Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question

3 Answers 3

up vote 12 down vote accepted

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.

share|improve this answer

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.

share|improve this answer
3  
+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

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.

share|improve this answer
1  
+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. –  alex.jordan Feb 7 '12 at 16:51
1  
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. –  Zarrax Feb 7 '12 at 16:53
5  
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
    
Thanks for the corrections. I've made appropriate edits. –  Ben Crowell 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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