The P Integrals

Two families of improper integrals appear so often that their behavior is worth memorizing: the $p$ -integrals $\int_{1}^{\infty} d x / x^{p}$ and $\int_{0}^{1} d x / x^{p}$ . The $p$ -test settles both at a glance, and almost every comparison you make later in this chapter is ultimately a comparison against one of them.

Item	Statement
Type I $p$ -integral	$\int_{1}^{\infty} \frac{d x}{x^{p}}$ converges $⟺ p > 1$ , with value $\frac{1}{p - 1}$
Type II $p$ -integral	$\int_{0}^{1} \frac{d x}{x^{p}}$ converges $⟺ p < 1$ , with value $\frac{1}{1 - p}$
Threshold	$p = 1$ in both cases, and $p = 1$ diverges in both cases
Slogan at infinity	The tail must decay faster than $1 / x$ , so $p > 1$
Slogan at the origin	The blow-up must be milder than $1 / x$ , so $p < 1$
Antiderivative	$\frac{x^{1 - p}}{1 - p}$ when $p \neq 1$ ; $\ln x$ when $p = 1$
Reading off $p$	Rewrite the integrand as $x^{- p}$ : for $\sqrt{x} = x^{1 / 2}$ we get $p = - \frac{1}{2}$
Shifted tail	$\int_{a}^{\infty} \frac{d x}{(x - c)^{p}}$ with $a > c$ converges $⟺ p > 1$ , value $\frac{(a - c)^{1 - p}}{p - 1}$
Shifted left endpoint	$\int_{a}^{b} \frac{d x}{(x - a)^{p}}$ converges $⟺ p < 1$ , value $\frac{(b - a)^{1 - p}}{1 - p}$
Shifted right endpoint	$\int_{a}^{b} \frac{d x}{(b - x)^{p}}$ converges $⟺ p < 1$ , value $\frac{(b - a)^{1 - p}}{1 - p}$
Both ends at once	$\int_{0}^{\infty} \frac{d x}{x^{p}}$ diverges for every real $p$
Logarithm in numerator	$\int_{1}^{\infty} \frac{\ln x}{x^{p}} d x$ converges $⟺ p > 1$ , value $\frac{1}{(p - 1)^{2}}$
Logarithmic analogue	$\int_{2}^{\infty} \frac{d x}{x (\ln x)^{p}}$ converges $⟺ p > 1$
Classic trap	A large $p$ helps at infinity and hurts at $0$ ; check which end is improper first

The Type I $p$ -Test

The first family lives on the infinite interval $[1, \infty)$ . The integrand $1 / x^{p}$ is perfectly well behaved everywhere on that interval, so the only question is whether the tail thins out fast enough for the total area to be finite. Here $p$ is any real number, positive, negative or zero.

The $p$ -test, infinite interval. For every real number $p$ ,

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int_1^\infty \frac{dx}{x^p} \;=\; \begin{cases} \dfrac{1}{p-1} & \text{if } p>1, \\ \infty & \text{if } p\leq 1. \end{cases}}

In other words, $\int_{1}^{\infty} d x / x^{p}$ converges if and only if $p > 1$ .

Proof

Case $p \neq 1$ . On the finite interval $[1, b]$ the power rule applies, and $x^{- p}$ has antiderivative $x^{1 - p} / (1 - p)$ :

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

Everything now hangs on the sign of the exponent $1 - p$ .

If $p > 1$ , then $1 - p < 0$ , so $b^{1 - p} = 1 / b^{p - 1} \to 0$ as $b \to \infty$ , and

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

If $p < 1$ , then $1 - p > 0$ , so $b^{1 - p} \to \infty$ and the integral diverges.

Case $p = 1$ . The power rule breaks down, because the antiderivative of $1 / x$ is $\ln x$ , not a power. The computation of $\int_{1}^{\infty} d x / x$ in Type I: Infinite Limits of Integration gives $lim_{b \to \infty} \ln b = \infty$ , so this case diverges too.

The picture below shows four members of the family on $[1, \infty)$ . All four curves pass through the same point $(1, 1)$ , so near the left endpoint they are indistinguishable. What separates them is entirely a matter of how fast they fall away.

On $[1, \infty)$ the curves with $p > 1$ (solid, blue) decay fast enough to enclose finite area, shown shaded; the curves with $p \leq 1$ (dashed, red) do not. The critical threshold is exactly $p = 1$ .

