Wednesday, 25 December 2013

calculus - Compute limlimitsxto+inftydfraclnxintx0frac|sint|trmdt.





Compute

lim





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.



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