Define $f,g,h:(2,\mathrm{\infty})\times \mathbb{R}\to \mathbb{R}$ by (${1}_{I}$ is the indicator function of the set I)

$\begin{array}{rl}f(R,x)& =\frac{{1}_{[1,R]}(x)}{\sqrt{1-{x}^{2}/{R}^{2}}}(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x})\phantom{\rule{thinmathspace}{0ex}},\\ g(R,x)& =\frac{{1}_{[1,R/2]}(x)}{\sqrt{1-{x}^{2}/{R}^{2}}}(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x})\phantom{\rule{thinmathspace}{0ex}},\\ h(R,x)& =\frac{{1}_{(R/2,R]}(x)}{\sqrt{1-{x}^{2}/{R}^{2}}}(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x})\phantom{\rule{thinmathspace}{0ex}}.\end{array}$

Then $f=g+h$. For $R>2$ and $x\in \mathbb{R}$ we have

$g(R,x)\le \frac{2}{\sqrt{3}}(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x}){1}_{[1,\mathrm{\infty})}(x)$

(the right-hand side is integrable on $\mathbb{R}$),

$\underset{R\to \mathrm{\infty}}{lim}g(R,x)=(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x}){1}_{[1,\mathrm{\infty})}(x)$

and

$h(R,x)=\frac{{1}_{(R/2,R]}(x)}{\sqrt{1-{x}^{2}/{R}^{2}}x\sqrt{{x}^{2}-1}(x+\sqrt{{x}^{2}-1})}\le \frac{4\cdot {1}_{[0,R]}(x)}{\sqrt{1-{x}^{2}/{R}^{2}}({R}^{2}-4{)}^{3/2}}\phantom{\rule{thinmathspace}{0ex}}.$

Therefore, we can apply the dominated convergence theorem to the integral over g, while the integral over h goes to zero:

$\underset{\mathbb{R}}{\int}h(R,x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\le \frac{4}{({R}^{2}-4{)}^{3/2}}\underset{0}{\overset{R}{\int}}\frac{\mathrm{d}x}{\sqrt{1-{x}^{2}/{R}^{2}}}=\frac{2\pi R}{({R}^{2}-4{)}^{3/2}}\stackrel{R\to \mathrm{\infty}}{\u27f6}0\phantom{\rule{thinmathspace}{0ex}}.$

This yields

$\underset{R\to \mathrm{\infty}}{lim}\underset{\mathbb{R}}{\int}f(R,x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=\underset{R\to \mathrm{\infty}}{lim}\underset{\mathbb{R}}{\int}g(R,x)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\stackrel{\text{DCT}}{=}\underset{1}{\overset{\mathrm{\infty}}{\int}}(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x})\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=\mathrm{log}(2)$

as claimed.