Convergence of a difficult integral using the limit comparison test

Here’s a great problem from an exam in my second-semester Advanced Calculus (introductory real analysis) course taught by Yuri Ledyaev at Western Michigan University:

Find the values of \(p\) for which the integral converges:
$$
\int_{1}^{\infty} \frac{\left(\tan\frac{1}{x}\right)^p}{x+x^2}
$$

To determine what test to use, it is best to recall that
$$
\lim_{\theta \to 0} \tan\theta \sim \theta
$$

or
$$
\lim_{\theta \to 0} \frac{\tan\theta}{\theta} = 1
$$

which is a fancy way of saying that at very small values of \(\theta\), \(\tan\theta\) behaves like \(\theta\). (In case you forget this, it is easy to recall by remembering that \(\tan\theta = \frac{\sin\theta}{\cos\theta}\), whose denominator approaches \(1\) as \(\theta\) approaches \(0\), so \(\tan\theta\) behaves like \(\sin\theta\) for very small values of \(\theta\). All semi-advanced math students should remember the small-angle approximation rule, i.e., for small values of \(\theta\), \(\sin\theta \approx \theta\).)

We can substitute \(1/x\) for \(\theta\) to see that
$$
\lim_{x \to \infty} \tan\frac{1}{x} \sim \lim_{x \to \infty} \frac{1}{x}
$$

The reason we’re interested in this is that we need to know what \(\frac{\left(\tan\frac{1}{x}\right)^p}{x+x^2}\) behaves like or looks like as \(x \rightarrow \infty\). Whatever our integrand looks like is what we’ll compare it to in the limit comparison test. So:

$$
\lim_{x \to \infty} \frac{\left(\tan\frac{1}{x}\right)^p}{x+x^2} \sim \lim_{x \to \infty} \frac{\left(\frac{1}{x}\right)^p}{x+x^2} \nonumber = \lim_{x \to \infty} \frac{1}{x^{p+1} + x^{p+2}} \nonumber \\
$$

In this case, it is a good bet to choose the term in the denominator with the greater exponent rather than the term with the lesser exponent for use in the limit comparison test, so we’ll choose \(x^{p+2}\). That is, for the limit comparison test, let \(f(x) = \frac{\left(tan\frac{1}{x}\right)^p}{x+x^2}\) and \(g(x) = \frac{1}{x^{p+2}}\).

The limit comparison test for integrals says that if \(f\) and \(g\) are both defined and positive on \([a, \infty)\) and integrable on \([a, b]\) for all \(b \geq a\), and if \(\lim_{x \to \infty} \frac{f(x)}{g(x)}\) exists and is not equal to \(0\), then the integrals \(\int_{a}^{\infty} f(x) dx\) and \(\int_{a}^{\infty} g(x) dx\) are equiconvergent.

Observe,

$$
\begin{eqnarray}
\lim_{x \to \infty} \frac{f(x)}{g(x)} &=& \lim_{x \to \infty} \frac{\frac{\left(\tan\frac{1}{x}\right)^p}{x+x^2}}{\frac{1}{x^{p+2}}} \\[3pt] \nonumber \\
&=& \lim_{x \to \infty} \frac{\left(\tan\frac{1}{x}\right)^p}{x+x^2} \cdot x^{p+2} \\[3pt] \nonumber \\
&\sim& \lim_{x \to \infty} \frac{\frac{1}{x^p}}{x+x^2} \cdot x^{p+2} \\[3pt] \nonumber \\
&=& \lim_{x \to \infty} \frac{x^2}{x+x^2} \\[3pt] \nonumber \\
&=& 1
\end{eqnarray}
$$

Thus, our choices of \(f(x)\) and \(g(x)\) satisfy the limit comparison test, meaning \(\int_{a}^{\infty} f(x) dx\) converges when \(\int_{a}^{\infty} g(x) dx\) converges.

When does \(\int_{a}^{\infty} g(x) dx\) converge? When \(p+2 > 1 \) (by the p-series rule). Thus, both \(\int_{a}^{\infty} f(x) dx\) and \(\int_{a}^{\infty} g(x) dx\) converge when \(p>-1\) and diverge when \(p \leq -1\).

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