calculus - Compute $limlimits_{x to +infty}dfrac{ln x}{ int_0^x frac{|sin t|}{t}{rm d}t}$.

Compute

$$\lim\limits_{x \to +\infty}\dfrac{\ln x}{\displaystyle \int_0^x \dfrac{|\sin t|}{t}{\rm d}t}.$$

Maybe, we can solve it by L'Hospital's rule, but there still exists a difficulty here. Though $x \to +\infty$ implies $\ln x \to +\infty$, we do not know the limit of the denominator. How to solve it?

Answer

Thanks to @Alex B.'s hint , I complete the solution. Please correct me if I'm wrong.

For any $x>0$, we can choose some $n \in \mathbb{N}$ such that $n \pi\leq x<(n+1)\pi$. Thus, we obtain
$$\int_0^{n\pi}\frac{|\sin t|}{t}{\rm d}t \leq \int_0^x \frac{|\sin t|}{t}{\rm d}t<\int_0^{(n+1)\pi}\frac{|\sin t|}{t}{\rm d}t.$$

On one hand, notice that
\begin{align*}
\int_0^{n \pi} \frac{|\sin t|}{t}{\rm d}t&=\int_0^\pi \frac{|\sin t|}{t}{\rm d}t+\sum_{k=1}^{n-1}\int_{k\pi}^{(k+1) \pi} \frac{|\sin t|}{t}{\rm d}t\\
&> \sum_{k=1}^{n-1}\int_{k\pi}^{(k+1) \pi} \frac{|\sin t|}{t}{\rm d}t\\
& > \sum_{k=1}^{n-1}\int_{k\pi}^{(k+1) \pi} \frac{|\sin t|}{(k+1)\pi}{\rm d}t\\
&=\sum_{k=1}^{n-1}\frac{\int_{k\pi}^{(k+1)\pi}|\sin t|{\rm d}t}{(k+1)\pi}\\
&=\frac{2}{\pi}\sum_{k=2}^{n}\frac{1}{k}.
\end{align*}

On the other hand, likewise,
\begin{align*}
\int_0^{(n+1) \pi} \frac{|\sin t|}{t}{\rm d}t&=\int_0^\pi \frac{|\sin t|}{t}{\rm d}t+\sum_{k=1}^{n}\int_{k\pi}^{(k+1) \pi} \frac{|\sin t|}{t}{\rm d}t\\
&< \int_0^\pi {\rm d}t+\sum_{k=1}^{n}\int_{k\pi}^{(k+1) \pi} \frac{|\sin t|}{k\pi}{\rm d}t\\
&=\pi+\sum_{k=1}^{n}\frac{\int_{k\pi}^{(k+1)\pi}|\sin t|{\rm d}t}{k\pi}\\
&=\pi+\frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}.
\end{align*}

Therefore
$$\frac{2}{\pi}\sum_{k=2}^{n}\frac{1}{k} <\int_0^x \frac{|\sin t|}{t}{\rm d}t<2+\frac{2}{\pi}\sum_{k=1}^{n}\frac{1}{k}.$$

Since
$$\ln n\pi\leq \ln x<\ln(n+1)\pi,$$
we have
$$\dfrac{\ln n\pi}{\pi+\dfrac{2}{\pi}\sum\limits_{k=1}^{n}\dfrac{1}{k}}<\dfrac{\ln x}{\int_0^x \dfrac{|\sin t|}{t}{\rm d}t}<\dfrac{\ln(n+1)\pi}{\dfrac{2}{\pi}\sum\limits_{k=2}^{n}\dfrac{1}{k}}.$$

Applying the subsitution as follows
$$\sum_{k=1}^n \frac{1}{k}=\ln n+\gamma+\varepsilon_n,$$
(in fact, we only need to recall that $\sum\limits_{k=1}^n \dfrac{1}{k}$ and $\ln n$ are equivalent infinities), we can readily infer that the limits of the both sides in the last expression are both equal to $\dfrac{\pi}{2}$ under the process $n \to \infty$（i.e. $x \to +\infty$）. Hence, according to the squeeze theorem, we can conclude that
$$\frac{\ln x}{\int_0^x \frac{|\sin t|}{t}{\rm d}t} \to \frac{\pi}{2}(x \to +\infty),$$which is what we want to evaluate.