In This Answer, I wrote

"It is straightforward to show that $\displaystyle \lim_{L\to \infty}\int_0^\infty \frac{\sin(Lx)}{x}\,\cos(x^3/3)\,dx=\frac\pi2$."

For completeness, I've included the "straightforward approach" that I had in mind in that which now follows.

First, for any $\nu>0$ we can write

$$\begin{align} \int_0^\infty \frac{\sin(Lx)}{x}\,\cos(x^3/3)\,dx-\frac\pi2&=\int_0^{\nu} \frac{\sin(Lx)}{x}\,\left(\cos(x^3/3)-1\right)\,dx\\\\ &+\int_{\nu}^\infty \frac{\sin(Lx)}{x}\,(\cos(x^3/3)-1)\,dx\tag1\\\\ &=\int_0^{\nu L} \frac{\sin(x)}{x}\,\left(\cos\left(\frac{x^3}{3L^3}\right)-1\right)\,dx\\\\ &+\int_{\nu L}^\infty \frac{\sin(x)}{x}\,\left(\cos\left(\frac{x^3}{3L^3}\right)-1\right)\,dx \end{align}$$

The second integral on the right-hand side of $(1)$ converges as an improper Riemann integral, which can be shown be integration by parts with $u=\frac{\sin(Lx)}{x^3}$ and $v=\sin(x^3/3)$. And integration by parts with $u=\frac{\cos(x^3/3)}{x}$ and $v=\frac{\cos(Lx)}{L}$ facilitates showing that the limit as $L\to \infty$ is $0$.

Second, given $\epsilon>0$, there exists a $\delta>0$ such that $|\cos(x^3/3L^3)-1|<\epsilon$ whenever $|x|<\delta$. We take $\nu <\min(\delta, (3\pi)^{1/3})$. Since $\cos(x^3/3L^3)-1$ is decreasing for $x\in [0,\nu L]$, the second mean value theorem guarantees that there exists a number $\xi \in (0,\nu L)$ such that

$$\begin{align} \left|\int_0^{\nu L} \frac{\sin(x)}{x}\,\left(\cos\left(\frac{x^3}{3L^3}\right)-1\right)\,dx\right|&=\left|\left(\cos\left(\frac{\nu^3}{3}\right)-1\right)\right|\,\left|\int_\xi^{\nu L}\frac{\sin(x)}{x}\,dx\right|\\\\ &<\epsilon \pi/2 \end{align}$$

whence we see that $\lim_{L\to \infty}\int_0^{\nu L} \frac{\sin(x)}{x}\,\left(\cos\left(\frac{x^3}{3L^3}\right)-1\right)\,dx=0$. And we are done!

While the given approach is a standard one, I am interested in seeing "better, stronger, faster" approaches. For example, direct use of the Dominated Convergence theorem to $\int_0^\infty \frac{\sin(x)}{x}\cos(x^3/3L^3)\,dx$ doesn't seem to apply here. The Riemann-Lebesgue Lemma fails since $\frac{\cos(x^3/3)}{x}$ is not $L^1$. The function $\cos(x^3/3)$ does not have compact support on $\mathbb{R}$. And integration by parts schemes haven't appeared to be illuminating. So, is there a "better, stronger, faster" approach?

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    $\begingroup$ How is it obvious that the second term of $\text{(1)}$ vanishes as $L \to \infty$? $\endgroup$ – Sangchul Lee Jul 18 '17 at 1:02
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    $\begingroup$ The integrand also changes with $L$, so the vanishing of the second term does not come free as opposed to the case with fixed integrand. As an artificial example, we have $$ \int_{\nu L}^{\infty} \frac{1}{x} \cos\left(\frac{x^3}{3L^3}\right) \, dx = \int_{\nu}^{\infty} \frac{1}{x} \cos\left(\frac{x^3}{3}\right) \, dx $$ and hence it does not converge to $0$ as $L\to\infty$ for many choices of $\nu > 0$. The vanishing of the second term is attributed to the high-oscillation as in Riemann-Lebesgue lemma, which makes proof less obvious. $\endgroup$ – Sangchul Lee Jul 18 '17 at 1:30
  • $\begingroup$ @sangchullee The RL doesn't apply to the original problem since $\left |\frac{\cos(x^3/3)}{x}\right|$ is not $L^1(\nu,\infty)$. $\endgroup$ – Mark Viola Jul 18 '17 at 1:54
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    $\begingroup$ Yes, that is why I mentioned 'as in Riemann-Lebesgue lemma', rather than directly attributing to the lemma itself. The main reason for vanishing of $$ \int_{v}^{\infty} \frac{\sin (Lx)}{x}\cos\left(\frac{x^3}{3}\right) \, dx $$ is the cancellation by high-oscillation of $\sin(Lx)$, but improving this intuition to the actual proof requires some work. $\endgroup$ – Sangchul Lee Jul 18 '17 at 2:01
  • $\begingroup$ @SangchulLee Integration by parts with $u=\frac{\cos(x^3/3)}{x}$ and $v=-\frac{\cos(Lx)}{L}$ seems to work here. $\endgroup$ – Mark Viola Jul 18 '17 at 2:32

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