12
$\begingroup$

I'm interested in the convergence of integrals of the form $$\int_0^{+\infty} \sin g(x) \ dx$$

where $g$ is nonnegative, increasing and growing without bound as $x \to +\infty$ (hence $\sin g(x)$ oscillates as $x \to +\infty$). For example, it's known that $$\int_0^{+\infty} \sin x^p \ dx$$ converges for $p>1$. Similarly, $\int_0^{+\infty} \sin(\exp x) \ dx$ converges.

Here are three questions (they are related enough that I didn't think it was worth making three separate questions).

Consider functions $g:[0, +\infty) \to [0, +\infty)$ which are continuous, strictly increasing and unbounded.

$(i)$ Suppose we add the additional hypothesis that $g$ is strictly convex. Does $\int_0^{+\infty} \sin g(x) \ dx$ converge?

$(ii)$ Can we characterize those $g$'s for which $\int_0^{+\infty} \sin g(x) \ dx$ converges?

$(iii)$ Is there a $g$ such that $\int_0^{+\infty} \sin g(x) \ dx$ converges absolutely?

$\endgroup$
8
$\begingroup$

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:

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

$\endgroup$
2
$\begingroup$

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 $[0,+\infty)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.