Improper integrals: convergence tests

Section 1.25 Improper integrals: convergence tests

Theorem 1.25.1. Direct comparison test.

Let \(f\) and \(g\) be nonnegative functions on an interval \(I\text{,}\) and suppose \(f(x)\leq g(x)\) for all \(x\) in \(I\text{.}\) If the integral of \(g\) over \(I\) converges, then the integral of \(f\) over \(I\) converges. Using logical notation:

\begin{equation} \begin{array}{c} \text{ integral of \(g\) over } \\ \text{\(I\) converges} \end{array} \implies \begin{array}{c} \text{ integral of \(f\) over} \\ \text{\(I\) converges} \end{array}\text{.}\tag{1.25.1} \end{equation}

Equivalently, using the contrapositive of (1.25.1), we have

\begin{equation} \begin{array}{c} \text{ integral of \(f\) over} \\ \text{\(I\) diverges} \end{array} \implies \begin{array}{c} \text{ integral of \(g\) over} \\ \text{\(I\) diverges} \end{array}\text{.}\tag{1.25.2} \end{equation}

Example 1.25.2. Direct comparison test.

Decide whether \(\displaystyle\int_2^\infty \frac{1}{x^5+\sqrt{x+3}}\, dx\) converges.

Solution.

Observe that

\begin{equation*} f(x)=\frac{1}{x^5+\sqrt{x+3}}\leq g(x)=\frac{1}{x^5}\text{.} \end{equation*}

Since \(\int_2^\infty g(x)\, dx\) converges by the \(p\)-test, Theorem 1.25.1 implies that \(\int_2^\infty f(x)\, dx\) converges.

Example 1.25.3. Direct comparison test.

Decide whether \(\displaystyle\int_1^\infty \frac{2+\sin x}{x}\, dx\) converges.

Solution.

Observe that we have

\begin{equation*} g(x)=\frac{1}{x}\leq f(x)=\frac{2+\sin x}{x}\text{.} \end{equation*}

Since \(\int_1^\infty g(x)\, dx\) diverges by the \(p\)-test, Theorem 1.25.1 implies that \(\int_1^\infty f(x)\, dx\) diverges.

Theorem 1.25.4. Limit comparison test.

Let \(f\) and \(g\) be continuous and positive on the interval \(I\text{.}\)

If \(I=[a,\infty)\) and \(\displaystyle\lim_{x\to\infty}\frac{f(x)}{g(x)}=L\) with \(0\lt L \lt \infty\text{,}\) then

\begin{equation*} \int_a^\infty f(x)\, dx \text{ converges } \iff \int_a^\infty g(x)\, dx \text{ converges }\text{.} \end{equation*}
If \(I=(-\infty,a]\) and \(\displaystyle\lim_{x\to-\infty}\frac{f(x)}{g(x)}=L\) with \(0\lt L \lt \infty\text{,}\) then

\begin{equation*} \int_{-\infty}^a f(x)\, dx \text{ converges } \iff \int_{-\infty}^a g(x)\, dx \text{ converges }\text{.} \end{equation*}
If \(I=(a,b]\) and \(\displaystyle\lim_{x\to a^+}\frac{f(x)}{g(x)}=L\) with \(0\lt L \lt \infty\text{,}\) then

\begin{equation*} \int_{a}^b f(x)\, dx \text{ converges } \iff \int_a^b g(x)\, dx \text{ converges }\text{.} \end{equation*}
If \(I=[a,b)\) and \(\displaystyle\lim_{x\to b^-}\frac{f(x)}{g(x)}=L\) with \(0\lt L \lt \infty\text{,}\) then

\begin{equation*} \int_{a}^b f(x)\, dx \text{ converges } \iff \int_a^b g(x)\, dx \text{ converges }\text{.} \end{equation*}

Procedure 1.25.5. Limit comparison test: type I.

Assume \(f\) is positive and continuous on \([a,\infty)\text{.}\) To decide whether \(\int_a^\infty f(x)\, dx\) converges using the limit comparison test, proceed as follows.

Find a simpler continuous function \(g\) that is limit-comparable to \(f\text{:}\) i.e., a function \(g\) satisfying

\begin{equation*} \lim_{x\to \infty}\frac{f(x)}{g(x)}=c > 0\text{.} \end{equation*}
Useful intuition to help find the function \(g\text{:}\) as \(x\to \infty\text{,}\) larger positive powers of \(x\) dominate smaller positive powers of \(x\text{,}\) exponential functions dominate power functions, and positive powers of \(x\) dominate logarithmic functions.
Conclude that \(\int_a^\infty f(x)\, dx\) converges if and only if \(\int_a^\infty g(x)\, dx\) converges.

