1.12 Improper Integrals

There are two kinds of impropreity in an integral: an infinite discontinuity in the integrand, and an infinite limit of integration.

Any real number in $[1,\infty)$ or $(-\infty,-1]$, and $b = \pm \infty$.

If $b= \pm \infty$, then our integral is improper because one limit is not a real number.

Furthermore, our integral will be improper if its domain of integration contains either of its infinite discontinuities, $x=1$ and $x=-1$. Since one limit of integration is 0, the integral is improper if $b \geq 1$ or if $b \leq -1$.

Below, we've graphed $\frac{1}{x^2-1}$ to make it clearer why values of $b$ in $(-1,1)$ are the only values that don't result in an improper integral when the other limit of integration is $a=0$.

The integrand is continuous for all $x$.

$b = \pm\infty$

Since the integrand is continuous for all real $x$, the only kind of impropriety available to us is to set $b = \pm\infty$.

What matters is which function is bigger for large values of $x$, not near the origin.

The red function is $f(x)$, and the blue function is $g(x)$.

For large values of $x$, $|\text{red function}|\leq \text{(blue function)}$ and $0 \leq \text{(blue function)}$. If the blue function's integral converged, then the red function's integral would as well (by the comparison test, Theorem 1.12.17 in the CLP-2 text). Since one integral converges and the other diverges, the blue function is $g(x)$ and the red function is $f(x)$.

Read both the question and Theorem 1.12.17 in the

CLP-2 text very carefully.

False. For example, the functions $f(x)=e^{-x}$ and $g(x)=1$ provide a counterexample.

False. The inequality goes the “wrong way" for Theorem 1.12.17 in the CLP-2 text: the area under the curve $f(x)$ is finite, but the area under $g(x)$ could be much larger, even infinitely larger.

For example, if $f(x)=e^{-x}$ and $g(x)=1$, then $0 \leq f(x) \leq g(x)$ and $\displaystyle\int_{1}^{\infty} f(x) \,\dee{x}$ converges, but $\displaystyle\int_{1}^{\infty} g(x) \,\dee{x}$ diverges.

(a) What if $h(x)$ is negative? What if it's not?
(b) What if $h(x)$ is very close to $f(x)$ or $g(x)$, rather than right in the middle?
(c) Note $|h(x)| \leq 2f(x)$.

First: is the integrand unbounded, and if so, where?

Second: when evaluating integrals, always check to see if you can use a simple substitution before trying a complicated procedure like partial fractions.

The integral diverges.

The denominator is zero when $x=1$, but the numerator is not, so the integrand has a singularity (infinite discontinuity) at $x=1$. Let's replace the limit $x=1$ with a variable that creeps toward 1.

To evaluate this integral we use the substitution $u=x^5$, $\dee{u}=5x^4\dee{x}$. When $x=0$ we have $u=0$, and when $x=t$ we have $u=t^5$, so

The limit does not exist, so the integral diverges.

Is the integrand bounded?

The integral diverges.

The denominator of the integrand is zero when $x=-1$, but the numerator is not. So, the integrand has a singularity (infinite discontinuity) at $x=-1$. This is the only “source of impropriety" in this integral, so we only need to make one break in the domain of integration.

Let's start by considering the left limit.

Since this limit diverges, the integral diverges. (A similar argument shows that the second integral diverges. Either one of them diverging is enough to conclude that the original integral diverges.)

See Example 1.12.21 in the

CLP-2 text. Rather than antidifferentiating, you can find a nice comparison.

The integral does not converge.

First, let's identify all “sources of impropriety." The integrand has a singularity when $4x^2-x=0$, that is, when $x(4x-1)=0$, so at $x=0$ and $x=\frac{1}{4}$. Neither of these are in our domain of integration, so the only “source of impropriety" is the unbounded domain of integration.

We could antidifferentiate this function (it looks like a nice candidate for a trig substitution), but is seems easier to use a comparison. For large values of $x$, the term $x^2$ will be much larger than $x$, so we might guess that our integral behaves similarly to $\int_1^\infty \frac{1}{\sqrt{4x^2}}~\dee{x}=\int_1^\infty \frac{1}{2x}~\dee{x}$.

For all $x\ge 1$, $\sqrt{4x^2-x}\le\sqrt{4x^2}=2x$. So, $\frac{1}{\sqrt{4x^2-x}} \geq \frac{1}{2x}$. Note $\int_1^\infty \frac{1}{2x}~\dee{x}$ diverges:

So:

By the comparison test, Theorem 1.12.17 in the CLP-2 text, the integral does not converge.

Which of the two terms in the denominator is more important when $x\approx 0$? Which one is more important when $x$ is very large?

The integral converges.

The integrand is positive everywhere. So, either the integral converges to some finite number, or it is infinite. We want to generate a guess as to which it is.

