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Prove $\displaystyle\lim_{n \rightarrow \infty} n \int_{\frac{1}{n}}^{1} \frac{\cos(x+\frac{1}{n})-\cos(x)}{x^{\frac{3}{2}}}\,dx$ exists.

I want to use Dominated convergence theorem to show the limit, but I'm stuck on finding the bound $g(x)$ here. Once I get $g(x)$, I will just exchange the limit and integral.

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Applying Cauchy's Mean Value Theorem twice says that there are $0\lt h_1,h_2\lt h$ so that $$ \begin{align} \left|\frac1h\left(\frac{\cos(x+h)-\cos(x)}h+\sin(x)\right)\right| &=\left|\frac{\cos(h)-1}{h^2}\cos(x)+\frac{h-\sin(h)}{h^2}\sin(x)\right|\\ &=\left|-\frac{\cos(h_1)}2\,\cos(x)+\frac{\sin(h_2)}2\sin(x)\right|\\[4pt] &\le1 \end{align} $$ For $x\in[h,1]$, $$ \begin{align} \left|\frac{\cos(x+h)-\cos(x)}h+\sin(x)\right|\frac1{x^{3/2}} &\le\left|\frac1h\left(\frac{\cos(x+h)-\cos(x)}h+\sin(x)\right)\right|\frac1{x^{1/2}}\\ &\le\frac1{x^{1/2}} \end{align} $$ Therefore, $$ {\large\chi}_{[h,1]}(x)\left(\frac{\cos(x+h)-\cos(x)}{hx^{3/2}}+\frac{\sin(x)}{x^{3/2}}\right) $$ is dominated by $\frac1{x^{1/2}}\in L^1[0,1]$. Thus, by Dominated Convergence, we have, with $h=\frac1n$, $$ \begin{align} \lim_{n\to\infty}n\int_{1/n}^1\frac{\cos\left(x+\frac1n\right)-\cos(x)}{x^{3/2}}\,\mathrm{d}x &=\lim_{h\to0}\int_h^1\frac{\cos(x+h)-\cos(x)}{hx^{3/2}}\,\mathrm{d}x\\ &=\lim_{h\to0}\int_0^1{\large\chi}_{[h,1]}(x)\frac{\cos(x+h)-\cos(x)}{hx^{3/2}}\,\mathrm{d}x\\ &=\int_0^1\lim_{h\to0}\,{\large\chi}_{[h,1]}(x)\frac{\cos(x+h)-\cos(x)}{hx^{3/2}}\,\mathrm{d}x\\ &=-\int_0^1\frac{\sin(x)}{x^{3/2}}\,\mathrm{d}x \end{align} $$

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  • $\begingroup$ In the first line, shouldn't it be $... - \frac{h + \sin h}{h^2}\sin x$ since $\cos(x+h) = \cos x \cos h - \sin x \sin h$. Also I don't see why $\frac{h-\sin h}{h^2} = \sin(h_2)/2$ -- I would think $ = - \cos(h_2)h/6$ from extended MVT. And if it is, in fact, $h+ \sin h$ in the numerator, then the Taylor expansion does not seem to help. $\endgroup$ – RRL Feb 4 '16 at 19:39
  • $\begingroup$ @RRL: Thanks for noticing. I have fixed a few +/- typos. $\endgroup$ – robjohn Feb 4 '16 at 21:26
  • $\begingroup$ @RRL: sorry, I fixed the typos, but forgot to address your question about CMVT. Two applications of CMVT give $$\begin{align} \frac{\cos(h)-1}{h^2} &=\frac{-\sin(h_{1a})}{2h_{1a}}\\ &=\frac{-\cos(h_1)}{2} \end{align}$$ where $0\lt h_1\lt h_{1a}\lt h$ and $$\begin{align} \frac{h-\sin(h)}{h^2} &=\frac{1-\cos(h_{2a})}{2h_{2a}}\\ &=\frac{\sin(h_2)}{2} \end{align}$$ where $0\lt h_2\lt h_{2a}\lt h$ $\endgroup$ – robjohn Feb 5 '16 at 2:59
  • $\begingroup$ I see now - thanks - very nice +1. $\endgroup$ – RRL Feb 5 '16 at 4:12
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Hint: Show that $$ \int_0^1 \frac{\sin x}{x^{3/2}}dx \quad\text{exists } $$ And $$ \lim_{n \rightarrow \infty} n \int_{\frac{1}{n}}^{1} \frac{\cos(x+\frac{1}{n})-\cos(x)}{x^{\frac{3}{2}}}dx=-\int_0^1 \frac{\sin x}{x^{3/2}}dx $$

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  • $\begingroup$ Thanks. I know how to calculate after I can switch the integral and the limit. My question is how can you switch them? I have difficulty finding the bound function required in dominated convergence theorem. $\endgroup$ – wen Feb 4 '16 at 16:36
  • $\begingroup$ @wen : $\cos(x+1/n) - \cos(x) = -2\sin(1/(2n)) \sin(x+1/(2n))$, so everything is absolutely convergent $\endgroup$ – reuns Feb 4 '16 at 21:35

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