Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How could I solve this integral:

$$\int_0^\infty \exp[-ix(ax^{2}+bx+c)+gx] \,\mathrm{d}x $$

where $a, \ b, \ c$ and $ g $ are real constants, analytically?

And there is any way to solve this integral:

$$ \int_0^\infty \exp\left[-ix\left(a+\sum_0^\infty b_{n}x^{n}\right)+c\right] \, \mathrm{d}x $$

where $a, \ b_n, \ c$ and $ c $ are real constants, without using numerical methods?

share|improve this question
Neither integral has a closed form. In the second integral, the integrand can be just about anything you like, since the series can be the series expansion of any analytic function. –  joriki Jul 6 '11 at 12:22
yeah, the integrand could really be anything... but what I meant was just a linear combination of $x^{n}$, just like this: $x^{2}+x+1$. –  Rodrigo Thomas Jul 6 '11 at 15:40
I see -- then you should put a finite limit on the sum. (Also $b_0$ is redundant with $a$.) Even the integral for a general quadratic exponent has no closed form (it can be expressed in terms of the error function) -- you get a closed form only for a linear exponent or for a purely quadratic one without linear term. By the way, for this sort of thing Wolfram Alpha is quite helpful; with a simple integral like this, if WA doesn't give you a closed form, that's a very strong indication that there isn't one. –  joriki Jul 6 '11 at 16:16
@joriki I already tried to solve both integrals using Mathematica, and the doesn't return a closed form. Thank you! –  Rodrigo Thomas Jul 6 '11 at 16:50

1 Answer 1

For g=0, these are generalized Fresnel integrals as outlined in http://arxiv.org/abs/1211.3963 including cubic, quartic phases and so on as defined in the article.

share|improve this answer
No they are not; their phase is cubic, not quadratic. –  Ron Gordon Nov 19 '13 at 20:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.