan 18, 2017 at 15:40
• @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 $\frac{1}{z^\alpha (1+z)^\beta}$. Jan 18, 2017 at 15:56