The theorem also tells you the exact value, not just a yes or no answer. Since $\int_{1}^{\infty} d x / x^{p} = 1 / (p - 1)$ , the area is small when $p$ is large and blows up as $p$ approaches $1$ from above: at $p = 2$ the area is $1$ , at $p = 1.1$ it is $10$ , at $p = 1.01$ it is $100$ . The convergence is real but it degrades without bound as you slide toward the threshold.

The Type II p-Test

The second family lives on the finite interval $(0, 1]$ . Now the tail is not the problem: the interval is bounded. The problem is at the left endpoint, where $1 / x^{p}$ blows up whenever $p > 0$ . This is a Type II improper integral, an infinite discontinuity at an endpoint, and the question is whether the spike at $0$ is narrow enough for the area beside it to be finite.

The $p$ -test, endpoint singularity. For every real number $p$ ,

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int_0^1 \frac{dx}{x^p} \;=\; \begin{cases} \dfrac{1}{1-p} & \text{if } p<1, \\ \infty & \text{if } p\geq 1. \end{cases}}

In other words, $\int_{0}^{1} d x / x^{p}$ converges if and only if $p < 1$ .

Proof

Case $p \neq 1$ . The singularity is at the left endpoint $a = 0$ , so by the definition of a Type II improper integral we start from $a + ϵ = ϵ$ and let $ϵ \to 0^{+}$ . On $[ϵ, 1]$ the integrand is continuous and the power rule applies:

\begin{aligned} \int_\epsilon^1 x^{-p}\,dx &= \left[\frac{x^{1-p}}{1-p}\right]_\epsilon^1 \\ &= \frac{1-\epsilon^{1-p}}{1-p}. \end{aligned}

Again the sign of $1 - p$ decides everything, but this time $ϵ$ is heading to $0$ rather than to $\infty$ , so the roles are swapped.

If $p < 1$ , then $1 - p > 0$ , so $ϵ^{1 - p} \to 0$ as $ϵ \to 0^{+}$ , and

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

If $p > 1$ , then $1 - p < 0$ , so $ϵ^{1 - p} = ϵ^{- (p - 1)} \to \infty$ and the integral diverges.

Case $p = 1$ . The antiderivative is again $\ln x$ , and the evaluation of $\int_{0}^{1} d x / x$ in Type II: Discontinuous Integrands gives $lim_{ϵ \to 0^{+}} (- \ln ϵ) = \infty$ . This case diverges as well.

Here is the corresponding picture. Four exponents are again drawn, two convergent and two divergent, but now on $(0, 1]$ , and the order of the curves is reversed: on $(0, 1)$ a larger $p$ makes $x^{- p}$ larger, so the big exponents are the tall ones.

On $(0, 1]$ the singularities at $0$ with $p < 1$ (solid, blue) are integrable, and the shaded regions have finite area even though they are unbounded in height; those with $p \geq 1$ (dashed, red) are not integrable. The threshold is again $p = 1$ , but the condition is reversed compared with the Type I case.

Notice what the value $1 / (1 - p)$ says: as $p$ climbs toward $1$ from below the area grows without bound, exactly mirroring what happened on the other side in the Type I test. At $p = \frac{1}{2}$ the area is $2$ ; at $p = 0.9$ it is $10$ ; at $p = 0.99$ it is $100$ .

Why the Two Conditions Are Reversed

Students routinely memorize one of these tests and then misapply it to the other. The fix is to stop memorizing inequalities and start remembering what each test is measuring.

Summary of the $p$ -integrals.

\begin{aligned} \int_1^\infty \frac{dx}{x^p} &\ \text{ converges} \;\Longleftrightarrow\; p>1, \\ \int_0^1 \frac{dx}{x^p} &\ \text{ converges} \;\Longleftrightarrow\; p<1. \end{aligned}

The threshold is $p = 1$ in both cases, but the conditions are exactly reversed. Decay that is too slow at infinity causes divergence for Type I; a singularity that is too strong at $0$ causes divergence for Type II.

Here is the sentence to keep:

At infinity, the tail must decay faster than $1 / x$ , so we need $p > 1$ . At the origin, the blow-up must be milder than $1 / x$ , so we need $p < 1$ . In both cases $p = 1$ itself fails.

The curve $y = 1 / x$ is the common reference point, and it is the one curve that is bad at both ends. Every improper $p$ -integral is a comparison against it: on $[1, \infty)$ you must beat $1 / x$ by decaying more steeply, and on $(0, 1]$ you must beat it by blowing up more gently. The number line below shows the two conditions side by side.

The two convergence sets are open rays that approach the same point $p = 1$ from opposite sides and never include it. The solid red dot at $p = 1$ on each strip records that the endpoint belongs to the divergent side both times.

