Comparison Tests For Improper Integrals

Most improper integrals cannot be evaluated in closed form, and most of the time we do not need the value: we only need to know whether the area is finite. The comparison tests settle that question by measuring the integrand against a simpler one whose behavior we already know, almost always a $p$ -integral or an exponential.

Item	Statement
Direct Comparison Test	if $0 \leq f \leq g$ on $[a, \infty)$ : $\int g$ converges $\Rightarrow$ $\int f$ converges; $\int f$ diverges $\Rightarrow$ $\int g$ diverges
What it never tells you	a divergent $\int g$ says nothing about $f$ ; a convergent $\int f$ says nothing about $g$
Limit Comparison Test	$L = lim f / g$ with $0 < L < \infty$ : the two integrals both converge or both diverge
Case $L = 0$	usable only when $\int g$ converges (then $\int f$ converges); inconclusive when $\int g$ diverges
Case $L = \infty$	usable only when $\int g$ diverges (then $\int f$ diverges); inconclusive when $\int g$ converges
How to pick $g$	keep the dominant term of the numerator and of the denominator, then read off the power of $x$
Which term dominates	as $x \to \infty$ the highest power wins; as $x \to 0^{+}$ the lowest power wins
$p$ -test, Type I	$\int_{1}^{\infty} x^{- p} d x$ converges $⟺ p > 1$
$p$ -test, Type II	$\int_{0}^{1} x^{- p} d x$ converges $⟺ p < 1$
Exponential benchmark	$\int_{a}^{\infty} e^{- k x} d x = e^{- k a} / k$ converges for every $k > 0$
Handy size facts	$0 < \ln x < x$ for $x > 1$ ; $\ln x < 2 \sqrt{x}$ for $x \geq 1$ ; $e^{- x^{2}} \leq e^{- x}$ for $x \geq 1$
Small- $x$ substitutes	$\sin x \approx x$ , $\tan x \approx x$ , $1 - e^{- x} \approx x$ , $1 - \cos x \approx x^{2} / 2$
Absolute convergence	$\int_{a}^{\infty} \| f \| d x$ converges $\Rightarrow$ $\int_{a}^{\infty} f d x$ converges
Two values worth knowing	$\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ and $\int_{0}^{\infty} \frac{\sin x}{x} d x = \frac{π}{2}$

The Direct Comparison Test

Many improper integrals cannot be evaluated in closed form. The comparison tests let us determine convergence or divergence by relating the integral to a simpler one whose behavior is known, typically a $p$ -integral from the $p$ -test or an exponential.

The first test is the one suggested by the picture of area: a region that sits inside a region of finite area must itself have finite area.

Direct Comparison Test. Suppose $0 \leq f (x) \leq g (x)$ for all $x \geq a$ .

If $\int_{a}^{\infty} g (x) d x$ converges, so does $\int_{a}^{\infty} f (x) d x$ .
If $\int_{a}^{\infty} f (x) d x$ diverges, so does $\int_{a}^{\infty} g (x) d x$ .

An analogous statement holds for Type II improper integrals, where the trouble is a singularity at an endpoint rather than an infinite limit of integration.

In symbols, the whole test is the following pair of one-way implications.

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{0\le f\le g \text{ on } [a,\infty)\ \Longrightarrow\ \begin{cases} \displaystyle\int_a^\infty g\ \text{converges}\ \Rightarrow\ \int_a^\infty f\ \text{converges},\\ \displaystyle\int_a^\infty f\ \text{diverges}\ \Rightarrow\ \int_a^\infty g\ \text{diverges}.\end{cases}}

Read the two lines as "finite area on top forces finite area underneath" and "infinite area underneath forces infinite area on top". Everything else about the two functions is irrelevant.

The blue region lies inside the red one. If the red region has finite total area, so does the blue one; if the blue region has infinite area, so does the red one.

Part (a) says: if the larger function has finite total area, so does the smaller. Part (b) says: if the smaller function has infinite area, so does the larger. Note that the hypothesis $f \geq 0$ is essential. The test compares areas, and an integrand that changes sign can converge through cancellation instead of through smallness; that situation is handled by absolute and conditional convergence.

Show that $\int_{1}^{\infty} \frac{d x}{x^{2} + \sqrt{x}}$ converges.

Solution

For $x \geq 1$ the extra term $\sqrt{x}$ in the denominator only makes the fraction smaller:

x^{2} + \sqrt{x} \geq x^{2} ⟹ 0 \leq \frac{1}{x^{2} + \sqrt{x}} \leq \frac{1}{x^{2}} .

Since $\int_{1}^{\infty} d x / x^{2}$ converges (the $p$ -test with $p = 2 > 1$ ), the Direct Comparison Test gives convergence.

The test gives no value, but it does give a bound: the integral is at most $\int_{1}^{\infty} x^{- 2} d x = 1$ . Numerically it is about $0.7471$ .

Show that $\int_{2}^{\infty} \frac{d x}{\ln x}$ diverges.

Solution

For $x > 1$ we have $0 < \ln x < x$ , and taking reciprocals reverses the inequality:

\frac{1}{\ln x} > \frac{1}{x} > 0.

Since $\int_{2}^{\infty} d x / x$ diverges (the $p$ -test with $p = 1$ ), the Direct Comparison Test gives divergence.

The point of the comparison is that $\ln x$ grows so slowly that $1 / \ln x$ decays even more slowly than $1 / x$ , and $1 / x$ already decays too slowly to enclose finite area.

When the Direct Comparison Test Says Nothing

Both parts of the test run in one direction only, and the two "reverse" statements are false:

If $\int_{a}^{\infty} g d x$ diverges, nothing follows about $\int_{a}^{\infty} f d x$ . The smaller function may still enclose finite area.
If $\int_{a}^{\infty} f d x$ converges, nothing follows about $\int_{a}^{\infty} g d x$ . The larger function may still enclose infinite area.

The following picture shows why. Both $\frac{1}{2 x}$ and $\frac{1}{x^{2}}$ lie below $\frac{1}{x}$ on $[1, \infty)$ , and $\int_{1}^{\infty} d x / x$ diverges. Yet $\int_{1}^{\infty} \frac{d x}{2 x}$ diverges while $\int_{1}^{\infty} \frac{d x}{x^{2}} = 1$ converges. Sitting below a divergent integral distinguishes nothing at all.

Two functions with the same divergent upper bound, one with finite area and one with infinite area.

An inconclusive comparison is still information: it means the chosen $g$ was too generous (or too mean) to detect the true size of $f$ . Go back and match $f$ more closely.

Choosing a Comparison Function

In practice, the only hard part of a comparison argument is deciding what to compare with. There is a reliable recipe.

Locate the trouble. A comparison is a local statement, so decide where the integral is improper: at $+ \infty$ , at $- \infty$ , or at a point where the integrand blows up. If there is more than one bad point, split the interval so that each piece has exactly one.
Keep only the dominant term of the numerator and only the dominant term of the denominator. As $x \to \infty$ the dominant term is the one with the highest power of $x$ ; as $x \to 0^{+}$ it is the one with the lowest power.
Simplify the resulting quotient to a single power $x^{- p}$ (or to a single exponential $e^{- k x}$ ). Discard constant factors: they never affect convergence.
Read off $p$ and apply the $p$ -test. That is your prediction.
Prove it. Either build an explicit inequality for the Direct Comparison Test, or compute $L = lim f / g$ for the Limit Comparison Test below. The limit is the proof; steps 2 to 4 only tell you what to aim at.

