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$$\int^\infty_0\cos x^3dx$$ I think no, because $\cos x^3$ keeps jumping between $-1$ and $1$. How to justify this rigorously?

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The integral converges. – user2345215 Apr 2 '14 at 20:47
Wolfram Alpha result – Felix Castor Apr 2 '14 at 20:49
Ok. How to prove it then? No computer proving... – user130916 Apr 2 '14 at 20:51
up vote 7 down vote accepted

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 $M\to\infty$.

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It is not obvious to me that $\int_1^M\frac{\sin(x^3)}{6x^3}dx$ converges. – user130916 Apr 2 '14 at 21:08
Recall that if an integral converges absolutely, then it converges. We have $\left|\frac{\sin(x^3)}{6x^3}\right|\le \frac{1}{6x^3}$. Easily, $\int_1^\infty \frac{1}{6x^3}\,dx$ converges, so by Comparison our integral converges absolutely. – André Nicolas Apr 2 '14 at 21:12
@user130916: Thank you, corrected. (I was paying no attention to constants, but that's no excuse!) – André Nicolas Apr 2 '14 at 23:03
I don't see it as coming out of the above solution, would be interested in your argument. But the Ratio Test takes care of the series for $\cos(x^3)$. – André Nicolas Jan 2 '15 at 11:13
Just substitute $n^3$ for $x$ in the standard Maclaurin expansion of $\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$. – André Nicolas Jan 3 '15 at 7:00

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

$$\cos{\frac{\pi}{6}} \frac13 \int_0^{\infty} du \, u^{-2/3} \, e^{-u} = \frac{\sqrt{3}}{6} \Gamma \left (\frac13 \right ) $$

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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 $n>1$.

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For the case $n=2$, see Fresnel integrals. – Lucian Apr 3 '14 at 6:17

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 $\frac{\Gamma(\frac{1}{3})}{2\sqrt{3}}$.

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