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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?

Mark Viola
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    How is it obvious that the second term of $\text{(1)}$ vanishes as $L \to \infty$? – Sangchul Lee Jul 18 '17 at 01:02
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    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. – Sangchul Lee Jul 18 '17 at 01:30
  • @sangchullee The RL doesn't apply to the original problem since $\left |\frac{\cos(x^3/3)}{x}\right|$ is not $L^1(\nu,\infty)$. – Mark Viola Jul 18 '17 at 01:54
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    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. – Sangchul Lee Jul 18 '17 at 02:01
  • @SangchulLee Integration by parts with $u=\frac{\cos(x^3/3)}{x}$ and $v=-\frac{\cos(Lx)}{L}$ seems to work here. – Mark Viola Jul 18 '17 at 02:32
  • @MarkViola: There is a recent posting related to the integral of the Airy function over $[0,\infty)$ where the OP uses contour integration methods (steepest descent like method) that transforms the bad improper double integral into a perfectly nice double Lebesgue integral where the full power if Fubini-Tonelli's theorem can be used. – Mittens May 16 '23 at 06:01
  • @OliverDíaz Hi Oliver. That is really cool. Thank you for alerting me. – Mark Viola May 16 '23 at 13:56

1 Answers1

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Not better, not faster and just a bit late for an interesting problem.

$$I_L=\int_0^\infty \frac{\sin(Lx)}{x}\cos \left(\frac{x^3}{3}\right)\,dx$$

Using Mathematica $$I_L=\frac{L}{90\ 3^{5/6}}\Bigg(45\ 3^{2/3} \Gamma \left(\frac{1}{3}\right)\,A - L^4 \,\,\Gamma \left(-\frac{4}{3}\right)B\Bigg)$$ where $$A=\, _1F_4\left(\frac{1}{6};\frac{2}{6},\frac{3}{6},\frac{5}{6}, \frac{7}{6};\frac{L^6}{6^4}\right)\quad \text{and}\quad B=\, _1F_4\left(\frac{5}{6};\frac{7}{6},\frac{9}{6},\frac{10}{6}, \frac{11}{6};\frac{L^6}{6^4}\right)$$

Including the front factor, we face the difference between $$\frac{3-i \sqrt{3}}{12} \pi+\frac{\sqrt{\frac{\pi }{3}} }{4 L^{3/4}}e^{\frac{2 L^{3/2}}{3}}\Bigg(1+\frac{41}{48 L^{3/2}} +\frac{9241}{4608 L^3}+\frac{5075225}{663552 L^{9/2}}+O\left(\frac{1}{L^6}\right) \Bigg)$$ and $$-\frac{3+i \sqrt{3}}{12} \pi+ \frac{\sqrt{\frac{\pi }{3}} }{4 L^{3/4}}e^{\frac{2 L^{3/2}}{3}}\Bigg(1+\frac{41}{48 L^{3/2}} +\frac{9241}{4608 L^3}+\frac{5075225}{663552 L^{9/2}}+O\left(\frac{1}{L^6}\right) \Bigg)$$ So, at least for this level of (tedious) expansion, $$I_L= \frac \pi 2$$

I repeated the expansion up to $O\left(\frac{1}{L^{150}}\right)$ for the same result.

As a sanity check, I computed the derivative of $I_l$ with respect to $L$; the expression is $$\frac{I_L}{dL}=\frac{\pi \sqrt{L}}{6}\left(J_{-\frac{1}{3}}\left(\frac{2 L^{3/2}}{3}\right)+J_{\frac{1}{3}}\left(\frac{2 L^{3/2}}{3}\right)+I_{-\frac{1}{3}}\left(\frac{2 L^{3/2}}{3}\right)-I_{\frac{1}{3}}\left(\frac{2 L^{3/2}}{3}\right)\right)$$ which is $$\frac{I_L}{dL}=\frac{\pi}{2} (\text{Ai}(L)+\text{Ai}(-L))$$ which is asymptotic to $$\frac{\sqrt \pi}{4 L^{1/4}}\Bigg(e^{-\frac{2 L^{3/2}}{3}}+2 \sin \left(\frac{2L^{3/2}}{3}+\frac{\pi }{4}\right) \Bigg)+\frac{5 \sqrt{\pi }}{96 L^{7/4}}\sin \left(\frac{2 L^{3/2}}{3}-\frac{\pi }{4}\right)+\cdots$$

Claude Leibovici
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