Before applying the rule, replace any transcendental factor by its small- $x$ substitute if the trouble is at the origin: $\sin x \approx x$ , $\tan x \approx x$ , $1 - e^{- x} \approx x$ , $1 - \cos x \approx \frac{1}{2} x^{2}$ , $\ln (1 + x) \approx x$ . Each of these is exact in the limiting sense that the ratio of the two sides tends to a finite nonzero number, which is precisely what the Limit Comparison Test needs.

Integrand	Trouble at	Dominant terms	Take $g (x) =$	Verdict
$\frac{1}{x^{2} + \sqrt{x}}$	$x \to \infty$	$\frac{1}{x^{2}}$	$x^{- 2}$	$p = 2 > 1$ : converges
$\frac{x}{3 x^{4} + 5 x^{2} + 1}$	$x \to \infty$	$\frac{x}{3 x^{4}}$	$x^{- 3}$	$p = 3 > 1$ : converges
$\frac{x^{2} - 1}{\sqrt{x^{6} + 16}}$	$x \to \infty$	$\frac{x^{2}}{x^{3}}$	$x^{- 1}$	$p = 1$ : diverges
$\frac{x + 1}{\sqrt{x^{4} - x}}$	$x \to \infty$	$\frac{x}{x^{2}}$	$x^{- 1}$	$p = 1$ : diverges
$\frac{\sin x}{x^{3 / 2}}$	$x \to 0^{+}$	$\frac{x}{x^{3 / 2}}$	$x^{- 1 / 2}$	$p = \frac{1}{2} < 1$ : converges

Warning. The rule fails when the dominant terms cancel. For $f (x) = \sqrt{x + 1} - \sqrt{x}$ the naive reading gives $\sqrt{x} - \sqrt{x} = 0$ , which is useless. Rationalize first: $\sqrt{x + 1} - \sqrt{x} = \frac{1}{\sqrt{x + 1} + \sqrt{x}} \approx \frac{1}{2 \sqrt{x}}$ , so the right comparison is $g (x) = x^{- 1 / 2}$ with $p = \frac{1}{2}$ , and $\int_{1}^{\infty} (\sqrt{x + 1} - \sqrt{x}) d x$ diverges. Whenever a subtraction kills the leading behavior, do the algebra before applying the rule.

The Limit Comparison Test

Building an explicit inequality can be awkward: for $\frac{x + 1}{\sqrt{x^{4} - x}}$ it is not obvious what constant multiple of $1 / x$ will bound it. The Limit Comparison Test removes that work. Instead of an inequality it asks only for a ratio, and constants no longer matter.

Limit Comparison Test. Suppose $f (x) \geq 0$ and $g (x) > 0$ for all $x \geq a$ , and write $L$ for the limit of the ratio $f / g$ .

If $lim_{x \to \infty} \frac{f (x)}{g (x)} = L$ with $0 < L < \infty$ , then $\int_{a}^{\infty} f d x$ and $\int_{a}^{\infty} g d x$ both converge or both diverge.
If $L = 0$ and $\int_{a}^{\infty} g d x$ converges, then $\int_{a}^{\infty} f d x$ converges.
If $L = \infty$ and $\int_{a}^{\infty} g d x$ diverges, then $\int_{a}^{\infty} f d x$ diverges.

An analogous statement holds for Type II integrals, with $x \to a^{+}$ replacing $x \to \infty$ .

Part (a) is the one used almost every time:

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{0<\lim_{x\to\infty}\frac{f(x)}{g(x)}=L<\infty\quad\Longrightarrow\quad \int_a^\infty f\,dx\ \text{and}\ \int_a^\infty g\,dx\ \text{behave alike}}

A finite nonzero $L$ means that far out, $f$ is essentially a constant multiple of $g$ , and multiplying an integrand by a positive constant cannot turn finite area into infinite area.

Proof of part (a)

Let $ε = L / 2$ . Since $f (x) / g (x) \to L$ , there exists $M \geq a$ such that for all $x \geq M$ ,

| \frac{f (x)}{g (x)} - L | < \frac{L}{2},

which, after multiplying through by $g (x) > 0$ , gives

\frac{L}{2} g (x) < f (x) < \frac{3 L}{2} g (x) (x \geq M) .

If $\int_{M}^{\infty} g d x$ converges, the upper bound and the Direct Comparison Test imply $\int_{M}^{\infty} f d x$ converges. If $\int_{M}^{\infty} g d x$ diverges, the lower bound implies $\int_{M}^{\infty} f d x$ diverges. On $[a, M]$ both integrals are finite (since $f$ and $g$ are continuous there), so the same conclusion holds for $\int_{a}^{\infty}$ .

The choice $ε = L / 2$ is what makes the lower bound $L - ε = L / 2$ strictly positive; any $ε < L$ would do just as well.

For $x \geq M$ , the function $f$ is squeezed between $\frac{L}{2} g$ and $\frac{3 L}{2} g$ (the green band). Since both bounds are proportional to $g$ , convergence of $\int_{a}^{\infty} g (x) d x$ forces convergence of $\int_{a}^{\infty} f (x) d x$ , and divergence of $\int_{a}^{\infty} g (x) d x$ forces divergence of $\int_{a}^{\infty} f (x) d x$ .

The Two Inconclusive Cases

Parts (b) and (c) each cover only one half of a situation, and the missing halves really are undecidable.

$L = 0$ with $\int_{a}^{\infty} g d x$ divergent is inconclusive. Take $g (x) = 1 / x$ on $[2, \infty)$ , whose integral diverges. For $f (x) = 1 / x^{2}$ the ratio $f / g = 1 / x \to 0$ and $\int_{2}^{\infty} f d x$ converges. For $f (x) = \frac{1}{x \ln x}$ the ratio $f / g = 1 / \ln x \to 0$ as well, yet $\int_{2}^{\infty} \frac{d x}{x \ln x}$ diverges, because the substitution $u = \ln x$ turns it into $\int_{\ln 2}^{\infty} d u / u$ . Same $L$ , same divergent $g$ , opposite conclusions.
$L = \infty$ with $\int_{a}^{\infty} g d x$ convergent is inconclusive. Take $g (x) = 1 / x^{2}$ on $[1, \infty)$ , whose integral converges. For $f (x) = 1 / x$ the ratio $f / g = x \to \infty$ and $\int_{1}^{\infty} f d x$ diverges. For $f (x) = x^{- 3 / 2}$ the ratio $f / g = \sqrt{x} \to \infty$ too, yet $\int_{1}^{\infty} x^{- 3 / 2} d x = 2$ converges.

The pattern is easy to remember. $L = 0$ means $f$ is eventually far smaller than $g$ , so it is useful only for proving convergence. $L = \infty$ means $f$ is eventually far larger than $g$ , so it is useful only for proving divergence. A finite nonzero $L$ is the only case that works in both directions, which is why step 2 of the recipe aims for it.

