Integrals of the form $\int_0^{+\infty} \sin g(x) \ dx$

I have no idea for the full generality, but at least I have a complete answer when $g$ is convex.

Proposition. Assume that $g : [a,\infty) \to \Bbb{R}$ is increasing, convex and unbounded. Then

(1) $\int_{a}^{\infty} \sin g(x) \, dx$ does not converge absolutely.

(2) $\int_{a}^{\infty} \sin g(x) \, dx$ converges if and only if $g'_{+}(x) \to \infty$, where $g'_{+}$ is the right-hand derivative.

(3) $\int_{a}^{\infty} \sin g(x) \, dx$ converges in Cesaro sense. That is, the limit $$\lim_{R\to\infty} \frac{1}{R} \int_{a}^{R} \int_{a}^{r} \sin g(x) \, dx dr$$ exists.

Step 1. A bit of reduction and some observations

First, since $g$ is convex, it is continuous possibly except at $0$. Using the fact that $g$ is also increasing, we know that $g$ is indeed continuous everywhere on $[0,\infty)$. Then the condition implies that there exists $a' \in [a, \infty)$ such that $g$ is constant on $[a, a']$ and $g$ is strictly increasing on $[a', \infty)$. Since none of statements (1)-(3) is affected on modification of $g$ on a finite interval, we may truncate the interval from the left and assume that $g$ is strictly increasing.

The previous paragraph shows that it suffices to consider $g$ which is strictly increasing, continuous, convex and unbounded. Then its inverse $h : [g(0), \infty) \to [a, \infty)$ is a well-defined function which is strictly increasing, continuous, concave and unbounded. Thus its right-hand derivative $h'_+(x)$ is a decreasing function such that

$$ h'_+(x) = \lim_{h \to 0^+} \frac{h(x+h) - h(x)}{h} = \lim_{k \to 0^+} \frac{k}{g(h(x)+k) - g(h(x))} = \frac{1}{g'_+(h(x))}. $$

So $h'_+(x) \to 0$ if and only if $g'_{+}(x) \to \infty$. Moreover, for any continuous function $\varphi$ on $[a, b]$ we have the following formula

$$ \int_{a}^{b} \varphi(g(x)) \, dx = \int_{g(a)}^{g(b)} \varphi(y) \, dh(y) = \int_{g(a)}^{g(b)} \varphi(y) h'_+(y) \, dy. $$

Step 2. Actual computation.

• We first resolve (1). Let us write $\rho(x) = h'_+(x)$ for brevity. Choose $m \in \Bbb{Z}$ so that $m\pi > g(a)$. Then along $R_n = h(n\pi)$ with $n > m$,

  \begin{align*} \int_{a}^{R_n} \left|\sin g(x) \right| \, dx &\geq \int_{m\pi}^{n\pi} \rho(x) \left|\sin x \right| \, dx = \sum_{k=m}^{n-1} \int_{0}^{\pi} \rho(x+k\pi) \sin x \, dx \\ &\hspace{2em} \geq \sum_{k=m}^{n-1} 2\rho((k+1)\pi) \geq \sum_{k=m}^{n-1} \frac{2}{\pi} \int_{(k+1)\pi}^{(k+2)\pi} \rho(x) \, dx \\ &\hspace{4em} \geq \frac{2}{\pi} [h((n+1)\pi) - h((m+1)\pi)] \xrightarrow[n\to\infty]{} \infty \end{align*}

  and hence (1) follows.

• Next we resolve part (2). Let $m \in \Bbb{Z}$ be as before and define $F$ by

  $$F(r) = \int_{m\pi}^{r} \rho(x) \sin x \, dx.$$

  In view of the previous section, we can investigate the convergence of $F(r)$ as $r\to\infty$ instead. Also, since $F(r)$ always lies between $F(n\pi)$ and $F((n+1)\pi)$ whenever $x \in [n\pi, (n+1)\pi]$, it suffices to investigate the convergence of $F(n\pi)$ as $n\to\infty$. Now for $n > m$,

  $$ F(n\pi) = \sum_{k=m}^{n-1} (-1)^k \int_{0}^{\pi} \rho(x+k\pi) \sin x \, dx $$

  First, the general term satisfies

  $$ \left| (-1)^k \int_{0}^{\pi} \rho(x+k\pi) \sin x \, dx \right| \geq \rho((k+1)\pi) \int_{0}^{\pi} \sin x \, dx = 2\rho((k+1)\pi). $$

  So if $F(n\pi)$ converges as $n\to\infty$, then $\rho(x)$ also converges to $0$ as $x\to\infty$. By our previous remark, this implies $g'_+(x) \to \infty$ as $x\to\infty$.

  Conversely, assume that $g'_+(x) \to \infty$ as $x\to\infty$ so that $\rho(x) \to 0$ as $x\to\infty$. Then

  $$ F(n\pi) = \int_{0}^{\pi} \left( \sum_{k=m}^{n-1} (-1)^k \rho(x+k\pi) \right) \sin x \, dx $$

  and the inner term converges uniformly by the Dirichlet test. This implies the convergence of $F(n\pi)$ and hence the convergence of $F(x)$ as $x\to\infty$.

• Finally we resolve part (3). Let $\rho_{\infty} = \lim_{x\to\infty} \rho(x)$, which exists by the monotonicity of $\rho$. Then we can write

  $$ \int_{a}^{r} \sin g(x) \, dx = \underbrace{\int_{g(a)}^{g(r)} (\rho(x) - \rho_{\infty}) \sin x \, dx}_{=:A} + \rho_{\infty} \cos g(a) - \underbrace{\vphantom{\int_{g}} \rho_{\infty} \cos g(r)}_{=:B}. $$

  Now the term $A$ converges as $r\to\infty$ by (2). So its Cesaro mean also converges. In order to investigate the Cesaro mean of the term $B$, we have to look at

  $$ \frac{1}{R} \int_{a}^{R} \cos g(r) \, dr = \frac{1}{R} \int_{g(a)}^{g(R)} \rho(x) \cos x \, dx. $$

  Using the similar 'alternating series trick' as in part (2), we can check that $\int_{g(a)}^{g(R)} \rho(x) \cos x \, dx$ is uniformly bounded in $R$. Putting altogether, we obtains not only the Cesaro convergence but also its value:

  $$ \lim_{R\to\infty} \frac{1}{R} \int_{a}^{R} \int_{a}^{r} \sin g(x) \, dx dr = \int_{g(a)}^{\infty} (\rho(x) - \rho_{\infty}) \sin x \, dx + \rho_{\infty} \cos g(a). $$

For (iii). Let $(c_n)$ be a sequence of positive numbers, $0 < c_n < \pi$ for every $n$, and let $$ h(x) := \sum_{n=0}^\infty \frac{\pi}{c_n} \chi_{[n\pi, n\pi + c_n]}(x), \qquad g(x) := \int_0^x h(t)\, dt, $$ where $\chi_A$ is the characteristic function of the set $A$. (This function $g$ is not strictly increasing, but the construction can be easily modified in this sense.)

Since $\int_{n\pi}^{(n+1)\pi} g'(x)\, dx = \pi$, we have that $g(n\pi) = n\pi$ and $$ \int_{n\pi}^{(n+1)\pi} |\sin g(x)|\, dx = \int_0^{c_n} \sin\left(\frac{\pi}{c_n}\, t\right)\, dt = \frac{2 c_n}{\pi}. $$ Hence, if $\sum_n c_n$ converges, then $\sin g(x)$ is absolutely integrable on $[0,+\infty)$.