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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?

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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.

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

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