2022-06-30

I would like to see a rigorous proof that:
$\underset{R\to \mathrm{\infty }}{lim}{\int }_{1}^{R}\left(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x}\right)\frac{dx}{\sqrt{1-{x}^{2}/{R}^{2}}}={\int }_{1}^{\mathrm{\infty }}\left(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x}\right)dx.$

Korotnokby

Define $f,g,h:\left(2,\mathrm{\infty }\right)×\mathbb{R}\to \mathbb{R}$ by (${1}_{I}$ is the indicator function of the set I)
$\begin{array}{rl}f\left(R,x\right)& =\frac{{1}_{\left[1,R\right]}\left(x\right)}{\sqrt{1-{x}^{2}/{R}^{2}}}\left(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x}\right)\phantom{\rule{thinmathspace}{0ex}},\\ g\left(R,x\right)& =\frac{{1}_{\left[1,R/2\right]}\left(x\right)}{\sqrt{1-{x}^{2}/{R}^{2}}}\left(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x}\right)\phantom{\rule{thinmathspace}{0ex}},\\ h\left(R,x\right)& =\frac{{1}_{\left(R/2,R\right]}\left(x\right)}{\sqrt{1-{x}^{2}/{R}^{2}}}\left(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x}\right)\phantom{\rule{thinmathspace}{0ex}}.\end{array}$
Then $f=g+h$. For $R>2$ and $x\in \mathbb{R}$ we have
$g\left(R,x\right)\le \frac{2}{\sqrt{3}}\left(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x}\right){1}_{\left[1,\mathrm{\infty }\right)}\left(x\right)$
(the right-hand side is integrable on $\mathbb{R}$),
$\underset{R\to \mathrm{\infty }}{lim}g\left(R,x\right)=\left(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x}\right){1}_{\left[1,\mathrm{\infty }\right)}\left(x\right)$
and
$h\left(R,x\right)=\frac{{1}_{\left(R/2,R\right]}\left(x\right)}{\sqrt{1-{x}^{2}/{R}^{2}}x\sqrt{{x}^{2}-1}\left(x+\sqrt{{x}^{2}-1}\right)}\le \frac{4\cdot {1}_{\left[0,R\right]}\left(x\right)}{\sqrt{1-{x}^{2}/{R}^{2}}\left({R}^{2}-4{\right)}^{3/2}}\phantom{\rule{thinmathspace}{0ex}}.$
Therefore, we can apply the dominated convergence theorem to the integral over g, while the integral over h goes to zero:
$\underset{\mathbb{R}}{\int }h\left(R,x\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\le \frac{4}{\left({R}^{2}-4{\right)}^{3/2}}\underset{0}{\overset{R}{\int }}\frac{\mathrm{d}x}{\sqrt{1-{x}^{2}/{R}^{2}}}=\frac{2\pi R}{\left({R}^{2}-4{\right)}^{3/2}}\stackrel{R\to \mathrm{\infty }}{⟶}0\phantom{\rule{thinmathspace}{0ex}}.$
This yields
$\underset{R\to \mathrm{\infty }}{lim}\underset{\mathbb{R}}{\int }f\left(R,x\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=\underset{R\to \mathrm{\infty }}{lim}\underset{\mathbb{R}}{\int }g\left(R,x\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x\stackrel{\text{DCT}}{=}\underset{1}{\overset{\mathrm{\infty }}{\int }}\left(\frac{1}{\sqrt{{x}^{2}-1}}-\frac{1}{x}\right)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x=\mathrm{log}\left(2\right)$
as claimed.

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