When $x$ is small, $\sqrt{x}>x^2$, so we might guess that our integral behaves like the integral of $\frac{1}{\sqrt{x}}$ when $x$ is near to 0. On the other hand, when $x$ is large, $\sqrt{x}<x^2$, so we might guess that our integral behaves like the integral of $\frac{1}{x^2}$ as $x$ goes to infinity. This is the hunch that drives the following work:

Note that $\frac{1}{x^2+\sqrt{x}}$ is defined and continuous for all $x>0$, $\frac{1}{\sqrt{x}}$ is defined and continuous for $x>0$, and $\frac{1}{x^2}$ is defined and continuous for $x \ge 1$. So, the integral converges by the comparison test, Theorems 1.12.17 and ? in the CLP-2 text.

Remember to break the integral into two pieces.

The integral diverges.

There are two “sources of impropriety": the two (infinite) limits of integration. So, we break our integral into two pieces.

These are easy enough to antdifferentiate.

Since the limits don't exist, the integral diverges. (It happens that both limits don't exist; even if only one failed to exist, the integral would still diverge.)

Remember to break the integral into two pieces.

The integral diverges.

There are two “sources of impropriety": the $-\infty$ and the $+\infty$. So, we break our integral into two pieces.

Since the limits don't exist, the integral diverges. (It happens that both limits don't exist; even if only one failed to exist, the integral would diverge.)

Remark: it's very tempting to think that this integral should converge, because as an odd function the area to the right of the $x$-axis “cancels out" the area to the left when the limits of integration are symmetric. One justification for not using this intuition is given in Example 1.12.11 in the CLP-2 text. Here's another: In Question 10 we saw that $\int_{-\infty}^\infty \cos x~\dee{x}$ diverges. Since $\sin x = \cos (x-\pi/2)$, the area bounded by sine and the area bounded by cosine over an infinite region seem to be the same–only shifted by $\pi/2$. So if $\int_{-\infty}^\infty \sin x~\dee{x}=0$, then we ought to also have $\int_{-\infty}^\infty \cos x~\dee{x}=0$, but we saw in Question 10 this is not the case.

The easiest test in this case is limiting comparison, Theorem 1.12.22 in the CLP-2 text.

The integral diverges.

First, we check that the integrand has no singularities. The denominator is always positive when $x \ge 10$, so our only “source of impropriety" is the infinite limit of integration.

We further note that, for large values of $x$, the integrand resembles $\dfrac{x^4}{x^5} = \dfrac{1}{x\vphantom{\frac12}}$. So, we have a two-part hunch: that the integral diverges, and that we can show it diverges by comparing it to $\displaystyle\int_{10}^\infty \frac{1}{x}~\dee{x}$.

In order to use the comparison test, we'd need to show that $\displaystyle\frac{x^4-5x^3+2x-7}{x^5+3x+8} \geq \frac{1}{\vphantom{\frac12}x}$. If this is true, it will be difficult to prove–and it's not at all clear that it's true. So, we will use the limiting comparison test instead, Theorem 1.12.22 in the CLP-2 text, with $g(x)= \dfrac{1}{x}$, $f(x)=\dfrac{x^4-5x^3+2x-7}{x^5+3x+8}$, and $a=10$.

By the limiting comparison test, we conclude $\displaystyle\int_{10}^\infty f(x)~\dee{x}$ diverges.

Not all discontinuities cause an integral to be improper–only infinite discontinuities.

The integral diverges.

Our domain of integration is finite, so the only potential “sources of impropriety" are infinite discontinuities in the integrand. To find these, we factor.

A removable discontinuity doesn't affect the integral.

Use the substitution $u=x-10$, $\dee{u}=\dee{x}$. When $x=0$, $u=-10$, and when $x=10$, $u=0$.

This is a $p$-integral with $p=1$. From Example 1.12.9 and Theorem 1.12.20 in the CLP-2 text, we know it diverges.

Which of the two terms in the denominator is more important when $x$ is very large?

The integral diverges.

You might think that, because the integrand is odd, the integral converges to $0$. This is a common mistake– see Example 1.12.11 in the CLP-2 text, or Question 11 in this section. In the absence of such a shortcut, we use our standard procedure: identifying problem spots over the domain of integration, and replacing them with limits.

There are two “sources of impropriety,” namely $x\to +\infty$ and $x\to -\infty$. So, we split the integral in two, and treat the two halves separately. The integrals below can be evaluated with the substitution $u=x^2+1$, $\frac{1}{2}\dee{u} = x\dee{x}$.

Both halves diverge, so the whole integral diverges.

Once again: after we found that one of the limits diverged, we could have stopped and concluded that the original integrand diverges. Don't make the mistake of thinking that $\infty-\infty=0$. That can get you into big trouble. $\infty$ is not a normal number. For example $2\infty=\infty$. So if $\infty$ were a normal number we would have both $\infty-\infty=0$ and $\infty-\infty=2\infty-\infty =\infty$.

Which of the two terms in the denominator is more important when $x\approx 0$? Which one is more important when $x$ is very large?

The integral converges.

We don't want to antidifferentiate this integrand, so let's use a comparison. Note the integrand is positive when $x>0$.

