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Prove that $\ln(x^2+1) \arctan x$ is uniformly continuous in $\mathbb R$.

My attempts at proving this have not yielded anything worth sharing. I'd appreciate any help or a push in the right direction.

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Hint: Your function has a bounded derivative. – David Mitra Feb 2 '13 at 21:00
@DavidMitra Thank you very much!! You should convert your comment to an answer. Also, when proving that the derivative is bounded, is it enough to say that the function is continuous in $\mathbb R$ (meaning that there can't be a vertical asymptote) and that $\lim_{x\to\infty}f(x)=\lim_{x\to-\infty}f(x)=0$? – Richard Feb 2 '13 at 21:17
Yes, that would work. – David Mitra Feb 2 '13 at 21:34
up vote 1 down vote accepted

Hint: Since $f(x)=\ln(x^2+1)\arctan x$ is an odd function, it is enough to show that it is uniformly continuous on $[0,\infty)$.

Towards this end, note that, for $x\ge 0$ $$ f'(x) ={2x\arctan x\over 1+x^2}+{\ln(1+x^2)\over 1+x^2} \le {\pi x\over 1+x^2}+{\ln(1+x^2)\over 1+x^2}\ \ \buildrel{x\rightarrow\infty}\over\longrightarrow\ \ 0. $$ Use this and the fact that $f' $ is continuous to deduce that $f'(x)$ is bounded on $[0,\infty)$. The uniform continuity of $f$ over $[0,\infty)$ follows from this.

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