Does $\int_0^{+\infty}\cos(x^3)\,\mathrm{d}x$ converge absolutely or conditionally? $$\int\limits_0^{+\infty}\cos(x^3)\,\mathrm{d}x$$
I think that the integral absolutely does not converge, since the function does not tend to zero, but conditionally it converges, since if you look at the graph of the function, the areas of the cosine humps decrease, and the signs alternate, and then the Leibniz integral converges, but I think this solution is not strict and I would like to get a normal one, but there are no ideas.
 A: The integral is convergent (conditionally) as shown here.
The integral is not absolutely convergent. We have
$$\int_0^R|\cos x^3|\, dx \geqslant \int_0^R\cos^2 x^3 \, dx = \int_0^R\left(\frac{1}{2}\cos 2x^3 + \frac{1}{2}\right)\, dx \\ =  \frac{1}{2}\int_0^R\cos 2x^3 \, dx + \frac{R}{2},$$
and the RHS diverges to $+\infty$ as $R \to +\infty$ since $\int_0^\infty\cos 2x^3 \, dx = 2^{-1/3}\int_0^\infty \cos u^3 \, du$ is convergent.
A: The integral converges, but you should look in an integral table (or perform computer algebra) to find:
$$\frac{\Gamma \left(\frac{1}{3}\right)}{2 \sqrt{3}} \approx 0.773343$$
Here's a graph of the integral up to $t$:

A: Consider
$$I=\int \cos( x^3)\,dx=-\frac{1}{6} x \left(E_{\frac{2}{3}}\left(-i x^3\right)+E_{\frac{2}{3}}\left(i
   x^3\right)\right)$$
$$J=\int_0^t \cos( x^3)\,dx=\frac{\Gamma \left(\frac{1}{3}\right)}{2 \sqrt{3}}-\frac{1}{6} t \left(E_{\frac{2}{3}}\left(-i t^3\right)+E_{\frac{2}{3}}\left(i t^3\right)\right)$$ For large values of $t$, the last term is
$$t \left(E_{\frac{2}{3}}\left(-i t^3\right)+E_{\frac{2}{3}}\left(i t^3\right)\right)\sim \frac{i e^{-i t^3}}{6 t^2}$$
$$K=\int_0^\infty \cos( x^3)\,dx=\frac{\Gamma \left(\frac{1}{3}\right)}{2 \sqrt{3}}$$
A: $$I=\int\limits_0^{+\infty}\cos(x^3)dx=\frac{1}{3}\int\limits_0^{+\infty}\frac{\cos(t)}{t^{\frac{2}{3}}}dt=\frac{1}{3}\Re\int\limits_0^{+\infty}\frac{e^{it}}{t^{\frac{2}{3}}}dt$$
The variable change: $t\to{s}$ $ (t={is}=se^{\frac{\pi{i}}{2}})$.
The change is valid: due to Jordan lemma integral in the complex plane along a big quarter-circle $\to0$ as its radius $R\to\infty$. Integral along a small quarter-circle around $s=0$  also $\to0$ at $r\to0$ - due to the integrand asymptotics near zero. The change of the variable (and, therefore, the limits of integration) does not affect the integral value.
$$I=\frac{1}{3}\Re\,\,e^{\frac{\pi{i}}{6}}\int\limits_0^{+\infty}\frac{e^{-s}}{s^{\frac{2}{3}}}ds=\frac{1}{3}\cos(\frac{\pi}{6})\Gamma\Bigl(\frac{1}{3}\Bigr)=\frac{1}{2\sqrt3}\Gamma\Bigl(\frac{1}{3}\Bigr)$$
