Integrate $e^{-x^4+x^2}$

Question

I am looking for pointers on how one might approach the following definite integral:

$$ \int_{-\infty}^\infty e^{-x^4 + x^2}\, dx$$

Or more generally:

$$ \int_{-\infty}^\infty e^{-x^4 + \alpha x^2}\, dx, \quad \alpha > 0$$

Mathematica does return the following result, which seems correct based on numerical verification:

$$ \frac{\pi e^{\frac{\alpha ^2}{8}} \sqrt{\alpha } \left(I_{\frac{1}{4}}\left(\frac{\alpha ^2}{8}\right)+I_{-\frac{1}{4}}\left(\frac{\alpha ^2}{8}\right)\right)}{2 \sqrt{2}} $$

Here $I_a$ is the modified Bessel function of the first kind, which I am not very familiar with, though I can see its definition as the solution of a differential equation.

Is there anything I can do, other than browse formula tables like this one (p. 21), to see how one may arrive to this result or perhaps how to arrive to a different (and potentially more useful) representation?

Answer 1

In general, $~\displaystyle\int_0^\infty\exp\Big(-\sqrt[N]x\Big)~dx~=~N!~,~$ so even a relatively simple looking expression

like $~\displaystyle\int_0^\infty\exp\Big(-x^4\Big)~dx~=~\Big(\tfrac14\Big)!~=~\Gamma\bigg(\frac54\bigg)~$ cannot be expressed in terms of elementary

functions, let alone a slightly more complex one, like $~\displaystyle\int_0^\infty\exp\Big(-x^4+ax^2\Big)~dx,~$ for whose

evaluation even more obscure special functions are required. A first step, in this case, would

be to employ the parity of the integrand, by rewriting $~\displaystyle\int_{-\infty}^\infty~=~2\displaystyle\int_0^\infty$

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Basically, just like $~\displaystyle\int_0^\infty\exp\Big(-\sqrt[n]x\Big)~dx~$ cannot be expressed in terms of elementary

functions, but requires the creation of a completely new function, called factorial, to

help express its value, yielding the more general result $~\displaystyle\int_0^\infty x^{m-1}\cdot\exp\Big(-\sqrt[n]x\Big)~dx$

$=~n~\Gamma(mn),~$ which, by replacing the lower limit with an arbitrary value becomes

inexpressible even in terms of the latter, thus requiring the creation of yet another

special function to help parse its value, finally yielding $~\displaystyle\int_\ell^\infty x^{m-1}\cdot\exp\Big(-\sqrt[n]x\Big)~dx$

$=~n~\Gamma\Big(mn,~\sqrt[n]\ell\Big),~$ so this latter expression also becomes equally useless when asked

to evaluate $~\displaystyle\int_\ell^\infty(x+u)^{m-1}\cdot\exp\Big(-\sqrt[n]x\Big)~dx,~$ which, for $~m=n=\dfrac12$ and $~u~=-\ell$

$=~\dfrac a2~,~$ becomes our original integral.

Answer 2

You can take the taylor series around x=0 which is the sum from 0 to infinity of $(-x^4+x^2)^/k!$ take as many terms as you want for accuracy and integrate the polynomial.

You'll get $x+x^3/3-x^5/10-5x^7/42+x^9/216+41x^{11}/3120+O(x^13)$ Integrate this over a suitably large domain which graphically looks like from -2 to 2, improve the number of terms in the expansion to improve accuracy.

Answer 3

It is worth graphing these functions (and playing with the parameters). For $x>1$ they are both monotone a similar rate of growth to an exponential. The difference is over the range $[0,1]$, where $I_{\frac{1}{4}}$ is more like $x^{\frac{1}{4}}$ and $I_{\frac{-1}{4}}$ is more like $\frac{1}{x^{\frac{1}{4}}}$.

But the main point I want to make is that our classification of some functions as elementary is somewhat arbitrary and mainly historical accident. If it was done from scratch on a more rational basis, many other functions, including these two, would be classified as elementary. They have power series, interesting relations, smooth curves etc. Plus of course it is just as easy to get their values, plot them, manipulate them etc with the proper software, as it is with cosine or log.

The first two graphs show the two Bessel functions against the corresponding powers of $x$ in the range $[0,1]$. In each case the brown curve is the Bessel function.

In the next two you see the range $x>1$: the blue curve is the exponential, the green curve is a polynomial, and the brown curve is the Bessel function.