Convergence of $\int_{-\infty}^{\infty} \sin p(x)\,dx$ where $p$ is a polynomial with $\deg p>1$

I thought about this problem today, and tried to solve it. Let $ax^n$ be the leading term of $p$.

I can prove that $\displaystyle\int_{-\infty}^{\infty}\sin ax^n\,dx$ converges (below), and I argued that since $p(x)\sim ax^n$ for sufficiently large $x$, we have $\sin p(x)\sim \sin ax^n$, and so the convergence of $\displaystyle\int_{-\infty}^{\infty}\sin p(x)\,dx$ is determined by that of $\displaystyle \int_{-\infty}^{\infty}\sin ax^n\,dx$.

I am not sure whether this is correct though, and would appreciate some help if it is not. Many thanks!

$n>1$ throughout.

Call $ax^n$ the leading term of $p$. For sufficiently large $x,\;p(x)\sim ax^n$, and thus $\sin p(x)\sim \sin ax^n$

Without loss of generality, assume $a>0$ and restrict ourselves to $\mathbb{R}^+$. It is sufficient to consider:

$$\int_{0}^{\infty} \sin ax^n\,dx$$

For this, let $t=ax^n:$

$$\int_0^{\infty} \sin x^n\,dx=\frac{1}{\sqrt[n]{a}n}\int_0^{ \infty} t^{\frac{1}{n}-1}\sin t\,dt$$

Now, break up the real line into segments of length $\pi,$ and let $t=w+k\pi:$

$$\sum_{k\geq 0} \frac{1}{\sqrt[n]{a}n} \int_{k\pi}^{(k+1)\pi} t^{\frac{1}{n}-1}\sin t\,dt=\sum_{k\geq 0} \frac{(-1)^k}{\sqrt[n]{a}n}\underbrace{\int_0^{\pi}\frac{\sin w}{(w+k\pi)^{ 1-\frac{1}{n}}}\,dw}_{s_k}$$ We now work our way towards the Leibniz test:

$$0<s_k<\int_0^{\pi}\frac{dw}{(w+k\pi)^{1-\frac{1}{n}}}=n\sqrt[n]{\pi}\Big((k+1)^{\frac{1}{n}}-k^{\frac{1}{n}}\Big)\to 0\;\;\text{as}\;k\to \infty$$

And of course, since $\sin w>0$ for $w\in (0,\pi):$

$$\Big(w+(k+1)\pi\Big)^{\frac{1}{n}-1}<\Big(w+k\pi\Big)^{\frac{1}{n}-1}\Rightarrow s_{k+1}<s_k$$

Thus $s_k$ is a null sequence and is monotonically decreasing hence $\displaystyle\sum_{k\geq 0} (-1)^k s_k$ converges, and it follows that the integral in question converges also.

• Why is it the case that $\sin p(x)\sim \sin ax^n$? Commented Aug 23, 2013 at 14:02
• @detnvvp My reasoning was that this follows from $p\sim ax^n$ for large $x$. However, I am unsure of whether this reasoning is correct, hence my question. Commented Aug 23, 2013 at 14:05

Suppose that $f(x)$ is a function where $\lim_{x \to \infty}f(x)=\infty$ and $f'(x)$ is increasing and positive on $[N, \infty)$, for some $N>0$. Note that a polynomial of degree $>1$ with positive leading coefficient satisfies this condition. Suppose that $x_1>N$ such that $f(x_0)=2 k \pi$ for some $k \in \mathbb{N}$. We wish to find a condition so that $$\int_{x_0}^{\infty} \sin \left(f(x)\right) dx$$ converges. First, note that $f$ is strictly increasing to $\infty$, so for each $m \in \mathbb{N}$, there is an unique $x_m$ such that $$f(x_m)= (2k+m)\pi.$$ If $x$ is between $x_m$ and $x_{m+1}$, then $\int_{x_0}^x \sin(f(t))dt$ is between $\int_{x_0}^{x_m} \sin(f(t))dt$ and $\int_{x_0}^{x_{m+1}} \sin(f(t))dt$. As such, it suffices to show that the following series converges: $$\sum_{m=1}^{\infty} \int_{x_m}^{x_{m+1}} \sin(f(t))dt.$$ By our defintion of $x_m$, this series is alternating in sign. Using the substition $u=f(t)$ yields $$\int_{x_m}^{x_{m+1}} \sin(f(t))dt = \int_{(2k+m)\pi}^{(2k+m+1)\pi} \frac{\sin(u)}{f'(f^{-1}(u))} du.$$

Note that the conditions on $f$ force that the function $f'(f^{-1}(u))$ is increasing. Hence $$|\int_{x_m}^{x_{m+1}} \sin(f(t))dt | > |\int_{x_{m+1}}^{x_{m+2}} \sin(f(t))dt|$$ for each $m$. The last condition we need for convergence is only $$\lim_{m\to\infty} \int_{(2k+m)\pi}^{(2k+m+1)\pi} \frac{\sin(u)}{f'(f^{-1}(u))} du =0,$$ so the last condition we would want is that $f'(x)\to \infty$, which is satisfied by our polynomial.

• Thanks a lot for this. Could you possibly comment on whether my solution is correct (this was my main question)? Commented Aug 25, 2013 at 2:35
• One issue for your argument is that you would need to specifically deal with the lower order terms. In particular, you say that $\sin(p(x)) \sim \sin(a x^n)$ which is false. Notice that $\sin(x^2+\pi)=-\sin(x^2)$ has the opposite sign of $\sin(x^2)$. Commented Aug 25, 2013 at 17:57

Select an $N$ such that $p'(x) \neq 0$ for $|x| > N$. It suffices to show that $\int_N^{\infty} \sin p(x)\,dx$ and $\int_{-\infty}^{-N} \sin p(x)\,dx$ are convergent. We do the first integral as the second is done the same way. For $M > N$ write $$\int_N^M \sin p(x)\,dx = \int_N^M p'(x)\sin p(x) {1 \over p'(x)}\,dx$$ Integrate by parts, integrating $p'(x)\sin p(x)$ to $-\cos p(x)$ and differentiating ${1 \over p'(x)}$. we get $$\int_N^M p'(x)\sin p(x) {1 \over p'(x)}\,dx = -{\cos p(M) \over p'(M)} + {\cos p(N) \over p'(N)} - \int_N^M \cos p(x){p''(x) \over p'(x)^2}\,dx$$ Now let $M \rightarrow \infty$. Since $|\cos(p(M))| \leq 1$ and $p'(M) \rightarrow \infty$ as $M \rightarrow \infty$, the first term goes to zero. Also, the degree of the denominator of ${\displaystyle {p''(x) \over p'(x)^2}}$ exceeds the degree of the numerator by the degree of $p(x)$, which is at least $2$, so the integral converges absolutely, using that $|\cos p(x)| \leq 1$. So the integral converges. In particular, we have $$\int_N^{\infty} \sin p(x) \,dx = {\cos p(N) \over p'(N)} - \int_N^{\infty} \cos p(x){p''(x) \over p'(x)^2}\,dx$$ Here the right-hand integral converges absolutely.

• Right. Hence the first sentence of your post... Sorry for the noise.
– Did
Commented Aug 23, 2013 at 17:12
• Thanks you. Is there any chance you could comment on whether my solution is correct (this was my main question)? Commented Aug 25, 2013 at 2:35