I'm trying to prove that:
$$\int_1^x \frac{| \sin t|}{t}dt \sim \frac{2}{\pi} \ln x \quad \mathrm{as} \quad x \to \infty$$
I was trying to find a function $g$ such that $\frac{|\sin t|}{t} \underset{x \to \infty}{\sim} g$, but I don't know if it is a good way to start. Also, L'Hôpital's rule does not work in this case.Do you have any hints?
Thank you!
Fourier series give a way: $$ \left|\sin x\right| = \frac{2}{\pi}-\frac{4}{\pi}\sum_{n\geq 1}\frac{\cos(2nx)}{4n^2-1}=\frac{4}{\pi}\sum_{n\geq 1}\frac{1-\cos(2nx)}{4n^2-1}\tag{1} $$ leads to $$ \int_{1}^{x}\frac{\left|\sin t\right|}{t}\,dt =\frac{2}{\pi}\,\log(x)-\frac{4}{\pi}\sum_{n\geq 1}\frac{1}{4n^2-1}\int_{1}^{x}\frac{\cos(2nt)}{t}\,dt\tag{2}$$ where the remainder term bounded by an absolute constant, due to:
$$ \int_{1}^{x}\frac{\cos(2nt)}{t}\,dt = \underbrace{\left.\frac{\sin(2nt)}{2nt}\right|_{1}^{x}}_{\text{bounded}}+\underbrace{\int_{1}^{x}\frac{\sin(2nt)}{2nt^2}\,dt}_{\text{bounded}}\tag{3}$$We are simply stating that in the original integral we are allowed to replace $\left|\sin t\right|$ with its mean value without affecting the main term of the asymptotics, by integration by parts and Dirichlet's test.
First we split integral $I=[1,x]$ into the subintervals $I_n = [(n-1)\pi, n\pi] \cap I$ for $n=1,..., n^*=\lceil x/\pi \rceil$. In each interval, we have $$ \frac{1}{n\pi} \leq \frac1 t \leq \frac{1}{(n-1)\pi} \quad \text{for } t \in I_n.$$ Using the fact that $\int_0^\pi\!dt\,\sin t =2$, we have the estimate $$\frac{2}{\pi}\log x\sim\frac{2}{\pi} \log(n^*/2)=\frac{2}{\pi} \int_2^{n^*}\,\frac{dt}{t}\leq\sum_{n=2}^{n^*-1} \frac{2}{n\pi}\leq \int_\pi^x \!dt\,\frac{|\sin t|}{t} \leq \sum_{n=2}^{n^*} \frac{2}{(n-1)\pi} \leq \frac{2}{\pi} \int_1^{n^*}\,\frac{dt}{t}=\frac{2}{\pi} \log(n^*) \sim \frac{2}{\pi}\log x.$$
Now as $$\int_1^x \!dt\,\frac{|\sin t|}{t}=\int_1^\pi \!dt\,\frac{|\sin t|}{t} + \int_\pi^x \!dt\,\frac{|\sin t|}{t} $$ with the first term a constant, we have the result.