I'm having troubles evaluating the following limits: $$\lim_{n\to\infty}\int_0^1 \frac{1-\sin(\frac{n}{x})}{\sqrt{x^2+1/n}}dx\;.$$ $$\lim_{n\to\infty}\int_1^{2011} \frac{1-\sin(\frac{n}{x})}{\sqrt{x^2+1/n}}dx\;.$$ I tried,for instance, to use the dominated convergence theorem, for instance noticing that $$|\frac{1-\sin(\frac{n}{x})}{\sqrt{x^2+1/n}}|\le \frac{2}{\sqrt{x^2+1/n}}$$ but then in 0 the integral diverges... Thank you in advance!
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$$ \begin{align} \lim_{n\to\infty}\int_0^1\frac{1-\sin(\frac{n}{x})}{\sqrt{x^2+1/n}}\mathrm{d}x &=\lim_{n\to\infty}\int_1^\infty\frac{1-\sin(nx)}{\sqrt{1+x^2/n}}\frac1x\,\mathrm{d}x\tag{1}\\ &=\lim_{n\to\infty}\int_{1/\sqrt{n}}^\infty\frac{1-\sin(n^{3/2}x)}{\sqrt{1+x^2}}\frac1x\,\mathrm{d}x\tag{2}\\ &=\lim_{n\to\infty}\int_{1/\sqrt{n}-\delta}^\infty\frac{1+\sin(n^{3/2}x)}{\sqrt{1+(x+\delta)^2}}\frac1{x+\delta}\,\mathrm{d}x\tag{3}\\ &=\lim_{n\to\infty}\int_{1/\sqrt{n}}^\infty\frac{1+\sin(n^{3/2}x)}{\sqrt{1+x^2}}\frac1x\,\mathrm{d}x\\ &+O\left(\frac1n\right)\tag{4}\\ &=\lim_{n\to\infty}\int_{1/\sqrt{n}}^\infty\frac1{\sqrt{1+x^2}}\frac1x\,\mathrm{d}x\\ &+O\left(\frac1n\right)\tag{5}\\ \end{align} $$ where $\delta=\pi n^{-3/2}$ and $(5)$ is the average of $(2)$ and $(4)$. The $O\left(\frac1n\right)$ in $(4)$ comes from the size of $\delta$ and the derivative of $\frac1{\sqrt{1+x^2}}\frac1x$. The integral in $(5)$ grows like $\frac12\log(n)$. Thus, $$ \lim_{n\to\infty}\int_0^1\frac{1-\sin(\frac{n}{x})}{\sqrt{x^2+1/n}}\mathrm{d}x=\infty\tag{6} $$