The Role of p=1

The reason the threshold sits at $p = 1$ , and the reason it fails on both sides, is visible in the antiderivative. For $p \neq 1$ ,

\int x^{- p} d x = \frac{x^{1 - p}}{1 - p},

which is a power of $x$ . Whether that power is bounded near $\infty$ or near $0$ is decided by the sign of the exponent $1 - p$ , and that sign flips exactly at $p = 1$ :

To control the tail we need $x^{1 - p} \to 0$ as $x \to \infty$ , that is $1 - p < 0$ , that is $p > 1$ .
To control the spike we need $x^{1 - p} \to 0$ as $x \to 0^{+}$ , that is $1 - p > 0$ , that is $p < 1$ .

At $p = 1$ the antiderivative is not a power at all, it is $\ln x$ , and $\ln x$ is unbounded at both ends: $\ln b \to \infty$ as $b \to \infty$ and $\ln ϵ \to - \infty$ as $ϵ \to 0^{+}$ . That single fact is why $p = 1$ fails both tests. The logarithm is the borderline object here: it grows more slowly than every positive power of $x$ and more quickly than every negative power, so it sits precisely on the fence and falls off on the wrong side each time.

You can even watch the logarithm appear as the limiting case. Writing $t = 1 - p$ , the antiderivative evaluated from $1$ to $x$ is

\frac{x^{1 - p} - 1}{1 - p} = \frac{x^{t} - 1}{t} ⟶ \ln x as t \to 0,

so $\ln x$ really is the $p = 1$ member of the family, not an exception bolted on from outside.

How to Apply the p-Test

The test is easy; identifying $p$ correctly is where mistakes happen. Work through these steps every time.

Find out which end is improper. An infinite limit of integration is Type I. A point where the integrand blows up is Type II. If both occur, or if the singularity is in the interior, split the integral first and test each piece separately.
Write the integrand as a single power of $x$ . Convert every radical and every factor: $\sqrt{x} = x^{1 / 2}$ , $\sqrt[3]{x^{2}} = x^{2 / 3}$ , $x \sqrt{x} = x^{3 / 2}$ , $\frac{\sqrt[4]{x}}{x^{2}} = x^{- 7 / 4}$ .
Read off $p$ from the form $x^{- p}$ . The integrand $x^{k}$ corresponds to $p = - k$ . So $x^{- 3 / 2}$ gives $p = \frac{3}{2}$ , and $x^{1 / 2}$ gives $p = - \frac{1}{2}$ . The exponent you want is the one in the denominator.
Apply the correct half of the test. Tail at infinity: converges exactly when $p > 1$ . Singularity at the endpoint: converges exactly when $p < 1$ .
If it converges, quote the value. $\frac{1}{p - 1}$ for $\int_{1}^{\infty}$ , and $\frac{1}{1 - p}$ for $\int_{0}^{1}$ . There is no need to redo the limit computation.

Step 3 is the one that catches people. A negative $p$ is perfectly legal, and it always means divergence in the Type I test (since $p \leq 1$ ) and always means convergence in the Type II test (since $p < 1$ , and in fact the integral is not even improper there).

Worked Examples

Determine whether each integral converges. If so, find its value.

\begin{aligned} &\text{(a)}\;\int_1^\infty \frac{dx}{x^{3/2}} &&\text{(b)}\;\int_1^\infty \sqrt{x}\,dx \\ &\text{(c)}\;\int_0^1 \frac{dx}{x^{2/3}} &&\text{(d)}\;\int_0^1 \frac{dx}{x^{4/3}} \end{aligned}

Solution

The interval is $[1, \infty)$ and the integrand is already in the form $1 / x^{p}$ with $p = \frac{3}{2}$ . Since $p = \frac{3}{2} > 1$ , the Type I $p$ -test says the integral converges, and its value is
$\frac{1}{p - 1} = \frac{1}{\frac{3}{2} - 1} = \frac{1}{\frac{1}{2}} = 2.$
Write the integrand as a power in the denominator:
$\sqrt{x} = x^{1 / 2} = \frac{1}{x^{- 1 / 2}},$
so this is a Type I $p$ -integral with $p = - \frac{1}{2}$ . Since $p = - \frac{1}{2} \leq 1$ , the integral diverges. The sanity check is immediate: $\sqrt{x}$ grows without bound instead of decaying, so there is no chance of finite area.
The interval is $(0, 1]$ and the integrand blows up at $x = 0$ , so this is Type II with $p = \frac{2}{3}$ . Since $p = \frac{2}{3} < 1$ , the integral converges, and its value is
$\frac{1}{1 - p} = \frac{1}{1 - \frac{2}{3}} = \frac{1}{\frac{1}{3}} = 3.$
Type II again, this time with $p = \frac{4}{3}$ . Since $p = \frac{4}{3} \geq 1$ , the integral diverges.