Worked Examples

The first two examples are Type II: the interval is finite and the integrand blows up at an endpoint. Nothing changes except that the limit defining $L$ is taken as $x$ approaches the bad endpoint.

Determine whether $\int_{0}^{1} \frac{\sin x}{x^{3 / 2}} d x$ converges or diverges.

Solution

This is a Type II integral; the integrand is unbounded as $x \to 0^{+}$ . Near $x = 0$ we have $\sin x \approx x$ , so

\frac{\sin x}{x^{3 / 2}} \approx \frac{x}{x^{3 / 2}} = x^{- 1 / 2} .

That suggests $g (x) = x^{- 1 / 2}$ , and the Limit Comparison Test confirms it:

\begin{aligned} \lim_{x\to 0^{+}}\frac{\sin x/x^{3/2}}{x^{-1/2}} &= \lim_{x\to 0^{+}}\frac{\sin x}{x} \\ &= 1. \end{aligned}

Since $L = 1$ lies in $(0, \infty)$ and $\int_{0}^{1} x^{- 1 / 2} d x$ converges (the $p$ -test with $p = 1 / 2 < 1$ ), the Limit Comparison Test gives convergence.

Note that $\sin x$ is positive on $(0, 1]$ , so the hypothesis $f \geq 0$ holds and the test applies. Numerically the integral is about $1.935$ .

Determine whether $\int_{0}^{1} \frac{d x}{\sqrt{x - x^{2}}}$ converges, and if so find its value.

The integrand is unbounded at both endpoints, so the interval is split at $x = \frac{1}{2}$ and each half is tested separately.

Solution

Write $x - x^{2} = x (1 - x)$ . The integrand is unbounded at both $x = 0$ and $x = 1$ , so no single comparison can handle the whole interval. Split at $x = \frac{1}{2}$ :

\int_{0}^{1} \frac{d x}{\sqrt{x (1 - x)}} = \int_{0}^{1 / 2} \frac{d x}{\sqrt{x (1 - x)}} + \int_{1 / 2}^{1} \frac{d x}{\sqrt{x (1 - x)}} .

Near $x = 0$ . Since $1 - x \to 1$ , we have $\sqrt{x (1 - x)} \approx \sqrt{x}$ . With $g (x) = x^{- 1 / 2}$ :

lim_{x \to 0^{+}} \frac{1 / \sqrt{x (1 - x)}}{x^{- 1 / 2}} = lim_{x \to 0^{+}} \frac{1}{\sqrt{1 - x}} = 1.

Since $\int_{0}^{1 / 2} x^{- 1 / 2} d x$ converges, the first piece converges.

Near $x = 1$ . The substitution $u = 1 - x$ converts the second piece into $\int_{0}^{1 / 2} d u / \sqrt{u (1 - u)}$ , the same form as the first. So the second piece also converges, and the whole integral converges.

Value. Completing the square in the radicand:

x - x^{2} = \frac{1}{4} - {(x - \frac{1}{2})}^{2} .

Substitute $x - \frac{1}{2} = \frac{1}{2} \sin θ$ , so that $d x = \frac{1}{2} \cos θ d θ$ , and the limits $x = 0$ and $x = 1$ become $θ = - π / 2$ and $θ = π / 2$ :

\begin{aligned} \int_0^1 \frac{dx}{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^{2}}} &= \int_{-\pi/2}^{\pi/2}\frac{\frac{1}{2}\cos\theta\,d\theta}{\frac{1}{2}|\cos\theta|} \\ &= \int_{-\pi/2}^{\pi/2} d\theta \\ &= \pi. \end{aligned}

The absolute value disappears because $\cos θ \geq 0$ on $[- π / 2, π / 2]$ . The substitution has turned an improper integral into a proper one, which is legitimate precisely because we first checked that both halves converge.

Test for convergence:

$\int_{1}^{\infty} \frac{x d x}{3 x^{4} + 5 x^{2} + 1}$ ;
$\int_{2}^{\infty} \frac{x^{2} - 1}{\sqrt{x^{6} + 16}} d x$ .

Solution

(a) For large $x$ , the denominator $3 x^{4} + 5 x^{2} + 1$ is dominated by $3 x^{4}$ , so the integrand behaves like $x / (3 x^{4}) = 1 / (3 x^{3})$ . We apply the Limit Comparison Test with $g (x) = 1 / x^{3}$ :

lim_{x \to \infty} \frac{x / (3 x^{4} + 5 x^{2} + 1)}{1 / x^{3}} = lim_{x \to \infty} \frac{x^{4}}{3 x^{4} + 5 x^{2} + 1} = \frac{1}{3} .

Since $0 < 1 / 3 < \infty$ and $\int_{1}^{\infty} x^{- 3} d x$ converges (the $p$ -test with $p = 3 > 1$ ), the integral in (a) converges. The constant $\frac{1}{3}$ is irrelevant to the verdict; only the fact that it is finite and nonzero matters.

(b) For large $x$ , $x^{2} - 1 \approx x^{2}$ and $\sqrt{x^{6} + 16} \approx x^{3}$ , so the integrand behaves like $x^{2} / x^{3} = 1 / x$ . We apply the Limit Comparison Test with $g (x) = 1 / x$ , then divide numerator and denominator by $x^{3}$ (legitimate since $\sqrt{x^{6}} = x^{3}$ for $x > 0$ ):

\begin{aligned} \lim_{x\to\infty}\frac{(x^2-1)/\sqrt{x^6+16}}{1/x} &= \lim_{x\to\infty}\frac{x(x^2-1)}{\sqrt{x^6+16}} \\ &= \lim_{x\to\infty}\frac{1-1/x^{2}}{\sqrt{1+16/x^{6}}} \\ &= 1. \end{aligned}

Since $L = 1 \neq 0$ and $\int_{2}^{\infty} d x / x$ diverges (the $p$ -test with $p = 1$ ), the integral in (b) diverges.

Two Important Convergent Integrals

The next two integrals are the most famous in this chapter. Neither can be evaluated by the Fundamental Theorem, because neither integrand has an elementary antiderivative, and yet both converge. Comparison is what proves it.

Show that $\int_{- \infty}^{\infty} e^{- x^{2}} d x$ converges.

Solution

We split the integral at $c = 0$ :

\int_{- \infty}^{\infty} e^{- x^{2}} d x = \int_{- \infty}^{0} e^{- x^{2}} d x + \int_{0}^{\infty} e^{- x^{2}} d x .

By symmetry ( $e^{- x^{2}}$ is an even function), it suffices to show the right half converges. We split again at $x = 1$ :

\int_{0}^{\infty} e^{- x^{2}} d x = \int_{0}^{1} e^{- x^{2}} d x + \int_{1}^{\infty} e^{- x^{2}} d x .

The first piece is a proper integral (the integrand is continuous on $[0, 1]$ ) and is therefore finite. For the second piece, observe that for $x \geq 1$ :

x^{2} \geq x ⟹ - x^{2} \leq - x ⟹ e^{- x^{2}} \leq e^{- x} .

