Convergence of $\int_0^{\infty} x \cos (x^6)\,dx$ I feel that $\int_0^{\infty} x \cos (x^6) dx$ is convergent using the regular first year definition of an integral. I have been trying to convince a university professor of this, but according to him I am wrong. Perhaps I am?
My idea is that you do a u-substitution and get an value proportional to $\int_0^{\infty} \cos(x^3) dx$ and this is convergent because it alternates faster and faster. So, the area of each hump is alternating and absolutely smaller. From this every sequence is Cauchy.
Thoughts?
Edit: It appears that I was right. Let me curb some naysayers by recalling that $\int_0^{\infty}f(x)dx := \lim_{n\to \infty} \int_0^n f(x)dx$ and this is well defined for all $n>0$. Hence, we are just taking a limit. The question is if this limit exists.
 A: In the OP, the intuition is good, but details are missing. The oscillations are indeed more and more rapid, but one has to verify that the positive and negative parts do almost cancel. 
The integrand is well-behaved on $[0,1]$ so we can work with $\int_1^\infty x\cos(x^6)\,dx$. Rewrite this as
$$\int_1^\infty \frac{1}{x^4}\cdot x^5\cos(x^6)\,dx,$$
and integrate by parts, letting $u=\frac{1}{x^4}$ and $dv=x^5\cos(x^6)\,dx$. Then we have $du=-\frac{4}{x^5}\,dx$ and we can take $v=\frac{1}{6}\sin(x^6)$.  We end up needing to show that the integral
$$\int_1^\infty \frac{\sin(x^6)}{x^5}\,dx$$
converges, which is clear since $\sin(x^6)$ is bounded. 
Remark: If we want to be more formal, we can let $I(B)$ be the integral from $0$ to $B$. Then $I(B)$ is the integral from $0$ to $1$ plus the integral from $1$ to $B$. Essentially the same argument shows that $\lim_{B\to\infty} I(B)$ exists. 
A: This is an improper integral and is defined as
$$
\lim_{M\to\infty}\int_0^Mx\cos(x^6)\,\mathrm{d}x\tag{1}
$$
Although $\limsup\limits_{x\to\infty}x\cos(x^6)=\infty$ and $\liminf\limits_{x\to\infty}x\cos(x^6)=-\infty$, the oscillation of the integrand becomes so fast, that the limit $(1)$ exists.
$$
\begin{align}
\int_0^Mx\cos(x^6)\,\mathrm{d}x
&=\frac16\int_0^{M^6}x^{-2/3}\cos(x)\,\mathrm{d}x\\
&=\frac16\int_0^{\pi/2}x^{-2/3}\cos(x)\,\mathrm{d}x+\frac16\int_{\pi/2}^{M^6}x^{-2/3}\cos(x)\,\mathrm{d}x\\
&=\frac12\int_0^{\pi/2}\cos(x)\,\mathrm{d}x^{1/3}+\frac16\int_{\pi/2}^{M^6}x^{-2/3}\,\mathrm{d}\sin(x)\\
&=\frac12\int_0^{\pi/2}x^{1/3}\sin(x)\,\mathrm{d}x\\
&+\frac16\left(M^{-4}\sin\left(M^6\right)-(\pi/2)^{-2/3}\right)
+\frac19\int_{\pi/2}^{M^6}x^{-5/3}\sin(x)\,\mathrm{d}x\tag{2}
\end{align}
$$
Taking the limit as $M\to\infty$ yields,
$$
\begin{align}
\int_0^\infty x\cos(x^6)\,\mathrm{d}x
&=\frac12\int_0^{\pi/2}x^{1/3}\sin(x)\,\mathrm{dx}\\
&+\frac19\int_{\pi/2}^\infty x^{-5/3}\sin(x)\,\mathrm{d}x
-\frac16(\pi/2)^{-2/3}\tag{3}
\end{align}
$$
and these integrals converge.

Using a bit of contour integration, we can get a closed form for the preceding integral:
$$
\frac{\sqrt3}{12}\Gamma\left(\frac13\right)\tag{4}
$$
A: Consider this integration as sum of fragments where $cos(x^6)$ crosses zero two times. you know that the
 $$\int_{a_n}^{b_n} x \cos (x^6) dx<\int_{a_n}^{b_n} |x| dx<\Sigma (c_n \frac{2\pi}{n^6})$$ 
where $c_n$ is maximum value of x in this period. The order of $c_n$ is $O(n^1)$
$x$ grows with power one but the fragment length shrinks with power 6. So this integrate is smaller than a convergent sum of an exponential series.
A: *

*As far as convergence is concerned, see Riemann-Lebesgue lemma.

*As for evaluation, use Euler's formula in conjunction with the definition of the $\Gamma$ function to arrive at $\displaystyle\int_0^\infty f\big(x^n\big)~dx=f\bigg(\frac\pi{2n}\bigg)\cdot\Gamma\bigg(1+\frac1n\bigg)$, where $f\in\big\{\cos,\sin\big\}$ and $n>1$.