Parts (a) and (d) are worth comparing: the same exponent test gives opposite verdicts on $[1, \infty)$ and on $(0, 1]$ , which is the reversal in action.

The next example is pure exponent bookkeeping, which is where the real work of the $p$ -test lies.

Classify each integral, and evaluate the convergent ones.

\begin{aligned} &\text{(a)}\;\int_1^\infty \frac{\sqrt[4]{x}}{x^2}\,dx &&\text{(b)}\;\int_1^\infty \frac{dx}{\sqrt[3]{x^2}} \\ &\text{(c)}\;\int_0^1 \frac{dx}{\sqrt[5]{x^3}} &&\text{(d)}\;\int_0^1 \frac{\sqrt{x}}{x^2}\,dx \end{aligned}

Solution

Combine the powers first: $\frac{\sqrt[4]{x}}{x^{2}} = \frac{x^{1 / 4}}{x^{2}} = x^{1 / 4 - 2} = x^{- 7 / 4}$ . So this is $\int_{1}^{\infty} d x / x^{7 / 4}$ with $p = \frac{7}{4} > 1$ : it converges, with value
$\frac{1}{p - 1} = \frac{1}{\frac{7}{4} - 1} = \frac{1}{\frac{3}{4}} = \frac{4}{3} .$
Here $\sqrt[3]{x^{2}} = x^{2 / 3}$ , so $p = \frac{2}{3}$ . On $[1, \infty)$ we need $p > 1$ , and $\frac{2}{3} \leq 1$ , so the integral diverges.
Here $\sqrt[5]{x^{3}} = x^{3 / 5}$ , so $p = \frac{3}{5}$ . On $(0, 1]$ we need $p < 1$ , and $\frac{3}{5} < 1$ , so it converges, with value
$\frac{1}{1 - p} = \frac{1}{1 - \frac{3}{5}} = \frac{1}{\frac{2}{5}} = \frac{5}{2} .$
Combine again: $\frac{\sqrt{x}}{x^{2}} = x^{1 / 2 - 2} = x^{- 3 / 2}$ , so $p = \frac{3}{2}$ . On $(0, 1]$ we need $p < 1$ , and $\frac{3}{2} \geq 1$ , so the integral diverges.

Parts (b) and (c) use the same style of exponent, $\frac{2}{3}$ against $\frac{3}{5}$ , and both are less than $1$ ; the verdicts differ only because one integral is a tail and the other is a spike.

The final example in this group is the trap the whole section is designed to prevent.

Does $\int_{0}^{\infty} \frac{d x}{x^{3 / 2}}$ converge? The exponent satisfies $p = \frac{3}{2} > 1$ , so the Type I test is passed. Is that enough?

Solution

No. This integral is improper at both ends: the upper limit is infinite, and the integrand blows up at the lower limit $x = 0$ . By definition we must split it at any convenient interior point, say $x = 1$ , and require both pieces to converge:

\int_{0}^{\infty} \frac{d x}{x^{3 / 2}} = \int_{0}^{1} \frac{d x}{x^{3 / 2}} + \int_{1}^{\infty} \frac{d x}{x^{3 / 2}} .

The right-hand piece has $p = \frac{3}{2} > 1$ and converges, to $1 / (\frac{3}{2} - 1) = 2$ . The left-hand piece has $p = \frac{3}{2} \geq 1$ , which fails the Type II test, so it diverges. One divergent piece is enough: the whole integral diverges.

The lesson is that the exponent that rescues the tail is exactly the exponent that ruins the origin. Exercise 5 below shows that this collision is not an accident of the number $\frac{3}{2}$ : no exponent whatsoever makes $\int_{0}^{\infty} d x / x^{p}$ converge.

Shifted p-Integrals

In practice the singularity is rarely at $x = 0$ and the tail rarely starts at $x = 1$ . What you actually meet is a power of $x - c$ for some constant $c$ . A single substitution reduces every such integral to one of the two standard cases, so it is worth recording the result once and using it directly.

Shifted $p$ -integrals. Let $c$ be a real number and let $p$ be any real number.