Since $\int_{1}^{\infty} e^{- x} d x = e^{- 1}$ converges, the Direct Comparison Test gives convergence of $\int_{1}^{\infty} e^{- x^{2}} d x$ . Therefore $\int_{0}^{\infty} e^{- x^{2}} d x$ converges, and by symmetry $\int_{- \infty}^{\infty} e^{- x^{2}} d x$ converges.

Note why we split at $x = 1$ : the inequality $e^{- x^{2}} \leq e^{- x}$ is false on $(0, 1)$ , where $x^{2} < x$ . Restricting the comparison to the range where it actually holds, and treating the leftover piece as a proper integral, is a standard move.

The exact value $\int_{- \infty}^{\infty} e^{- x^{2}} d x = \sqrt{π}$ is established in multivariable calculus using a double-integral technique. Here we have confirmed only that the integral converges. That is enough for the Gaussian to serve as a probability density, which is its main use.

Show that $\int_{0}^{\infty} \frac{\sin x}{x} d x$ converges.

Solution

The integrand $(\sin x) / x$ has a removable singularity at $x = 0$ , since $lim_{x \to 0^{+}} (\sin x) / x = 1$ . We extend the integrand continuously to $x = 0$ by setting its value there equal to $1$ . The integral is therefore improper only at $+ \infty$ .

We split the interval at $x = π$ :

\int_{0}^{\infty} \frac{\sin x}{x} d x = \int_{0}^{π} \frac{\sin x}{x} d x + \int_{π}^{\infty} \frac{\sin x}{x} d x .

The first piece is a proper integral (the extended integrand is continuous on $[0, π]$ ). For the second piece we use integration by parts. Let $u = 1 / x$ and $d v = \sin x d x$ , so $d u = - d x / x^{2}$ and $v = - \cos x$ . Then:

\begin{aligned} \int_\pi^b \frac{\sin x}{x}\,dx &= \Bigl[-\frac{\cos x}{x}\Bigr]_\pi^b - \int_\pi^b \frac{\cos x}{x^2}\,dx. \end{aligned}

As $b \to \infty$ : $- \cos b / b \to 0$ , while the value at the lower limit, $(- \cos π) / π = 1 / π$ , must be subtracted, so the whole bracket approaches $0 - 1 / π = - 1 / π$ . For the remaining integral, $| \cos x / x^{2} | \leq 1 / x^{2}$ and $\int_{π}^{\infty} d x / x^{2}$ converges, so by the Direct Comparison Test $\int_{π}^{\infty} (\cos x) / x^{2} d x$ converges. Therefore $\int_{π}^{\infty} (\sin x) / x d x$ converges, and so does $\int_{0}^{\infty} (\sin x) / x d x$ .

Integration by parts was the essential step: it traded the slowly decaying $1 / x$ for the rapidly decaying $1 / x^{2}$ , and only then was a comparison available.

Note that the comparison in the last example was applied to $| \cos x | / x^{2}$ , not to $(\sin x) / x$ itself, which changes sign infinitely often and so is outside the scope of the tests. The integral $\int_{0}^{\infty} \frac{\sin x}{x} d x$ equals $π / 2$ and is the standard example of conditional convergence: it converges, but $\int_{0}^{\infty} \frac{| \sin x |}{x} d x$ diverges.

Exercises

Determine convergence or divergence using the Direct Comparison Test.

$\int_{1}^{\infty} \frac{\sin^{2} x}{x^{2}} d x$
$\int_{1}^{\infty} e^{- x^{2}} d x$
$\int_{1}^{\infty} \frac{x^{2} + 1}{x^{3} + x + 1} d x$

Answer

(a) converges. (b) converges. (c) diverges.

Solution

(a) Since $0 \leq \sin^{2} x \leq 1$ for all $x$ :

0 \leq \frac{\sin^{2} x}{x^{2}} \leq \frac{1}{x^{2}} for all x \geq 1.

Since $\int_{1}^{\infty} d x / x^{2}$ converges ( $p = 2 > 1$ ), the Direct Comparison Test gives convergence.

(b) For $x \geq 1$ , we have $x^{2} \geq x$ , so $e^{- x^{2}} \leq e^{- x}$ . Therefore:

0 < e^{- x^{2}} \leq e^{- x} for all x \geq 1.

Since $\int_{1}^{\infty} e^{- x} d x = e^{- 1}$ converges, the Direct Comparison Test gives convergence.

(c) The integrand behaves like $x^{2} / x^{3} = 1 / x$ , so we expect divergence. We claim $(x^{2} + 1) / (x^{3} + x + 1) \geq 1 / (3 x)$ for all $x \geq 1$ .

This inequality is equivalent to $3 x (x^{2} + 1) \geq x^{3} + x + 1$ , which simplifies to:

3 x^{3} + 3 x \geq x^{3} + x + 1, i.e., 2 x^{3} + 2 x - 1 \geq 0.

For $x \geq 1$ : $2 x^{3} + 2 x - 1 \geq 2 (1) + 2 (1) - 1 = 3 > 0$ . The claim holds, so:

\frac{x^{2} + 1}{x^{3} + x + 1} \geq \frac{1}{3 x} for all x \geq 1.

Since $\int_{1}^{\infty} d x / (3 x) = \frac{1}{3} \int_{1}^{\infty} d x / x$ diverges ( $p = 1$ ), the Direct Comparison Test gives divergence.

Use the Limit Comparison Test to determine convergence or divergence.

$\int_{1}^{\infty} \frac{x + 1}{\sqrt{x^{4} - x}} d x$
$\int_{1}^{\infty} \frac{d x}{\sqrt{x^{3} + 2 x + 1}}$
$\int_{0}^{1} \frac{1 - \cos x}{x^{2}} d x$

Answer

(a) diverges. (b) converges. (c) converges.

Solution

(a) For large $x$ , the numerator $x + 1 \approx x$ and $\sqrt{x^{4} - x} \approx x^{2}$ , so the integrand behaves like $x / x^{2} = 1 / x$ . We compare with $g (x) = 1 / x$ :

\frac{(x + 1) / \sqrt{x^{4} - x}}{1 / x} = \frac{x (x + 1)}{\sqrt{x^{4} - x}} .

To evaluate this limit, we square the ratio (both sides are positive for $x > 1$ ) and divide numerator and denominator by $x^{4}$ :

\begin{aligned} \left(\frac{x(x+1)}{\sqrt{x^4-x}}\right)^{2} &= \frac{x^2(x+1)^2}{x^4-x} \\ &= \frac{x^4+2x^3+x^2}{x^4-x} \\ &= \frac{1 + 2/x + 1/x^2}{1 - 1/x^3} \\ &\to \frac{1+0+0}{1-0} = 1 \quad\text{as }x\to\infty. \end{aligned}

Since the squared ratio tends to $1$ , the ratio itself tends to $\sqrt{1} = 1$ . With $L = 1$ and $\int_{1}^{\infty} d x / x$ diverging ( $p = 1$ ), the Limit Comparison Test gives divergence.

