<s>[INST] Taking a limit inside an integral I would like to see a rigorous proof that:
$$\lim_{R\rightarrow \infty}\int_1^R \left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\frac{dx}{\sqrt{1-x^2/R^2}}=\int_1^\infty \left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)dx.$$
Note that the second integral does converge (to $\log 2$). This probably follows from some basic theorems in integration theory but I am rusty in that area...I cannot see how to do it with the dominated convergence or monotone convergence theorems. Some reference to the relevant theorems would be great. Thanks for any help.
 [/INST] I finally was able to prove it, so I am posting the answer to my question. It was a case when doing it from scratch (the definition of limit) was easier (for me) than trying to find the right limit theorem that would apply. It is still possible it would follow right away from some limit theorem I don't know...
In my proof I replaced $R$ with $1/k$ and did the limit as $k\rightarrow 0^+$.
Let $$I(k)=\int_1^{1/k}\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\frac{1}{\sqrt{1-k^2x^2}}dx$$
We show that
$$\lim_{k\rightarrow 0^+}I(k)=\int_1^\infty\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)dx.$$
Let $\epsilon \gt 0$ be given and find $R\gt 0$ such that
$$\frac{1}{\sqrt{1-1/R^2}}-1 \lt \epsilon .$$
Then for $k\lt 1/R$,
$$\left|I(k)-\int_1^\infty\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)dx\right|\leq \left|\int_1^{1/k}\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\left(\frac{1}{\sqrt{1-k^2x^2}}-1\right)dx\right|+\int_{1/k}^\infty \left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)dx$$
and the last integral on the right goes to zero as $k\rightarrow 0^+$. Using the identity
$$\frac{1}{\sqrt{1-k^2x^2}}-1=\frac{k^2x^2}{\sqrt{1-k^2x^2}\left(\sqrt{1-k^2x^2}+1\right)}$$
we can write the first integral on the right as $J_1(k)+J_2(k)$, where
$$J_1(k) =\int_1^R\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\frac{k^2x^2}{\sqrt{1-k^2x^2}\left(\sqrt{1-k^2x^2}+1\right)}dx\\
J_2(k)=\int_R^{1/k}\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\frac{k^2x^2}{\sqrt{1-k^2x^2}\left(\sqrt{1-k^2x^2}+1\right)}dx.$$
Note that for $1\leq x\leq R$,
$$\frac{k^2x^2}{\sqrt{1-k^2x^2}\left(\sqrt{1-k^2x^2}+1\right)}\leq \frac{k^2R^2}{\sqrt{1-k^2R^2}},$$
and for $R\leq x \leq 1/k$,
$$\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\frac{k^2x^2}{\sqrt{1-k^2x^2}\left(\sqrt{1-k^2x^2}+1\right)}\leq \left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)\frac{k^2x^2}{\sqrt{1-k^2x^2}}\\
=\left(\frac{1}{\sqrt{1-1/x^2}}-1\right)\frac{k^2x}{\sqrt{1-k^2x^2}} \leq \left(\frac{1}{\sqrt{1-1/R^2}}-1\right)\frac{k^2x}{\sqrt{1-k^2x^2}}. $$
Then
$$J_1(k)\leq \frac{k^2R^2}{\sqrt{1-k^2R^2}}\int_1^R\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)dx \rightarrow 0 \mbox{ as } k\rightarrow 0^+$$
and
$$J_2(k) \leq \left(\frac{1}{\sqrt{1-1/R^2}}-1\right)\int_R^{1/k}\frac{k^2x dx}{\sqrt{1-k^2x^2}}\leq \epsilon \int_0^{1-k^2R^2}\frac{du}{2\sqrt{u}}\leq \epsilon \int_0^1\frac{du}{2\sqrt{u}}=\epsilon$$
Since $\epsilon$ is arbitrary, we conclude that
$$I(k)\rightarrow \int_1^\infty\left(\frac{1}{\sqrt{x^2-1}}-\frac{1}{x}\right)dx.$$ as $k\rightarrow 0^+$.
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