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.

Solution 1:

I don't think any of the standard theorems directly apply here. We can, however, still make use of the dominated convergence theorem after splitting the integral at an appropriate point (similar to your own answer).

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.

Solution 2:

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 $k\rightarrow 0^+$.

Solution 3:

It is not hard to establish the following generalization:

Proposition. Let $f$ be a locally integrable function on $[0, \infty)$ such that

1. $\limsup_{x \to \infty} |xf(x)| < \infty$, and
2. $\lim_{R\to\infty} \int_{0}^{R} f(x) \, \mathrm{d}x$ exists and is finite.

Then we have

$$ \bbox[border:1px blue solid;padding:5pt;background-color:azure]{\lim_{R \to \infty} \int_{0}^{R} \frac{f(x)}{\sqrt{1-x^2/R^2}} \, \mathrm{d}x = \lim_{R\to\infty} \int_{0}^{R} f(x) \, \mathrm{d}x.} $$

Remark 1. This result resolves OP's question, since $f(x) = \Bigl( \frac{1}{\sqrt{x^2-1}} - \frac{1}{x} \Bigr) \mathbf{1}_{[1, \infty)}(x)$ satisfies the assumptions.

Remark 2. Note that $f$ need not be integrable on $[0, \infty)$. For example, this result is applicable to $f(x) = \frac{\sin x}{x}$, which is only improperly integrable on $[0, \infty)$.

Proof. By the assumption, there exist $x_0, C \in (0, \infty)$ such that $|f(x)| \leq \frac{C}{x}$ for all $x > x_0$. In light of this, we decompose $f = f_1 + f_2$, where

$$ f_1(x) = f(x)\mathbf{1}_{[0,x_0]}(x) \qquad\text{and}\qquad f_2(x) = f(x)\mathbf{1}_{(x_0,\infty)}(x). $$

Then it suffices to show that the claim holds true with $f_1$ and $f_2$, respectively. However, since $f_1$ is integrable and compactly supported on $[0, \infty)$, the first part of the claim is almost obvious by DCT. So, it remains to show that the claim is true for $f_2$.

To this end, write $F_2(x) = \int_{0}^{x} f_2(t) \, \mathrm{d}t$. Then by the assumption, $F_2(R)$ converges as $R\to\infty$. Moreover, by the Fundamental Theorem of Calculus (FToC) and the Fubini's Theorem,

\begin{align*} \epsilon_2(R) &:= \int_{0}^{R} \frac{f_2(x)}{\sqrt{1-x^2/R^2}} \, \mathrm{d}x - F_2(R) \\ &= \int_{0}^{R} f_2(x) \biggl( \frac{1}{\sqrt{1-x^2/R^2}} - 1 \biggr) \, \mathrm{d}x \\ &= \int_{0}^{R} f_2(x) \biggl( \int_{0}^{x} \frac{u}{R^2(1-u^2/R^2)^{3/2}} \, \mathrm{d}u \biggr) \, \mathrm{d}x \tag{$\because$ FToC} \\ &= \int_{0}^{R} \frac{u}{R^2(1-u^2/R^2)^{3/2}}(F_2(R) - F_2(u)) \, \mathrm{d}u \tag{$\because$ Fubini} \\ &= \int_{0}^{1} \frac{s}{(1-s^2)^{3/2}}(F_2(R) - F_2(Rs)) \, \mathrm{d}s. \tag{$u = Rs$} \end{align*}

However, since $|f_2(x)| \leq \frac{C}{x}$ for all $x > 0$, we can bound the integrand of the above integral by

$$ \left| \frac{s}{(1-s^2)^{3/2}}(F_2(R) - F_2(Rs)) \right| \leq \frac{s}{(1-s^2)^{3/2}} \int_{Rs}^{R} \frac{C}{x} \, \mathrm{d}x = \frac{Cs\log(1/s)}{(1-s^2)^{3/2}}. $$

Then by noting that this dominating function is integrable on $[0, 1]$, we may apply DCT to get

$$ \lim_{R\to\infty} \epsilon_2(R) \stackrel{\text{(DCT)}}= \int_{0}^{1} \lim_{R\to\infty} \frac{s(F_2(R) - F_2(Rs))}{(1-s^2)^{3/2}} \, \mathrm{d}s = 0. $$

Therefore the desired conclusion follows. $\square$