Comparison Test for Improper Integrals Examples 1

# Comparison Test for Improper Integrals Examples 1

Recall from The Comparison Test for Improper Integral Convergence/Divergence page that if $0 ≤ g(x) ≤ f(x)$ and if:

• If $\int_a^{\infty} f(x) \: dx$ converges, then $\int_a^{\infty} g(x) \: dx$ converges.
• If $\int_a^{\infty} g(x) \: dx$ diverges, then $\int_a^{\infty} f(x) \: dx$ diverges.

We will now look at some examples of applying the comparison test for improper integrals.

## Example 1

Determine whether $\int_2^{\infty} \frac{\sin ^2 x}{x^2} \: dx$ converges or diverges.

We first note that for all $x \in \mathbb{R}$ and $x ≥ 1$ that $-1 ≤ \sin x ≤ 1$, and so for all $x \in \mathbb{R}$ and $x ≥ 1$, $0 ≤ \sin ^2 x ≤ 1$. Thus we have that:

(1)
\begin{align} 0 ≤ \frac{\sin ^2 x}{x^2} ≤ \frac{1}{x^2} \end{align}

If we can show that $\int_2^{\infty} \frac{1}{x^2} \: dx$ converges, then we can invoke the comparison test. Notice:

(2)
\begin{align} \quad \quad \int_2^{\infty} \frac{1}{x^2} \: dx = \lim_{b \to \infty} \int_2^b x^{-2} \: dx = \lim_{b \to \infty} -\frac{1}{x} \biggr \rvert_2^b = \lim_{b \to \infty} - \frac{1}{b} + \frac{1}{2} = \frac{1}{2} \end{align}

Therefore $\int_2^{\infty} \frac{1}{x^2} \: dx$ converges, so by the comparison test, $\int_2^{\infty} \frac{\sin ^2 x}{x^2} \: dx$ converges.

## Example 2

Determine whether $\int_1^{\infty} \frac{x^3 +2x + 1}{\ln x} \: dx$ converges or diverges.

Determining whether this integral converges or diverges can be simplified by getting rid of "$\ln x$" in the denominator. Notice that $\ln x < x$ for all $x > 0$, in particular, $x ≥ 1$. Therefore $\frac{1}{\ln x} > \frac{1}{x}$ which implies that:

(3)
\begin{align} \frac{x^3 +2x + 1}{x} < \frac{x^3 + 2x + 1}{\ln x} \end{align}

Now let's evaluate $\int_1^{\infty} \frac{x^3 + 2x + 1}{x} \: dx = \int_1^{\infty} x^2 + 2 + \frac{1}{x} \: dx$.

(4)
\begin{align} \quad \quad \int_1^{\infty} x^2 + 2 + \frac{1}{x} \: dx = \lim_{b \to \infty} \int_1^b x^2 + 2 + \frac{1}{x} \: dx = \lim_{n \to \infty} \left ( \frac{x^3}{3} + 2x + \ln x \right ) \biggr \rvert_1^b = \lim_{b \to \infty} \left ( \frac{b^3}{3} + 2b + \ln b - \frac{7}{3} \right ) = \infty \end{align}

Therefore $\int_1^{\infty} \frac{x^3 + 2x + 1}{x} \: dx$ diverges, and so $\int_1^{\infty} \frac{x^3 + 2x + 1}{\ln x} \: dx$ also diverges.