This is a problem in an old qualifier test in analysis that I am trying to solve.
Show that
- $I_n=\int^\pi_0\frac{|\sin nt|}{t}\,dt \sim \frac{\pi}{2} \log n$ as $n\rightarrow\infty$
- $J_n=\int^\pi_0 \max_{1\leq k\leq n} |\sin(kt)|\frac{dt}{t}\sim \log(n)$ as $n\rightarrow\infty$
I worked out the first part by splitting the integral in several pieces: \begin{align} \int^\pi_0 \frac{|\sin nt|}{t}\,dt &= \int^{n\pi}_0 \frac{|\sin t|}{t}\,dt\\ &=\sum^{n-1}_{k=0}\int^{(k+1)\pi}_{k\pi}\frac{|\sin t|}{t}\,dt\\ &=\int^{\pi}_0\frac{\sin t}{t}\,dt +\sum^{n-1}_{k=1} \int^{(k+1)\pi}_{k\pi}\frac{|\sin t|}{t}\,dt \end{align} For $k\geq 2$, $$\frac{2}{\pi}\frac{1}{k+1}=\frac{1}{(k+1)\pi}\int^\pi_0|\sin t|\,dt \leq \int^{(k+1)\pi}_{k\pi}\frac{|\sin t|}{t}\,dt\leq \frac{1}{\pi k}\int^\pi_0|\sin t|\,dt=\frac{2}{\pi}\frac{1}{k}$$ Therefore $$ \int^\pi_0\frac{\sin t}{t}\,dt+\frac{\pi}{2}\sum^n_{k=2}\frac{1}{k}\leq I_n \leq \int^\pi_0\frac{\sin t}{t}\,dt+\frac{\pi}{2}\sum^{n-1}_{k=1}\frac1k $$ From this, part 1 follows.
The part I am not making much progress is the second one. Any hints/ideas will be appreciated. Thank you!
I encountered a similar problem before whose solution I am adapting for the second part of the OP's problem.
Suppose $a_k$ be a nonnegative monotone nonincreasing sequence such that $ka_k$ is nondecreasing (for example $a_k=1$ which will work for the OP's problem, $a_k=k^{-\alpha}$ with $1-\alpha\geq0$, and $a_k=\frac{1}{\log k}$ with $k\geq2$ which works for the problem I studied before).
Let $$M_n(x)=\max\limits_{1\leq k\leq n}a_k|\sin kx|$$ Since $|\sin t|\leq t$ for all $t\geq0$, $a_k\leq a_k|\sin kx|\leq ka_kx\leq na_n x$ for $1\leq k\leq n$ and so, $$0\leq M_n(x)\leq na_nx\qquad x\geq0$$ Then \begin{align} \int^\pi_0M_n(x)\,\frac{dx}{x}&=\int^{1/n}_0 M_n(x)\frac{dx}{x} +\int^1_{1/n}M_n(x)\frac{dx}{x}+\int^\pi_1 M_n(x)\frac{dx}{x}\\ &\leq a_n+\int^1_{1/n}M_n(x)\frac{dx}{x}+a_1\log\pi \end{align}
The trick now is to find a useful upper bound for $M_n(x)$. Under the assumptions on $a_k$, for each $k\in\mathbb{N}$, if $\frac1k\leq x$ \begin{align} M_n(x)\leq\max\{a_1x,2a_2x,\ldots,ka_kx,a_{k+1},\ldots,a_n\}=ka_kx \end{align} Hence \begin{align} \int^1_{1/n}M_n(x)\frac{dx}{x}&=\sum^n_{k=2}\int^{\frac{1}{k-1}}_{\frac{1}{k}}M_n(x)\frac{dx}{x}\leq \sum^n_{k=1}ka_k\big(\frac1{k-1}-\frac1k\big)\\ &\leq \sum^n_{k=2}\frac{a_k}{k-1}\leq\sum^{n-1}_{k=1}\frac{a_k}{k} \end{align} Putting things together, $$J_n\leq a_1(1+\log \pi)+\sum^{n-1}_{k=1}\frac{a_k}{k}$$
For a lower bound \begin{align} \int^\pi_0M_n(x)\,\frac{dx}{x}&\geq \int^{\pi/2}_0 M_n(x)\frac{dx}{x} \geq \sum^n_{k=1}\int^{\frac{\pi}{2k}}_{\frac{\pi}{2(k+1)}}M_n(x)\frac{dx}{x}\\ \end{align} For $\frac{\pi}{2(k+1)}\leq x\leq \frac{\pi}{2k}$, $$M_n(x)\geq a_k|\sin kx|\geq \frac{2}{\pi}ka_kx$$ Putting things together $$J_n\geq \frac{2}{\pi}\sum^n_{k=1}ka_k\big(\frac{\pi}{2k}-\frac{\pi}{2(k+1)}\Big)=\sum^n_{k=1}\frac{a_k}{k+1}\geq \sum^n_{k=1}\frac{a_k}{k}$$
In summary,
$$\sum^n_{k=1}\frac{a_k}{k}\leq \int^\pi_0M_n(x)\frac{dx}{x}\leq a_1(1+\log\pi) +\sum^{n-1}_{k=1}\frac{a_k}{k}$$
For the problem in the OP, letting $a_k\equiv1$, yields $$H_n\leq J_n\leq 1+\log\pi +H_{n-1}$$ where $H_n=\sum^n_{k=1}\frac1k$ is the harmonic series. The conclusion follows from the well known fact that $H_n=\log n+\gamma+O(1/n)$ for some constant $\gamma$.
Comment: when $a_k=\frac{1}{\log k}$ we get another interesting asymptotic sequence:
$$\int^\pi_0\max\limits_{2\leq k\leq n}\frac{|\sin kx|}{\log k}\frac{dx}{x}\sim \log(\log n)\quad\text{as}\quad n\rightarrow\infty$$