For any $x$, $|\sin x| \leq 1$, so $\dfrac{|\sin x|}{x^{3/2}+x^{1/2}} \leq \dfrac{1}{x^{3/2}+x^{1/2}}$.

Since $x=0$ and $x \to \infty$ both cause the integral to be improper, we need to break it into two pieces. Since both terms in the denominator give positive numbers when $x$ is positive, $\dfrac{1}{x^{3/2}+x^{1/2}} \leq \dfrac{1}{x^{3/2}}$ and $\dfrac{1}{x^{3/2}+x^{1/2}} \leq \dfrac{1}{x^{1/2}}$. That gives us two options for comparison.

When $x$ is positive and close to zero, $x^{1/2} \ge x^{3/2}$, so we guess that we should compare our integrand to $\frac{1}{x^{1/2}}$ near the limit $x=0$. In contrast, when $x$ is very large, $x^{1/2} \le x^{3/2}$, so we guess that we should compare our integrand to $\frac{1}{x^{3/2}}$ as $x$ goes to infinity.

Now we have all the data we need to apply the comparison test, Theorems 1.12.17 and ? in the CLP-2 text.

Therefore, our integral $\displaystyle\int_0^\infty \frac{|\sin x|}{x^{3/2}+x^{1/2}}~\dee{x}$ converges.

What are the “problem $x$'s” for this integral? Get a simple approximation to the integrand near each.

The integral converges.

The integrand is positive everywhere, so either the integral converges to some finite number or it is infinite. There are two potential “sources of impropriety” — a possible singularity at $x=0$ and the fact that the domain of integration extends to $\infty$. So we split up the integral.

Let's develop a hunch about whether the integral converges or diverges. When $x\approx 0$, $x^2$ and $x$ are both a lot smaller than 1, so we guess we should compare the integrand to $\frac{1}{x^{1/3}}$.

Note $\int_0^1 \frac{1}{x^{1/3}}~\dee{x}$ converges by Example 1.12.9 in the CLP-2 text (it's a $p$-type integral), so we guess $\int_0^1 \frac{x+1}{x^{1/3}(x^2+x+1)}~\dee{x}$ converges as well.

When $x$ is very large, $x^2$ is much bigger than $x$, which is much bigger than 1, so we guess we should compare the integrand to $\frac{1}{x^{4/3}}$.

Note $\int_1^\infty \frac{1}{x^{4/3}}~\dee{x}$ converges by Example 1.12.8 in the CLP-2 text (it's a $p$-type integral), so we guess $\int_1^\infty \frac{x+1}{x^{1/3}(x^2+x+1)}~\dee{x}$ converges as well.

Now it's time to verify our guesses with the limiting comparison test, Theorems 1.12.22 and ? in the CLP-2 text. Be careful: our “$\approx$" signs are not strong enough to use either the limiting comparison test or the comparison test, they are only enough to suggest a reasonable function to compare to.

We conclude $\displaystyle\int_0^\infty \frac{x+1}{x^{1/3}(x^2+x+1)}~\dee{x}$ converges.

To find the volume of the solid, cut it into horizontal slices, which are thin circular disks.

The true/false statement is equivalent to saying that the improper integral giving the volume of the solid when $a=0$ diverges to infinity.

false

To find the volume of the solid, we cut it into horizontal slices, which are thin circular disks. At height $y$, the disk has radius $x=\frac{1}{y}$ and thickness $\dee{y}$, so its volume is $\frac{\pi}{y^2}\dee{y}$. The base of the solid is at height $y=1$, and its top is at height $y=\frac{1}{a}$. So, the volume of the entire solid is:

If we imagine sliding $a$ closer and closer to 0, the volume increases, getting closer and closer to $\pi$ units, but never quite reaching it.

So, the statement is false. For example, if we set $M=4$, no matter which $a$ we choose our solid has volume strictly less than $M$.

Remark: we've seen before that $\int_0^1 \frac{1}{x}\,\dee{x}$ diverges. If we imagine the solid that would result from choosing $a=0$, it would have a scant volume of $\pi$ cubic units, but a silhouette (side view) of infinite area.

Review Example 1.12.8 in the

CLP-2 text. Remember the antiderivative of $\frac{1}{x}$ looks very different from the antiderivative of other powers of $x$.

$q=\frac{1}{5}$

Our goal is to decide when this integral diverges, and where it converges. We will leave $q$ as a variable, and antidifferentiate. In order to antidifferentiate without knowing $q$, we'll need different cases. The integrand is $x^{-5q}$, so when $-5q \neq -1$, we use the power rule (that is, $\int x^n~\dee{x} = \frac{x^{n+1}}{n+1}$) to antidifferentiate. Note $x^{(-5q)+1} = x^{1-5q} = \dfrac{1}{x^{5q-1}}$.

Therefore,

The first two cases are divergent, and so the largest such value is $q=\frac{1}{5}$. (Alternatively, we might recognize this as a “$p$-integral” with $p=5q$, and recall that the $p$-integral diverges precisely when $p\le1$.)