# Does $\int_{1}^{\infty}\sin(x\log x)dx$ converge?

I'm trying to find out whether $\int_{1}^{\infty}\sin(x\log x)dx$ converges, I know that $\int_{1}^{\infty}\sin(x)dx$ diverges but $\int_{1}^{\infty}\sin(x^2)dx$ converges, more than that, $\int_{1}^{\infty}\sin(x^p)dx$ converges for every $p>0$, so it should be converges in infinity. I'd really love your help with this.

Thanks!

• $\int_1^\infty \sin x \, dx$ diverges but $\int_1^\infty \sin (x^p)\, dx$ converges for every $p>0$? One of those statements must be wrong. Nov 25 '15 at 11:57

Since $x\log(x)$ is monotonic on $[1,\infty)$, let $f(x)$ be its inverse. That is, for $x\in[0,\infty)$ $$f(x)\log(f(x))=x\tag{1}$$ Differentiating implicitly, we get $$f'(x)=\frac{1}{\log(f(x))+1}\tag{2}$$ Then \begin{align} \int_1^\infty\sin(x\log(x))\;\mathrm{d}x &=\int_0^\infty\sin(x)\;\mathrm{d}f(x)\\ &=\int_0^\infty\frac{\sin(x)}{\log(f(x))+1}\mathrm{d}x\tag{3} \end{align} Since $\left|\int_0^M\sin(x)\;\mathrm{d}x\right|\le2$ and $\frac{1}{\log(f(x))+1}$ monotonically decreases to $0$, Dirichlet's test (Theorem 17.5) says that $(3)$ converges.

This is a version of the Van der Corput lemma, basically.

Note that it's enough to find some $n$ for which $\int_n^{\infty} \sin(x\log(x))\,dx$ converges. The key facts about $f(x) = x\log(x)$ that allow this are a) $\lim_{x \rightarrow \infty} f'(x) = \infty$ and b) $f''(x) > 0$ for $x$ large enough. Specifically, we write $$\int_n^{\infty} \sin(f(x))\,dx = \int_n^{\infty} f'(x) {\sin(f(x)) \over f'(x)}\,dx$$ $$= \lim_{N \rightarrow \infty} \int_n^{N} (f'(x) \sin(f(x)){1 \over f'(x)}\,dx$$ Integrating the integral on the right by parts you get $$\int_n^{N} (f'(x) \sin(f(x)){1 \over f'(x)}\,dx = -{\cos(f(N)) \over f'(N)} + {\cos(f(n)) \over f'(n)} + \int_n^N \cos(f(x)){d \over dx}{1 \over f'(x)}$$ $$= -{\cos(f(N)) \over f'(N)} + {\cos(f(n)) \over f'(n)} - \int_n^N \cos(f(x)) {f''(x) \over (f'(x))^2}$$ As $N$ goes to infinity the first term goes to zero since $f'(x)$ goes to $\infty$ as $x$ goes to $\infty$ and $|\cos(f(N))| \leq 1$. The third term is bounded in absolute value by $$\int_n^N\bigg|{f''(x) \over (f'(x))^2}\bigg|\,dx$$ Since $f''(x) > 0$ we can just take off the absolute values to get $$\int_n^N{f''(x) \over (f'(x))^2}\,dx$$ Integrating this becomes $${1 \over f'(N)} - {1 \over f'(n)}$$ Since $f'(N) \rightarrow \infty$ as $N$ goes to $\infty$ this converges as $N$ goes to infinity. Hence the integral $\int_n^{\infty} \cos(f(x)) {f''(x) \over (f'(x))^2}$ converges absolutely, and thus converges.

Hence we have shown the original integral converges.

• +1. This kind of answer reminds all of us why we fell in love with calculus after mastering it either as honors undergraduates or after relearning it in graduate school. Feb 7 '12 at 17:21

The graph of the integrand consists of a series of positive and negative humps, each one narrower than the last but with the same height. Therefore if you define the infinite series consisting of the positive and negative areas, it's an alternating series in which the terms decrease in magnitude and also approach zero, so it converges.

• +1 for the outline. Of course you need to show that the size of the humps is approaching zero, and not merely getting smaller with each step. Feb 7 '12 at 16:51
• also you need to know that the height of the function doesn't ever grow on average, so as to compensate for the decreased width... otherwise the integrals over the humps may not decrease. Feb 7 '12 at 16:53
• You might try $\int_0^\infty f(\sin(x \ln x))\ dx$ where $f(t) = t^3$ for $t\ge 0$ and $t$ for $t \le 0$. This also has humps of the same height and width, but it won't give you an alternating series. Feb 7 '12 at 17:41
• Thanks for the corrections. I've made appropriate edits.
– user13618
Feb 7 '12 at 18:20
• @Robert: Furthermore, it diverges since the average of $|\sin^3(x)|$ is $2/3$ the average of $|\sin(x)|$.
– robjohn
Feb 7 '12 at 18:44