Procedure 1.25.6. Limit comparison test: type II.

Assume \(f\) is positive and continuous on the interval \((0,a]\text{.}\) To decide whether \(\int_0^a f(x)\, dx\) converges using the limit comparison test, proceed as follows.

Find a simpler continuous function \(g\) that is limit-comparable to \(f\text{:}\) i.e., a function \(g\) satisfying

\begin{equation*} \lim_{x\to 0^+}\frac{f(x)}{g(x)}=c > 0\text{.} \end{equation*}
Useful intuition to help find the function \(g\text{:}\) as \(x\to 0^+\text{,}\) smaller positive powers of \(x\) dominate larger positive powers of \(x\text{.}\)
Conclude that \(\int_a^\infty f(x)\, dx\) converges if and only if \(\int_a^\infty g(x)\, dx\) converges.

Example 1.25.7. Limit comparison test.

Decide whether \(\displaystyle\int_0^\infty\frac{1}{\sqrt{x}+3x^5}\, dx\) converges.

Solution.

Note that the integral is doubly improper: the integrand \(f(x)\) is not defined at \(x=0\text{,}\) and the interval is infinite. Accordingly we look at the improper integrals \(\int_0^1 f(x)\, dx\) and \(\int_1^\infty f(x)\, dx\) separately. As it turns out, the converge question for both improper integrals can be decided using the direct comparison test, but we will use the limit comparison test just by way of illustration.

For the improper integral \(\int_0^1 f(x)\, dx\text{,}\) our intution is that as \(x\to 0^+\) the \(\sqrt{x}\) term dominates the \(3x^5\) term. As such, we expect \(f(x)\) to be limit-comparable to \(g(x)=\frac{1}{\sqrt{x}}\text{.}\) Since \(\int_0^1 g(x)\, dx\) converges by the \(p\)-test, it will then follow that \(\int_0^1 f(x)\, dx\) converges. It remains only to confirm that \(f\) and \(g\) are limit-comparable:

\begin{align*} \lim_{x\to 0^+}\frac{f(x)}{g(x)} \amp = \lim_{x\to 0^+}\frac{\sqrt{x}}{\sqrt{x}+3x^5}\\ \amp = \lim_{x\to 0^+}\frac{\sqrt{x}}{\sqrt{x}(1+3x^{9/2})}\\ \amp = \lim_{x\to 0^+}\frac{1}{1+3x^{9/2}}\\ \amp = 1\text{.} \end{align*}

For the integral \(\int_1^\infty f(x)\, dx\text{,}\) our intuition is that as \(x\to \infty\) the \(3x^5\) term dominates the \(\sqrt{x}\) term. Thus we suspect that \(f\) should be limit-comparable to \(g(x)=\frac{1}{x^5}\) (the constant 3 here is not important). Let’s confirm:

\begin{align*} \lim_{x\to \infty}\frac{f(x)}{g(x)} \amp = \lim_{x\to \infty} \frac{x^5}{\sqrt{x}+3x^5}\\ \amp = \lim_{x\to \infty}\frac{x^5}{x^5(1/x^{9/2}+3)}\\ \amp = \lim_{x\to \infty}\frac{1}{3+1/x^{9/2}}\\ \amp = \frac{1}{3+0}\\ \amp = \frac{1}{3}\text{.} \end{align*}

This shows \(f\) and \(g\) are limit-comparable. Since \(\int_1^\infty g(x)\, dx\) converges by the \(p\)-test, the limit comparison test implies \(\int_1^\infty f(x)\, dx\) converges.

Example 1.25.8. Limit comparison test.

Decide whether \(\displaystyle \int_0^{1/2}\frac{1}{\sqrt{x}-x}\, dx\text{.}\)

Solution.

