<s>[INST] $\int_{0}^{\infty} x \cdot \cos(x^3) dx$ convergence $$\int_{0}^{\infty} x \cdot \cos(x^3) dx$$
I only want to prove, that this integral converges, I don't need to calculate the exact value.
I don't know what to do with the cosinus, I can't get rid of it.
I know that the integral is equal to $$\frac{1}{3} \cdot \int_{0}^{\infty} \frac{\sin(x^3)}{x^2} dx$$
but here is also the problem, that I can't get rid of the sinus...
Any hints?
 [/INST] Thanks, to everyone!
$$\int_{0}^{\infty} x \cdot \cos(x^3) dx = \frac{1}{3} \cdot \int_{0}^{\infty} \frac{1}{x} \cdot 3 \cdot x^2 \cdot \cos(x^3) dx$$
and because of $\int 3 \cdot x^2 \cdot \cos(x^3) dx = \sin(x^3)$ we get

$$\int_{0}^{\infty} x \cdot \cos(x^3) dx=\frac{1}{3} \cdot \int_{0}^{\infty} \frac{\sin(x^3)}{x^2} dx$$

We can split the integral into 2 parts:
$$\frac{1}{3} \cdot \int_{0}^{\infty} \frac{\sin(x^3)}{x^2} dx = \frac{1}{3} \cdot \left(\int_{0}^{1} \frac{\sin(x^3)}{x^2} dx+\int_{1}^{\infty} \frac{\sin(x^3)}{x^2} dx\right)$$
$\int_{0}^{1} \frac{\sin(x^3)}{x^2} dx$ is finite, so we only need to prove, that $\int_{1}^{\infty} \frac{\sin(x^3)}{x^2} dx$ is finite, too.
Since $\frac{\sin(x^2)}{x^2} \le \frac{1}{x^2}$ for $x > 0$, hence 

$$\int_{1}^{\infty} \frac{\sin(x^3)}{x^2} dx \le \int_{1}^{\infty} \frac{1}{x^2} dx$$

$\int_{1}^{\infty} \frac{1}{x^2} dx$ is bounded, hence $\int_{1}^{\infty} \frac{\sin(x^3)}{x^2} dx$ is bounded, too.
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