One detail worth noticing: the integrand is also improper at the left endpoint, since $x^{4} - x = 0$ at $x = 1$ . That singularity is harmless. Near $x = 1$ we have $x^{4} - x = x (x - 1) (x^{2} + x + 1) \approx 3 (x - 1)$ , so the integrand behaves like $2 / \sqrt{3 (x - 1)}$ , a $p = \frac{1}{2}$ singularity, which is integrable. The divergence comes entirely from the behavior at infinity.

(b) For large $x$ , $x^{3} + 2 x + 1 \approx x^{3}$ , so $1 / \sqrt{x^{3} + 2 x + 1} \approx x^{- 3 / 2}$ . Compare with $g (x) = x^{- 3 / 2}$ :

\begin{aligned} \frac{1/\sqrt{x^3+2x+1}}{x^{-3/2}} &= \sqrt{\frac{x^3}{x^3+2x+1}} \\ &= \sqrt{\frac{1}{1+2/x^2+1/x^3}} \\ &\to \sqrt{1} = 1. \end{aligned}

Since $L = 1$ and $\int_{1}^{\infty} x^{- 3 / 2} d x$ converges ( $p = 3 / 2 > 1$ ), the Limit Comparison Test gives convergence.

(c) This looks like a Type II integral because $(1 - \cos x) / x^{2}$ appears undefined at $x = 0$ . We check whether the singularity is real or removable. Applying L'Hopital's rule:

lim_{x \to 0^{+}} \frac{1 - \cos x}{x^{2}} = lim_{x \to 0^{+}} \frac{\sin x}{2 x} = \frac{1}{2} lim_{x \to 0^{+}} \frac{\sin x}{x} = \frac{1}{2} .

The limit is finite, so the singularity at $x = 0$ is removable: if we define $f (0) = 1 / 2$ , then $f$ is continuous on $[0, 1]$ . Apply the Limit Comparison Test with $g (x) = 1$ (the constant function):

lim_{x \to 0^{+}} \frac{(1 - \cos x) / x^{2}}{1} = \frac{1}{2} .

Since $L = 1 / 2 \in (0, \infty)$ and $\int_{0}^{1} 1 d x = 1$ converges, the Limit Comparison Test gives convergence.

Determine whether $\int_{0}^{\infty} \frac{\sin x}{x} d x$ converges. Specifically, show that:

The integrand extends continuously to $x = 0$ , so there is no singularity there.
The integral over $[1, \infty)$ converges.

(This integral equals $π / 2$ and is known as the Dirichlet integral. Its convergence rests entirely on cancellation between the humps of $\sin x$ above and below the axis: the integral of the absolute value, $\int_{0}^{\infty} | \sin x | / x d x$ , is infinite. Its exact value requires techniques beyond the scope of this text.)

Answer

Converges.

Solution

(a) The function $(\sin x) / x$ is undefined at $x = 0$ . We compute the limit using L'Hopital's rule:

lim_{x \to 0^{+}} \frac{\sin x}{x} = lim_{x \to 0^{+}} \frac{\cos x}{1} = 1.

If we define $f (0) = 1$ and $f (x) = (\sin x) / x$ for $x > 0$ , then $f$ is continuous on $[0, \infty)$ . There is no singularity at $x = 0$ , and $\int_{0}^{\infty} (\sin x) / x d x$ is improper only at the upper limit.

(b) We show $\int_{1}^{\infty} (\sin x) / x d x$ converges by integrating by parts. Let $u = 1 / x$ and $d v = \sin x d x$ , so $d u = - d x / x^{2}$ and $v = - \cos x$ :

\begin{aligned} \int_1^b \frac{\sin x}{x}\,dx &= \left[-\frac{\cos x}{x}\right]_1^b - \int_1^b \frac{\cos x}{x^2}\,dx \\ &= -\frac{\cos b}{b} + \cos 1 - \int_1^b \frac{\cos x}{x^2}\,dx. \end{aligned}

As $b \to \infty$ :

The boundary term satisfies $| \cos b / b | \leq 1 / b \to 0$ .
For the remaining integral, since $| \cos x | \leq 1$ : $| \frac{\cos x}{x^{2}} | \leq \frac{1}{x^{2}} for all x \geq 1.$
By the Direct Comparison Test, $\int_{1}^{\infty} | \cos x | / x^{2} d x$ converges (with $\int_{1}^{\infty} d x / x^{2} = 1$ ). By the exercise below, in which convergence of $\int | f |$ is shown to force convergence of $\int f$ , the integral $\int_{1}^{\infty} (\cos x) / x^{2} d x$ also converges.

Therefore $\int_{1}^{\infty} (\sin x) / x d x$ converges. Combined with the fact that $\int_{0}^{1} (\sin x) / x d x$ is a proper integral (by part (a)), the full integral $\int_{0}^{\infty} (\sin x) / x d x$ converges.

When $\int_{a}^{\infty} | f (x) | d x$ converges, the integral $\int_{a}^{\infty} f (x) d x$ is said to converge absolutely. Prove that an absolutely convergent integral converges: that is, if $\int_{a}^{\infty} | f (x) | d x$ converges, then so does $\int_{a}^{\infty} f (x) d x$ .

Hint: Define $f^{+} = max (f, 0)$ and $f^{-} = max (- f, 0)$ , and note that $f = f^{+} - f^{-}$ , $| f | = f^{+} + f^{-}$ , and $0 \leq f^{\pm} \leq | f |$ .

Solution

Define the positive part and negative part of $f$ :

f^{+} (x) = max (f (x), 0) and f^{-} (x) = max (- f (x), 0) .

These satisfy:

\begin{aligned} f &= f^{+} - f^{-}, \\ |f| &= f^{+} + f^{-}, \\ 0 &\leq f^{+}(x)\leq|f(x)| \quad\text{and}\quad 0\leq f^{-}(x)\leq|f(x)| \quad\text{for all }x\geq a. \end{aligned}

Suppose $\int_{a}^{\infty} | f | d x$ converges. Since $0 \leq f^{\pm} \leq | f |$ , the Direct Comparison Test implies both $\int_{a}^{\infty} f^{+} d x$ and $\int_{a}^{\infty} f^{-} d x$ converge separately. This is exactly why the positive and negative parts are introduced: they are nonnegative, so the comparison tests apply to them even though they do not apply to $f$ .

Since both integrals are finite, we can write:

\int_{a}^{\infty} f d x = \int_{a}^{\infty} (f^{+} - f^{-}) d x = \int_{a}^{\infty} f^{+} d x - \int_{a}^{\infty} f^{-} d x .

This is the difference of two finite numbers, hence finite. Therefore $\int_{a}^{\infty} f d x$ converges.

Remark. An integral $\int_{a}^{\infty} f d x$ for which $\int_{a}^{\infty} | f | d x$ also converges is said to converge absolutely. An integral that converges but for which $\int_{a}^{\infty} | f | d x$ diverges is said to converge conditionally: it survives only because of cancellation. The Dirichlet integral of the previous exercise is of this second kind, since the integral of $| \sin x | / x$ is infinite while the integral of $(\sin x) / x$ is not.

Determine whether each integral converges or diverges. (These continue the exercises of Type I: Infinite Limits of Integration; both parts need the Direct Comparison Test.)

