# Taking a limit inside an integral

integrationlimits

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.

Define $$f,g,h \colon (2,\infty) \times \mathbb{R} \to \mathbb{R}$$ by ($$1_I$$ is the indicator function of the set $$I$$) \begin{align} f(R,x) &= \frac{1_{[1,R]}(x)}{\sqrt{1-x^2/R^2}} \left(\frac{1}{\sqrt{x^2-1}} - \frac{1}{x}\right) \, , \\ g(R,x) &= \frac{1_{[1,R/2]}(x)}{\sqrt{1-x^2/R^2}} \left(\frac{1}{\sqrt{x^2-1}} - \frac{1}{x}\right) \, , \\ h(R,x) &= \frac{1_{(R/2,R]}(x)}{\sqrt{1-x^2/R^2}} \left(\frac{1}{\sqrt{x^2-1}} - \frac{1}{x}\right) \, . \end{align} Then $$f = g + h$$. For $$R > 2$$ and $$x \in \mathbb{R}$$ we have $$g(R,x) \leq \frac{2}{\sqrt{3}} \left(\frac{1}{\sqrt{x^2-1}} - \frac{1}{x}\right) 1_{[1,\infty)} (x)$$ (the right-hand side is integrable on $$\mathbb{R}$$), $$\lim_{R \to \infty} g(R,x) = \left(\frac{1}{\sqrt{x^2-1}} - \frac{1}{x}\right) 1_{[1,\infty)} (x)$$ and $$h(R,x) = \frac{1_{(R/2,R]}(x)}{\sqrt{1-x^2/R^2}x \sqrt{x^2-1} (x + \sqrt{x^2-1})} \leq \frac{4 \cdot 1_{[0,R]}(x)}{\sqrt{1-x^2/R^2} (R^2 - 4)^{3/2}} \, .$$ Therefore, we can apply the dominated convergence theorem to the integral over $$g$$, while the integral over $$h$$ goes to zero: $$\int \limits_{\mathbb{R}} h(R,x) \, \mathrm{d} x \leq \frac{4}{(R^2-4)^{3/2}} \int \limits_0^R \frac{\mathrm{d} x}{\sqrt{1-x^2/R^2}} = \frac{2\pi R}{(R^2-4)^{3/2}} \overset{R \to \infty}{\longrightarrow} 0 \, .$$ This yields $$\lim_{R \to \infty} \int \limits_{\mathbb{R}} f(R,x) \, \mathrm{d} x = \lim_{R \to \infty} \int \limits_{\mathbb{R}} g(R,x) \, \mathrm{d} x \overset{\text{DCT}}{=} \int \limits_1^\infty \left(\frac{1}{\sqrt{x^2-1}} - \frac{1}{x}\right) \, \mathrm{d} x = \log(2)$$ as claimed.