# Why is this integral not absolutely integrable?

I would like to prove that the following integral is not absolutely convergent: $$\int\limits_{1}^\infty \frac{1}{x}\cdot \sin(\alpha x+c)\cdot \sin(\beta x+ c)dx.$$ $\alpha,\beta,c>0$ are real constants. I could manage to prove that this integral is convergent for $\alpha\neq \beta$. But I do not know why it is not absolutely convergent. Can anyone give me a hint?

Any help will be very appreciated.

Best wishes

• Here's the intuition: The sequence (-1)^n / n is convergent, but 1/n is not. This integral is convergent because you can think of it as 1/x (which is divergent) modified by the two sin functions so that parts of it are positive and parts are negative. When you take the absolute value, you then get all parts being positive or 0 which makes it closer to 1/x. – Nitin Oct 25 '16 at 12:41
• But actually, I'm surprised you managed to show that integral converged for all $\alpha, \beta$. I don't think it should converge when $\alpha = \beta$. – Nitin Oct 25 '16 at 12:47
• Yes, $\alpha \ne \beta$ is required for convergence – GEdgar Oct 25 '16 at 12:55
• I unfortunately haven't had any good ideas. If you reduce to the case that $a, b$ are rational then you can compute the integral of $|\sin{ax} \sin{bx}|$ over one period (call this value $C$ and the period $T$) then you can show that the integral is greater than $\sum_{n > 0} C / (1+Tn)$ which diverges. – Nitin Oct 25 '16 at 14:22
• @Nitin: there is a small issue: what is the period of $\left| \sin(\pi x)\sin(\sqrt{2} x)\right|$, for instance? – Jack D'Aurizio Oct 26 '16 at 3:47

If $\alpha=\beta$ the integral is not even conditionally convergent, hence we may assume $\alpha\neq\beta$.
By Fourier series or other means it is not difficult to show that for some $K_{\alpha,\beta}>0$ $$\int_{1}^{N}\left|\sin(\alpha x+c)\right|\cdot \left|\sin(\beta x+c)\right|\,dx = K_{\alpha,\beta} N + O(1)\tag{1}$$ as $N\to +\infty$, hence by integration by parts
$$\int_{1}^{N}\frac{\left|\sin(\alpha x+c)\right|\cdot \left|\sin(\beta x+c)\right|}{x}\,dx =O(1)+K_{\alpha,\beta}\int_{1}^{N}\frac{dx}{x}\gg\log(N)\tag{2}$$ and the integral in the LHS is divergent as $N\to +\infty$.
It is also possible to prove that as soon as $\alpha\neq\beta$, $$\int_{1}^{N}\frac{\left|\sin(\alpha x+c)\right|\cdot \left|\sin(\beta x+c)\right|}{x}\,dx = \frac{4}{\pi^2}\log(N)+O(1)\tag{3}$$ details are given in the comment below: in particular, the constant $K_{\alpha,\beta}$ appearing in $(1)$ does not really depend on $\alpha$ or $\beta$, it is just $\frac{4}{\pi^2}$, and $(3)$ follows by summation by parts.
• @DenilsonOrr: start by computing the Fourier cosine series of $\left|\sin x\right|$ over the interval $[0,2\pi]$. Please notify me once you have it, I'll guide you through. – Jack D'Aurizio Oct 26 '16 at 3:45
• @DenilsonOrr: correct. Uniform convergence and the possibility to exchange $\sum$ and $\int$ are ensured by the fact that both $a_k$ and $b_k$ behave like $\frac{1}{k^2}$. – Jack D'Aurizio Oct 26 '16 at 4:47
• So we have that $\left|\sin(\alpha x+c)\right|$ is a function with average value $\frac{2}{\pi}$ and $\left|\sin(\alpha x+c)\right|-\frac{2}{\pi}$ is almost ortogonal to $\left|\sin(\beta x+c)\right|-\frac{2}{\pi}$ on the interval $[1,N]$. As soon as $N$ is large enough, the integral over $[1,N]$ of $\left|\sin(\alpha x+c)\right|\cdot\left|\sin(\beta x+c)\right|$ hence behaves like $\frac{4}{\pi^2}(N-1)$ plus a bounded error. – Jack D'Aurizio Oct 26 '16 at 4:50
• Given that, $(3)$ simply follows by integration by parts. – Jack D'Aurizio Oct 26 '16 at 4:52