$\int_{1}^{\infty} \frac{2 + \cos x}{x} d x$
$\int_{1}^{\infty} \frac{d x}{x^{2} + x + 1}$

Answer

(a) diverges. (b) converges.

Solution

(a) Since $- 1 \leq \cos x \leq 1$ for all $x$ , we have $2 + \cos x \geq 2 - 1 = 1$ . Therefore:

\frac{2 + \cos x}{x} \geq \frac{1}{x} for all x \geq 1.

Since $\int_{1}^{\infty} d x / x$ diverges (by the $p$ -test with $p = 1$ ), the Direct Comparison Test gives divergence.

(b) For $x \geq 1$ , we have $x^{2} + x + 1 \geq x^{2}$ , so:

\frac{1}{x^{2} + x + 1} \leq \frac{1}{x^{2}} .

Since $\int_{1}^{\infty} d x / x^{2}$ converges ( $p = 2 > 1$ ), the Direct Comparison Test gives convergence.

Recall from the exercises of Type II: Discontinuous Integrands that $\int_{0}^{1} \frac{d x}{\sqrt{x} (1 + x)} = \frac{π}{2}$ . Using this, determine whether $\int_{0}^{\infty} \frac{d x}{\sqrt{x} (1 + x)}$ converges or diverges. If it converges, identify its type (Type I, Type II, or both) and find its value.

Answer

Converges; value $π$ ; both Type I and Type II.

Solution

This integral has a singularity at $x = 0$ (Type II) and an infinite upper limit (Type I). Split at $x = 1$ :

\int_{0}^{\infty} \frac{d x}{\sqrt{x} (1 + x)} = \int_{0}^{1} \frac{d x}{\sqrt{x} (1 + x)} + \int_{1}^{\infty} \frac{d x}{\sqrt{x} (1 + x)} .

The left piece was evaluated in the Type II exercises: it equals $π / 2$ and converges.

For the right piece, note that for $x \geq 1$ :

\sqrt{x} (1 + x) \geq \sqrt{x} \cdot x = x^{3 / 2},

so $1 / (\sqrt{x} (1 + x)) \leq 1 / x^{3 / 2}$ . Since $\int_{1}^{\infty} x^{- 3 / 2} d x$ converges ( $p = 3 / 2 > 1$ ), the Direct Comparison Test gives convergence.

The substitution $u = \sqrt{x}$ , so that $x = u^{2}$ and $d x = 2 u d u$ , extends to the full range:

\begin{aligned} \int_0^\infty \frac{dx}{\sqrt{x}\,(1+x)} &= 2\int_0^\infty \frac{du}{1+u^2} \\ &= 2\lim_{R\to\infty}\bigl[\arctan u\bigr]_0^R \\ &= 2\cdot\frac{\pi}{2} = \pi. \end{aligned}

The value $π$ splits as $\frac{π}{2} + \frac{π}{2}$ : exactly half the area sits over $[0, 1]$ and half over $[1, \infty)$ .

Show that $\int_{0}^{\infty} \frac{d x}{1 + x^{p}}$ converges if and only if $p > 1$ . Assume $p > 0$ .

Hint: Split at $x = 1$ . The left piece is a proper integral for all $p > 0$ . For the right piece, compare with $1 / x^{p}$ using the Limit Comparison Test.

Solution

Split the integral at $x = 1$ :

\int_{0}^{\infty} \frac{d x}{1 + x^{p}} = \int_{0}^{1} \frac{d x}{1 + x^{p}} + \int_{1}^{\infty} \frac{d x}{1 + x^{p}} .

Left piece: $\int_{0}^{1} d x / (1 + x^{p})$ . For $p > 0$ and $x \in [0, 1]$ , the denominator satisfies $1 + x^{p} \geq 1 > 0$ , so the integrand is continuous and bounded. Hence $\int_{0}^{1} d x / (1 + x^{p})$ is a proper Riemann integral, and it converges for all $p > 0$ .

Right piece: $\int_{1}^{\infty} d x / (1 + x^{p})$ . Apply the Limit Comparison Test with $g (x) = 1 / x^{p}$ , dividing numerator and denominator by $x^{p}$ :

\begin{aligned} \lim_{x\to\infty}\frac{1/(1+x^p)}{1/x^p} &= \lim_{x\to\infty}\frac{x^p}{1+x^p} \\ &= \lim_{x\to\infty}\frac{1}{1/x^p + 1} \\ &= \frac{1}{0+1} = 1. \end{aligned}

Since $L = 1 \in (0, \infty)$ and $\int_{1}^{\infty} d x / x^{p}$ converges if and only if $p > 1$ (by the $p$ -test for infinite intervals), the Limit Comparison Test gives: $\int_{1}^{\infty} d x / (1 + x^{p})$ converges if and only if $p > 1$ .

Conclusion. The left piece always converges (for $p > 0$ ). The right piece converges if and only if $p > 1$ . Therefore $\int_{0}^{\infty} d x / (1 + x^{p})$ converges if and only if $p > 1$ .

The first worked example of this section showed that $\int_{1}^{\infty} \frac{d x}{\sqrt{x} + x^{2}}$ converges, using the Direct Comparison Test. Reach the same conclusion with the Limit Comparison Test. Then show that the choice $g (x) = x^{- 1 / 2}$ leads nowhere, and explain which case of the test fails.

Answer

Converges. With $g (x) = 1 / x^{2}$ the limit is $L = 1$ . With $g (x) = x^{- 1 / 2}$ the limit is $L = 0$ while $\int_{1}^{\infty} g d x$ diverges, which is one of the inconclusive cases.

Solution

The denominator $\sqrt{x} + x^{2}$ is dominated by $x^{2}$ as $x \to \infty$ , so the recipe says to take $g (x) = 1 / x^{2}$ . Dividing numerator and denominator by $x^{2}$ :

\begin{aligned} \lim_{x\to\infty}\frac{1/(\sqrt{x}+x^{2})}{1/x^{2}} &= \lim_{x\to\infty}\frac{x^{2}}{\sqrt{x}+x^{2}} \\ &= \lim_{x\to\infty}\frac{1}{x^{-3/2}+1} \\ &= 1. \end{aligned}

Since $L = 1$ lies in $(0, \infty)$ and $\int_{1}^{\infty} x^{- 2} d x = 1$ converges ( $p = 2 > 1$ ), the Limit Comparison Test gives convergence.

Now try $g (x) = x^{- 1 / 2}$ , which keeps the wrong term of the denominator:

\begin{aligned} \lim_{x\to\infty}\frac{1/(\sqrt{x}+x^{2})}{x^{-1/2}} &= \lim_{x\to\infty}\frac{\sqrt{x}}{\sqrt{x}+x^{2}} \\ &= \lim_{x\to\infty}\frac{1}{1+x^{3/2}} \\ &= 0. \end{aligned}

Here $L = 0$ , and $\int_{1}^{\infty} x^{- 1 / 2} d x$ diverges ( $p = 1 / 2 < 1$ ). Part (b) of the Limit Comparison Test requires $\int g d x$ to converge before it says anything, so this comparison is inconclusive. The lesson is that $g$ must match the dominant term of the denominator, not merely some term of it.

