Does this improper integral converge? $$\int^\infty_0\cos x^3dx$$
I think no, because $\cos x^3$ keeps jumping between $-1$ and $1$. How to justify this rigorously?
 A: The part from $0$ to $1$ is harmless, so let's work from $1$ to $\infty$. Consider
$$\int_1^M \frac{1}{3x^2}3x^2 \cos(x^3)\,dx,$$
and integrate by parts, using $u=\frac{1}{3x^2}$ and $dv=3x^2\cos(x^3)\,dx$.
We get
$$\left. \frac{1}{3x^2}\sin(x^3)\right|_1^M -\int_1^M \frac{-2\sin(x^3)}{3x^3}\,dx.$$
Since $|\sin(x^3)|\le 1$, everything behaves nicely as $M\to\infty$. 
A: You can use complex analysis to derive the result convincingly.  Consider the integral
$$\oint_C dz \, e^{i z^3}$$
where $C$ is a circular wedge on the positive real, upper half plane, of radius $R$, forming an angle of $\pi/6$ with the real axis.  Thus, this contour integral is equal to
$$\int_0^R dx \, e^{i x^3} + i R \int_0^{\pi/6} d\theta \, e^{i \theta} \, e^{i R^3 e^{i 3 \theta}} + e^{i \pi/6}\int_R^0 dt \, e^{-t^3} $$
Note that, in the 3rd integral, $i e^{i 3 \pi/6} = i^2=-1$.
The second integral vanishes as $R \to \infty$.  To see this, note that its magnitude is bounded by
$$R \int_0^{\pi/6} d\theta \, e^{-R^3 \sin{3 \theta}} \le  R \int_0^{\pi/6} d\theta \, e^{-6 R^3 \theta/\pi} \le \frac{\pi}{6 R^2}$$
By Cauchy's theorem, the contour integral is zero.  Thus we can say that
$$\int_0^{\infty} dx \, e^{i x^3} = e^{i \pi/6} \int_0^{\infty} dt \, e^{-t^3}$$
which is clearly a convergent integral. The integral you seek is the real part of this.
To get a number for the integral, sub $t=u^{1/3}$, $dt = (1/3) t^{-2/3}$ and get for the original integral
$$\cos{\frac{\pi}{6}} \frac13 \int_0^{\infty} du \, u^{-2/3} \, e^{-u} = \frac{\sqrt{3}}{6} \Gamma \left (\frac13 \right ) $$
A: Generally speaking, $n!=\mathcal G\bigg(\dfrac1n\bigg)$, where $\displaystyle\mathcal G(n)=\int_0^\infty e^{-x^n}dx$ is a generalized Gaussian integral. Since $e^{ix}=\cos x+i\sin x$, we have $\cos(x^n)=\Re\Big(e^{ix^n}\Big)$ and $\sin(x^n)=\Im\Big(e^{ix^n}\Big)$, so our integral becomes $\displaystyle\int_0^\infty\cos(x^n)dx=\Gamma\Big(1+\tfrac1n\Big)\cdot\cos\frac\pi{2n}~$ and $~\displaystyle\int_0^\infty\sin(x^n)dx=\Gamma\Big(1+\tfrac1n\Big)\cdot\sin\frac\pi{2n}$ for $n>1$.
A: The integrand oscillates, but the residue of that adds up to a finite amount. 
One might try to evaluate this integral using the method of steepest descent. In any case, Wolfram gives result as $\frac{\Gamma(\frac{1}{3})}{2\sqrt{3}}$.
