I would like to be able to integrate the sine (and cosine of a cubic polynomial) e.g.

$\int_X^Y \sin{(ax^3 + bx^2 + cx + d)} {\rm d}x$

Does anyone know if an analytical solution to this exists? Using a symbolic mathematics package (e.g. sympy) I know that the integral

$\int_X^Y \sin{(az^3)} {\rm d}z$

can be found in terms of a $_1{}F_2$ hypergeomtric function, but I can't find a change of variables between the two to allow the full cubic to be solved easily.

I also know that for a quadratic a solution can be found in the form of Fresnel integrals, but I can't seem to apply my method to find a similar thing for a cubic.

The underlying reason for this question is that I'd like (if it is actually possible) to find an analytic Fourier transform of a sinusoidal signal with a cubic phase evolution, so the actual thing I'm trying to find an expression for is

$\int_X^Y \exp{[i\pi(ax^3 + bx^2 + cx + d)]} {\rm d}x$.


1 Answer 1


First of all, make an appropriate linear substitution $x=At+B$, so that the cubic polynomial

$ax^3+bx^2+cx+d$ becomes $t^3+\alpha t+\beta$. So you are left with integrating $\exp\Big[i(t^3+\alpha t+\beta)\Big]$.

But $e^{i\beta}$ is a constant, and can be therefore extracted outside the integral sign. Then, setting the

limits of integration to $(0,\infty)$, we have $$I~=~C+iS~=~\dfrac{e^{i\beta}}6~\bigg\{\dfrac{4\pi}{\sqrt[3]3}\cdot w_{12}\cdot\text{Bi}\bigg[\dfrac\alpha{\sqrt[3]3}\cdot w_3\bigg]-i\alpha^2\cdot~_1F_2\bigg[1~;~\dfrac43~,~\dfrac53~;~\bigg(\dfrac\alpha3\bigg)^3~\bigg]\bigg\},$$ where Bi is the Airy function of the second kind, and $w_k=\exp\bigg(\dfrac{2\pi i}k\bigg).~$ Since Airy functions

are expressible in terms of Bessel functions, it implicitly follows that, in order for the general

integral you wrote $($with integration limits X and $Y)$ to possess a closed form, “incomplete”

Airy, and thus “incomplete” Bessel functions, are needed. However, the latter do not “exist”

as an object of study, and therefore do not possess any consecrated or universally accepted

symbolic notation.

  • $\begingroup$ Brilliant, thanks very much. $\endgroup$ Oct 8, 2015 at 20:33

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