$\int_{0}^{\infty} x \cdot \cos(x^3) dx$ 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?
 A: 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.
A: Hint: See Riemann-Lebesgue lemma. As for its value, use the definition of the $\Gamma$ function together with Euler's formula.
A: Substitution $u=x^3$ and integration by parts gives
$$
\begin{align}
\int_0^\infty x\cos(x^3)\mathrm{d}x
&=\int_0^\infty u^{1/3}\cos(u)\frac{\mathrm{d}u}{3u^{2/3}}\\
&=\frac13\int_0^\infty u^{-1/3}\cos(u)\,\mathrm{d}u\\
&=\frac19\int_0^\infty u^{-4/3}\sin(u)\,\mathrm{d}u\tag{1}
\end{align}
$$
$(1)$ converges absolutely by comparison to
$$
\overbrace{\frac19\int_0^1 u^{-1/3}\,\mathrm{d}u}^{|\sin(u)|\le u} +\overbrace{\frac19\int_1^\infty u^{-4/3}\,\mathrm{d}u}^{|\sin(u)|\le1}\tag{2}
$$
In fact,
$$
\begin{align}
\frac13\int_0^\infty u^{-1/3}\cos(u)\,\mathrm{d}u
&=\frac13\mathrm{Re}\left(\int_0^\infty u^{-1/3}e^{iu}\,\mathrm{d}u\right)\\
&=\frac13\mathrm{Re}\left(e^{\pi i/3}\int_0^\infty u^{-1/3}e^{-u}\,\mathrm{d}u\right)\\
&=\frac13\cos\left(\frac\pi3\right)\Gamma\left(\frac23\right)\\[3pt]
&=\frac16\Gamma\left(\frac23\right)\tag{3}
\end{align}
$$
A: I doubt this is the best way, but it will do.
Firstly, notice that you have no problems on the interval $[0,1]$, and in fact the integral is bounded above by $1$ there. So we are left to consider
$$ \int_1^\infty x\cos(x^3)dx.$$
Perform the substitution $u = x^3$ to see that this is equivalent to the convergence of
$$ \int_1^\infty \frac{\cos(x)}{\sqrt[3] x} dx.$$
The numerator $\cos x$ oscillates positive and negative in perfectly regular sequences, positive in intervals of the form $x \in [2n\pi + 3\pi/2, 2n\pi + 5\pi/2]$, and negative in intervals of the form $x \in [2n\pi + \pi/2, 2n\pi + 3\pi/2]$. The denominator is always positive, and the fractions $\frac{1}{\sqrt[3] x}$ are always decreasing. So the area in each positive hump and each negative hump is decreasing, and going to zero. 
As the humps alternate between positive and negative signs, and the areas in each hump are monotonically decreasing, the integral will converge to some limit $L$. This is a mimicry of the proof of the alternating series test.
In fact, if you call $H_{n}$ the area of the $n$th hump, so that $H_{2n} > 0$ and $H_{2n + 1} < 0$, then as $H_n$ is alternating, $\lvert H_{n + 1}\rvert < \lvert H_n \rvert$, and $\lvert H_n \rvert \to 0$, then
$$ \int_1^\infty \frac{\cos x}{\sqrt[3]x} dx = \sum_{n \geq 1} H_n = L$$
exactly by the classical alternating series test. 
A: The integral from $0$ to $1$ is no problem, so we deal with the integral from $1$ to $\infty$.
Rewrite our integrand as $\frac{1}{x} x^2\cos(x^3)$, and integrate by parts from $1$ to $B$,   using $u=\frac{1}{x}$ and $dv=x^2\cos(x^3)\,dx$. 
Then $du=-\frac{1}{x^2}\,dx$ and we can take $v=\frac{1}{3}\sin(x^3)$. 
The integral from $1$ to $B$ is
$$\left. \frac{1}{3x}\sin(x^3)\right|_1^B +\int_1^B \frac{1}{3x^2}\sin(x^3)\,dx.   $$
The integral on the right behaves nicely as $B\to\infty$, since the integrand is bounded in absolute value by $\frac{1}{3x^2}$. 
A: If we know that the first integral is equal to the second, the rest is simple.
$$
\left|\frac{\sin{x^3}}{x^2}\right|\leq\frac{1}{x^2},
$$
but 
$$
\int_{1}^{\infty}\frac{1}{x^2}\,dx<\infty
$$
and the function is bounded on $[0,1]$, hence
$$
\int_{0}^{1}\frac{\sin{x^3}}{x^2}\,dx<\infty.
$$
