# Proving $\int_{1}^{\infty}\frac{\sin x}{\left(\log x\right)^{\frac{1}{2}}}dx$ converges

Prove that the following improper integral converges $$\int_{1}^{\infty}\frac{\sin x}{\left(\log x\right)^{\frac{1}{2}}}dx.$$

I see that you can show this using Dirichlet's Convergence Test, but how would you show it not using this test?

• It's somewhat nasty, but you can integrate by parts twice. Let $du$ be the numerator each time and $v$ be the denominator. – Potato Jul 23 '12 at 0:13

You have two issues to address. First, is $\int_1^{1.1}\frac{\sin x}{\sqrt{\log x}}\,dx$ convergent? Second, is $\int_{1.1}^{\infty}\frac{\sin x}{\sqrt{\log x}}\,dx$ convergent?
The first integral blows up at its left end point. But we can show that for the some constant $C$, $\frac{\sin x}{\sqrt{\log x}}<\frac{C}{\sqrt{x-1}}$ for $x$ in $(1,1.1)$. Then by direct comparison, this piece is convergent. The numerator $\sin(x)$ is clearly positive and less than $1$. So we'd need to show that $\log x>k(x-1)$ for some constant $k$, for all $x\in(1,1.1)$. You can do this in the standard way by noting both are equal at $x=0$ and comparing derivatives.
The second integral breaks up into a sequence of alternating integrals $$\int_{1.1}^{\pi}\frac{\sin x}{\sqrt{\log x}}\,dx+\sum_{n=1}^{\infty}\int_{n\pi}^{n\pi+\pi}\frac{\sin x}{\sqrt{\log x}}\,dx$$ By the alternating test for integrals, we need to check that $\lim_{n\to\infty}\left|\int_{n\pi}^{n\pi+\pi}\frac{\sin x}{\sqrt{\log x}}\,dx\right|=0$. This is true, since \begin{align} \left|\int_{n\pi}^{n\pi+\pi}\frac{\sin x}{\sqrt{\log x}}\,dx\right|&<\frac{1}{\sqrt{\log(n\pi)}}\left|\int_{n\pi}^{n\pi+\pi}\sin(x)\,dx\right|\\ &=\frac{2}{\sqrt{\log(n\pi)}}\end{align} Joriki reminded me that the alternating series test also requires that the absolute values of the integrals be decreasing. (Otherwise you could have say, the positive integrals going down like $1/n$ with the negative ones going down like $1/2^n$, and the whole thing diverges.) So we should check that \begin{align} \left|\int_{n\pi}^{n\pi+\pi}\frac{\sin x}{\sqrt{\log x}}\,dx\right|&>\left|\int_{n\pi+\pi}^{n\pi+2\pi}\frac{\sin x}{\sqrt{\log x}}\,dx\right|\end{align} which is true since the numerator function just repeats another period with the same values, while the denominator function becomes larger.