Existence of $\lim_{x\to0} \int_{x}^{1 \over x} {(\cos 2r-\cos r) \over r} dr$ I am trying to prove that $\lim_{x\to0} \int_{x}^{1 \over x} {(\cos 2r-\cos r) \over r} dr$ exists. Wolframalpha gives me the value ${\int_{0}^{\infty} {(\cos 2x-\cos x)\over x} dx} = -\ln2$, but I don't know where to start, based only on this information. The question is in the book "Real Analysis and Foundations" by S. Krantz.
 A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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$\ds{\lim_{x \to 0^{+}}\ \int_{x}^{1/x}{\cos\pars{2r} - \cos\pars{r} \over r}
\,\dd r:\ {\large ?}}$.

\begin{align}
&\lim_{x \to 0^{+}}\ \int_{x}^{1/x}{\cos\pars{2r} - \cos\pars{r} \over r}\,\dd r =
\int_{0}^{\infty}\bracks{\cos\pars{2r} - \cos\pars{r}}\ \overbrace{%
\int_{0}^{\infty}\expo{-rt}\,\dd t}^{\ds{1 \over r}}\,\dd r
\\[5mm] = &\
\Re\int_{0}^{\infty}\int_{0}^{\infty}
\bracks{\expo{\pars{-t + 2\ic}r} - \expo{\pars{-t + \ic}r}}\dd r\,\dd t =
\Re\int_{0}^{\infty}\pars{-\,{1 \over -t + 2\ic} + {1 \over -t + \ic}}\,\dd t
\\[5mm] = &\
\left. \Re\ln\pars{t - 2\ic \over t - \ic}
\right\vert_{\ t\ =\ 0}^{\ t\ \to\ \infty} =\
\bbox[10px,#ffe,border:1px dotted navy]{\ds{-\ln\pars{2}}}
\end{align}
A: Hint:
$\lim_{x\to0} \int_{x}^{1 \over x} {(\cos 2r-\cos r) \over r} dr$
$=\lim_{x\to0} \int_{x}^{1 \over x} (\cos 2r-\cos r) d(\ln r)$
$=\lim_{x\to0} \int_{x}^{1 \over x}  \cos 2r\, d(\ln r) - \int_{x}^{1 \over x} \cos r\,d(\ln r)$
Using integration by parts should help