For reference, the value of the integral is about $0.7471$ , comfortably below the bound $\int_{1}^{\infty} x^{- 2} d x = 1$ from the direct comparison.

Determine whether $\int_{1}^{\infty} \frac{\ln x}{x^{2} + 1} d x$ converges or diverges.

Answer

Converges.

Solution

The integrand is nonnegative on $[1, \infty)$ , since $\ln x \geq 0$ there, so the comparison tests apply. The difficulty is that $\ln x$ is not a power of $x$ . The way around it is the fact that $\ln x$ grows more slowly than every positive power of $x$ ; the version we need is

\ln x < 2 \sqrt{x} for all x \geq 1.

To see this, note that both sides agree in size at $x = 1$ ( $0 < 2$ ) and compare derivatives: $\frac{d}{d x} \ln x = 1 / x$ while $\frac{d}{d x} 2 \sqrt{x} = 1 / \sqrt{x}$ , and $1 / x \leq 1 / \sqrt{x}$ for $x \geq 1$ . So the right side starts larger and grows faster.

Now compare, using $x^{2} + 1 > x^{2}$ :

0 \leq \frac{\ln x}{x^{2} + 1} < \frac{2 \sqrt{x}}{x^{2}} = \frac{2}{x^{3 / 2}} for all x \geq 1.

Since $\int_{1}^{\infty} 2 x^{- 3 / 2} d x = 4$ converges ( $p = 3 / 2 > 1$ ), the Direct Comparison Test gives convergence.

Alternative, with the Limit Comparison Test. Take $g (x) = x^{- 3 / 2}$ . Then

\begin{aligned} \lim_{x\to\infty}\frac{\ln x/(x^{2}+1)}{x^{-3/2}} &= \lim_{x\to\infty}\frac{x^{3/2}\ln x}{x^{2}+1} \\ &= \lim_{x\to\infty}\frac{\ln x}{\sqrt{x}} \\ &= 0, \end{aligned}

the last step by L'Hopital's rule: $(1 / x) / (\frac{1}{2} x^{- 1 / 2}) = 2 / \sqrt{x} \to 0$ . Since $L = 0$ and $\int_{1}^{\infty} x^{- 3 / 2} d x$ converges, part (b) of the Limit Comparison Test gives convergence. This is the one situation in which $L = 0$ is genuinely useful.

The exact value happens to be Catalan's constant, about $0.9160$ , well inside the bound $4$ found above.

Determine whether $\int_{2}^{\infty} \frac{d x}{x \sqrt{x^{2} - 1}}$ converges. If it converges, find its value.

Answer

Converges; value $\frac{π}{6} \approx 0.5236$ .

Solution

First check where the integral is improper. The radicand $x^{2} - 1$ vanishes at $x = 1$ , but the interval starts at $x = 2$ , where $\sqrt{x^{2} - 1} = \sqrt{3}$ . So the integrand is continuous on $[2, \infty)$ and the only trouble is the infinite upper limit: this is a pure Type I integral.

For large $x$ , $\sqrt{x^{2} - 1} \approx x$ , so the integrand behaves like $1 / x^{2}$ . Apply the Limit Comparison Test with $g (x) = 1 / x^{2}$ :

\begin{aligned} \lim_{x\to\infty}\frac{1/\bigl(x\sqrt{x^{2}-1}\bigr)}{1/x^{2}} &= \lim_{x\to\infty}\frac{x^{2}}{x\sqrt{x^{2}-1}} \\ &= \lim_{x\to\infty}\frac{x}{\sqrt{x^{2}-1}} \\ &= \lim_{x\to\infty}\frac{1}{\sqrt{1-1/x^{2}}} \\ &= 1. \end{aligned}

Since $L = 1 \in (0, \infty)$ and $\int_{2}^{\infty} x^{- 2} d x$ converges ( $p = 2 > 1$ ), the integral converges.

For the value, recall that for $x > 1$

\frac{d}{d x} \arccos \frac{1}{x} = \frac{1}{x \sqrt{x^{2} - 1}},

which one checks by the chain rule: \frac{d}{dx}\arccos u=-u'/\sqrt{1-u^{2}} with $u = 1 / x$ gives $\frac{1 / x^{2}}{\sqrt{1 - 1 / x^{2}}} = \frac{1}{x \sqrt{x^{2} - 1}}$ . Therefore

\begin{aligned} \int_2^\infty \frac{dx}{x\sqrt{x^{2}-1}} &= \lim_{b\to\infty}\left[\arccos\frac{1}{x}\right]_2^b \\ &= \lim_{b\to\infty}\arccos\frac{1}{b} - \arccos\frac{1}{2} \\ &= \frac{\pi}{2} - \frac{\pi}{3} \\ &= \frac{\pi}{6}. \end{aligned}

The value $π / 6 \approx 0.5236$ is consistent with the comparison, which predicted a finite answer of roughly the size of $\int_{2}^{\infty} x^{- 2} d x = \frac{1}{2}$ .

Determine whether $\int_{0}^{1} \frac{e^{x}}{x} d x$ converges or diverges.

Answer

Diverges.

Solution

This is a Type II integral: the integrand is unbounded as $x \to 0^{+}$ , since the numerator tends to $1$ while the denominator tends to $0$ .

For $0 < x \leq 1$ we have $e^{x} \geq e^{0} = 1$ , because the exponential is increasing. Therefore

\frac{e^{x}}{x} \geq \frac{1}{x} > 0 for all x \in (0, 1] .

Since $\int_{0}^{1} d x / x$ diverges (the $p$ -test for an endpoint singularity, with $p = 1$ ), the Direct Comparison Test gives divergence.

The Limit Comparison Test reaches the same verdict: with $g (x) = 1 / x$ ,

lim_{x \to 0^{+}} \frac{e^{x} / x}{1 / x} = lim_{x \to 0^{+}} e^{x} = 1,

so $L = 1 \in (0, \infty)$ and the two integrals behave alike; since $\int_{0}^{1} d x / x$ diverges, so does the given integral.

The moral is that a factor tending to a finite nonzero limit, here $e^{x} \to 1$ , never changes the verdict. Only the power of $x$ matters.

Determine whether each integral converges or diverges.

$\int_{0}^{1} \frac{1 - e^{- x}}{x^{3 / 2}} d x$
$\int_{0}^{1} \frac{\tan x}{x^{3 / 2}} d x$

Answer

(a) converges. (b) converges.

Solution

(a) The integrand is unbounded as $x \to 0^{+}$ : the numerator tends to $0$ like $x$ , but the denominator tends to $0$ faster. Using the small- $x$ substitute $1 - e^{- x} \approx x$ , the integrand behaves like $x / x^{3 / 2} = x^{- 1 / 2}$ , so take $g (x) = x^{- 1 / 2}$ :

\begin{aligned} \lim_{x\to 0^{+}}\frac{\bigl(1-e^{-x}\bigr)/x^{3/2}}{x^{-1/2}} &= \lim_{x\to 0^{+}}\frac{1-e^{-x}}{x} \\ &= \lim_{x\to 0^{+}}\frac{e^{-x}}{1} \\ &= 1, \end{aligned}

the middle step by L'Hopital's rule (the limit is also just the derivative of $1 - e^{- x}$ at $x = 0$ ). Since $L = 1 \in (0, \infty)$ and $\int_{0}^{1} x^{- 1 / 2} d x = 2$ converges ( $p = 1 / 2 < 1$ ), the Limit Comparison Test gives convergence. Numerically the value is about $1.723$ .

