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 $\alpha\rightarrow +\infty$?

  • $\begingroup$ Why do you have the $dx $ before the function? Is it on purpose? $\endgroup$
    – RGS
    Jan 18, 2017 at 14:00
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    $\begingroup$ @RSerrao It's just one of those notation things $\endgroup$ Jan 18, 2017 at 14:01
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    $\begingroup$ i think there is no elementary antiderivative $\endgroup$ Jan 18, 2017 at 14:16
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    $\begingroup$ wolframalpha.com/input/… $\endgroup$
    – tired
    Jan 18, 2017 at 14:52
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    $\begingroup$ 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. $\endgroup$ Jan 18, 2017 at 15:22

1 Answer 1


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

  • $\begingroup$ 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) $\endgroup$
    – tired
    Jan 18, 2017 at 15:40
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    $\begingroup$ @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 $\frac{1}{z^\alpha (1+z)^\beta}$. $\endgroup$ Jan 18, 2017 at 15:56

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