# 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!

-

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. –  Mathemagician1234 Feb 7 '12 at 17:21
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. –  Robert Israel Feb 7 '12 at 17:41
@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