This example is designed to force us to use the limit comparison test: because of the minus sign in the denominator, we cannot make use of any useful inequality that would allow us to use the direct comparison test. Note that for \(x\in (0,1/2]\) we have \(\sqrt{x}> x\text{,}\) and thus the integrand is continuous on \((0,1/2]\text{.}\) Our intuition is that the \(\sqrt{x}\) term dominates as \(x\to 0^+\text{,}\) and thus that \(f(x)=\frac{1}{\sqrt{x}-x}\) is limit-comparable to \(g(x)=\frac{1}{\sqrt{x}}\) as \(x\to 0^+\text{.}\) Let’s confirm this:

\begin{align*} \lim_{x\to 0^+}\frac{f(x)}{g(x)} \amp = \lim_{x\to 0^+} \frac{\sqrt{x}}{\sqrt{x}-x} \\ \amp = \lim_{x\to 0^+}\frac{1}{1-\sqrt{x}}\\ \amp =\frac{1}{1-0}\\ \amp = 1\text{.} \end{align*}

This proves that \(f\) and \(g\) are limit-comparable as \(x\to 0^+\text{.}\) Since \(\int_0^{1/2} \frac{1}{\sqrt{x}}\, dx\) converges by the type-II \(p\)-test, we conclude that \(\int_0^{1/2}f(x)\, dx\) converges.

Example 1.25.9. Limit comparison test.

Let \(f(x)=ax^2+bx+c\) be any fixed irreducible quadratic polynomial with \(a>0\text{.}\) Decide whether \(\displaystyle\int_{-\infty}^\infty \frac{1}{f(x)}\, dx\) exists.

Solution.

First observe that since \(f(x)\) is irreducible, we have \(f(x)\ne 0\) for all \(x\in \R\text{.}\) Thus the given integral is improper only on account of the infinite interval. Furthermore, since \(a> 0\text{,}\) it follows that \(f(x)\) is positive eventually; since \(f(x)\) is never equal to zero, it must always be positive; this means we are in a situation where we can apply the limit comparison test. As usual, we consider the two improper integrals \(\int_{-\infty}^0 \frac{1}{f(x)}\, dx\) and \(\int_0^\infty \frac{1}{f(x)}\, dx\) separately. In both cases our intuition tells us that the \(ax^2\) term will dominate as \(x\to \pm \infty\text{,}\) and thus that the \(\frac{1}{f(x)}\) is limit-comparable to \(\frac{1}{x^2}\) as \(x\to -\infty\) and as \(x\to \infty\text{.}\) We confirm both these suspicions simultaneously below:

\begin{align*} \lim_{x\to -\infty} \frac{1/f(x)}{1/x^2} \amp = \lim_{x\to -\infty}\frac{x^2}{ax^2+bx+c} \amp \lim_{x\to \infty} \frac{1/f(x)}{1/x^2} \amp = \lim_{x\to \infty}\frac{x^2}{ax^2+bx+c} \\ \amp = \lim_{x\to -\infty}\frac{1}{a+b/x+c/x^2} \amp \amp = \lim_{x\to \infty}\frac{1}{a+b/x+c/x^2}\\ \amp =\frac{1}{a} \amp \amp =\frac{1}{a}\text{.} \end{align*}

Now consider first the improper integral

\begin{equation*} \int_0^\infty \frac{1}{f(x)}\, dx=\int_0^1\frac{1}{f(x)}\, dx +\int_1^\infty \frac{1}{f(x)}\, dx\text{.} \end{equation*}

The first integral \(\int_0^1 1/f(x)\, dx\) is not improper at all (\(1/f\) is continuous on \([0,1]\)) and so converges. Since \(\int_1^\infty 1/x^2\, dx\) converges and \(1/f\) is limit-comparable to \(1/x^2\) as \(x\to \infty\text{,}\) we see by the direct limit comparison test that the second integral \(\int_1^\infty 1/f(x)\, dx\) converges. We conclude that \(\int_0^\infty 1/f(x)\, dx\) converges.

A very similar analysis shows that \(\int_{-\infty}^{0}1/f(x)\, dx\) also converges. Note, however, that technically we cannot use the type-I \(p\)-test, since it is only stated for integrals over intervals of the form \([a,\infty)\text{.}\) Instead we must show directly that \(\int_{-\infty}^1 \frac{1}{x^2}\, dx\) converges, which is not hard to do.

In all, we see that both \(\int_{-\infty}^0 \frac{1}{f(x)}\, dx\) and \(\int_0^\infty \frac{1}{f(x)}\, dx\) converge, and hence \(\int_{-\infty}^{\infty}\frac{1}{f(x)}\, dx\) converges.

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