(b) On $[0, 1]$ the function $\tan x$ is continuous and nonnegative, since $1 < π / 2$ and the only trouble of $\tan$ is at $π / 2$ . So again the sole singularity is at $x = 0$ . With $\tan x \approx x$ near the origin, the integrand behaves like $x^{- 1 / 2}$ , and we take $g (x) = x^{- 1 / 2}$ once more:

\begin{aligned} \lim_{x\to 0^{+}}\frac{\tan x/x^{3/2}}{x^{-1/2}} &= \lim_{x\to 0^{+}}\frac{\tan x}{x} \\ &= \lim_{x\to 0^{+}}\frac{\sin x}{x}\cdot\frac{1}{\cos x} \\ &= 1\cdot 1 = 1. \end{aligned}

Since $L = 1$ and $\int_{0}^{1} x^{- 1 / 2} d x$ converges, the integral converges. Numerically it is about $2.175$ .

Both parts illustrate the same principle: near the origin, replace the transcendental factor by the power of $x$ it imitates, then let the $p$ -test decide.

Determine whether $\int_{1}^{\infty} \frac{d x}{x + e^{x}}$ converges or diverges.

Answer

Converges.

Solution

The exponential dominates the polynomial, so the comparison should be with $e^{- x}$ , not with a power of $x$ . For $x \geq 1$ both terms of the denominator are positive, so dropping $x$ only makes the denominator smaller and the fraction larger:

0 < \frac{1}{x + e^{x}} < \frac{1}{e^{x}} = e^{- x} for all x \geq 1.

Since

\int_{1}^{\infty} e^{- x} d x = lim_{b \to \infty} [- e^{- x}]_{1}^{b} = e^{- 1}

converges, the Direct Comparison Test gives convergence, together with the bound $\int_{1}^{\infty} \frac{d x}{x + e^{x}} < e^{- 1} \approx 0.3679$ . (The true value is about $0.2901$ .)

What not to do. Comparing with $1 / x$ gives $\frac{1}{x + e^{x}} < \frac{1}{x}$ , which is true but useless, since $\int_{1}^{\infty} d x / x$ diverges: this is the inconclusive case "smaller than something divergent". A limit comparison against a divergent power, meaning $g (x) = x^{- p}$ with $p \leq 1$ , is equally useless: $lim_{x \to \infty} \frac{x^{p}}{x + e^{x}} = 0$ , and $L = 0$ paired with a divergent $\int g$ proves nothing. A convergent power such as $g (x) = x^{- 2}$ does settle the question, through part (b) of the Limit Comparison Test, since $L = 0$ with $\int_{1}^{\infty} x^{- 2} d x$ convergent forces convergence. But exponential decay outruns every power, so the exponential is the sharper and more natural yardstick here.

Each part below starts with a comparison that fails. Explain exactly which hypothesis of which test breaks down, then choose a better $g$ and settle the question.

$\int_{1}^{\infty} \frac{d x}{x \sqrt{x + 1}}$ , first attempted with $g (x) = 1 / x$ and the Direct Comparison Test.
$\int_{1}^{\infty} \frac{d x}{\sqrt{x} (x + 1)}$ , first attempted with $g (x) = 1 / x^{2}$ and the Limit Comparison Test.

Answer

Both integrals converge. In (a) the failed comparison puts $f$ below a divergent $g$ , which proves nothing; use $g (x) = x^{- 3 / 2}$ . In (b) the failed comparison gives $L = \infty$ with a convergent $\int g$ , which proves nothing; use $g (x) = x^{- 3 / 2}$ again, obtaining $L = 1$ .

Solution

(a) The attempted comparison is correct as an inequality: for $x \geq 1$ we have $\sqrt{x + 1} > 1$ , so

0 < \frac{1}{x \sqrt{x + 1}} < \frac{1}{x} .

But $\int_{1}^{\infty} d x / x$ diverges, and the Direct Comparison Test never concludes anything from a divergent upper bound: a function below a divergent one may have finite or infinite area. The comparison is inconclusive.

The fix is to keep the dominant term of $\sqrt{x + 1}$ , which is $\sqrt{x}$ . Since $\sqrt{x + 1} > \sqrt{x}$ for $x \geq 1$ ,

0 < \frac{1}{x \sqrt{x + 1}} < \frac{1}{x \sqrt{x}} = \frac{1}{x^{3 / 2}} .

Now $\int_{1}^{\infty} x^{- 3 / 2} d x = 2$ converges ( $p = 3 / 2 > 1$ ), so the Direct Comparison Test gives convergence, with the bound $2$ . (The exact value is $2 \ln (1 + \sqrt{2}) \approx 1.7627$ , obtained with the substitution $u = \sqrt{x + 1}$ .)

(b) With $g (x) = 1 / x^{2}$ the ratio grows without bound:

\begin{aligned} \lim_{x\to\infty}\frac{1/\bigl(\sqrt{x}\,(x+1)\bigr)}{1/x^{2}} &= \lim_{x\to\infty}\frac{x^{2}}{\sqrt{x}\,(x+1)} \\ &= \lim_{x\to\infty}\frac{x^{3/2}}{x+1} \\ &= \infty. \end{aligned}

So $L = \infty$ . Part (c) of the Limit Comparison Test requires $\int_{1}^{\infty} g d x$ to diverge, but $\int_{1}^{\infty} x^{- 2} d x = 1$ converges. The hypothesis fails and nothing follows: $L = \infty$ only ever proves divergence.

The fix is to match the true size of the integrand. Since $\sqrt{x} (x + 1) \approx x^{3 / 2} \cdot 1$ for large $x$ , take $g (x) = x^{- 3 / 2}$ :

\begin{aligned} \lim_{x\to\infty}\frac{1/\bigl(\sqrt{x}\,(x+1)\bigr)}{x^{-3/2}} &= \lim_{x\to\infty}\frac{x^{3/2}}{\sqrt{x}\,(x+1)} \\ &= \lim_{x\to\infty}\frac{x}{x+1} \\ &= 1. \end{aligned}

Since $L = 1 \in (0, \infty)$ and $\int_{1}^{\infty} x^{- 3 / 2} d x$ converges, the integral converges. Its value is $π / 2$ , which follows from the exercise computing $\int_{0}^{\infty} \frac{d x}{\sqrt{x} (1 + x)} = π$ minus the value $π / 2$ over $[0, 1]$ .

In both parts the failed choice of $g$ was off by a half power of $x$ . That is the usual symptom: an inconclusive comparison almost always means the exponent was not matched exactly.

The Direct Comparison Test

When the Direct Comparison Test Says Nothing

Choosing a Comparison Function

The Limit Comparison Test

The Two Inconclusive Cases

Worked Examples

Two Important Convergent Integrals

Exercises

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