Shifted tail. If $a > c$ , then $\int_{a}^{\infty} \frac{d x}{(x - c)^{p}}$ converges if and only if $p > 1$ , and its value is then $\frac{(a - c)^{1 - p}}{p - 1}$ .
Singularity at the left endpoint. If $a < b$ , then $\int_{a}^{b} \frac{d x}{(x - a)^{p}}$ converges if and only if $p < 1$ , and its value is then $\frac{(b - a)^{1 - p}}{1 - p}$ .
Singularity at the right endpoint. If $a < b$ , then $\int_{a}^{b} \frac{d x}{(b - x)^{p}}$ converges if and only if $p < 1$ , and its value is then $\frac{(b - a)^{1 - p}}{1 - p}$ .

Proof

For the first two statements substitute $u = x - c$ (in the second, $c = a$ ), so that $d u = d x$ and the integrand becomes $u^{- p}$ .

Shifted tail. The limits become $u = a - c > 0$ and $u \to \infty$ , so

\int_{a}^{\infty} \frac{d x}{(x - c)^{p}} = \int_{a - c}^{\infty} \frac{d u}{u^{p}} .

For $p > 1$ the antiderivative $u^{1 - p} / (1 - p)$ tends to $0$ at infinity, giving

\int_{a - c}^{\infty} u^{- p} d u = {[\frac{u^{1 - p}}{1 - p}]}_{a - c}^{\infty} = \frac{- (a - c)^{1 - p}}{1 - p} = \frac{(a - c)^{1 - p}}{p - 1} .

For $p \leq 1$ the same computation gives $\infty$ , exactly as in the unshifted test. Note that only the starting point matters, not where the interval begins relative to $1$ : convergence at infinity is a statement about the tail alone.

Left endpoint. With $u = x - a$ the limits become $u = 0$ and $u = b - a > 0$ , so

\int_{a}^{b} \frac{d x}{(x - a)^{p}} = \int_{0}^{b - a} \frac{d u}{u^{p}},

and for $p < 1$ this equals ${[u^{1 - p} / (1 - p)]}_{0}^{b - a} = (b - a)^{1 - p} / (1 - p)$ , while for $p \geq 1$ it diverges.

Right endpoint. Substitute $u = b - x$ , so $d u = - d x$ and the limits $x = a$ , $x = b$ become $u = b - a$ , $u = 0$ . The two sign changes cancel and

\int_{a}^{b} \frac{d x}{(b - x)^{p}} = \int_{0}^{b - a} \frac{d u}{u^{p}},

which is the previous case verbatim.

In words: shifting the singularity or the starting point changes the value of a $p$ -integral but never changes whether it converges. The threshold is still $p = 1$ , and the direction of the inequality is still decided by which end of the interval is causing the trouble.

Evaluate each integral or show that it diverges.

\begin{aligned} &\text{(a)}\;\int_5^\infty \frac{dx}{(x-1)^3} &&\text{(b)}\;\int_1^4 \frac{dx}{\sqrt{x-1}} \end{aligned}

Solution

This is a shifted tail with $c = 1$ , $a = 5$ and $p = 3$ . Since $p = 3 > 1$ the integral converges. Substituting $u = x - 1$ makes it explicit:
\begin{aligned} \int_5^\infty \frac{dx}{(x-1)^3} &= \int_4^\infty u^{-3}\,du \\ &= \left[\frac{u^{-2}}{-2}\right]_4^\infty \\ &= 0 - \left(-\frac{1}{2\cdot 4^2}\right) \\ &= \frac{1}{32}. \end{aligned}
The formula $\frac{(a - c)^{1 - p}}{p - 1} = \frac{4^{- 2}}{2} = \frac{1}{32}$ agrees.
Here the integrand blows up at the left endpoint $x = 1$ , so this is a shifted Type II integral with $a = 1$ , $b = 4$ and $p = \frac{1}{2}$ . Since $p = \frac{1}{2} < 1$ it converges. Substituting $u = x - 1$ :
\begin{aligned} \int_1^4 \frac{dx}{\sqrt{x-1}} &= \int_0^3 u^{-1/2}\,du \\ &= \left[2u^{1/2}\right]_0^3 \\ &= 2\sqrt{3}. \end{aligned}
The formula $\frac{(b - a)^{1 - p}}{1 - p} = \frac{3^{1 / 2}}{1 / 2} = 2 \sqrt{3}$ agrees.

Common Pitfalls

Four traps to watch for.

Using the wrong half of the test. Ask first which end is improper. A big exponent is good news at infinity and bad news at a singularity.
Missing the second improper end. $\int_{0}^{\infty} d x / x^{p}$ and $\int_{1}^{\infty} d x / (x - 1)^{p}$ are improper at two places each, and each must be split before any test is applied. Both of them diverge for every $p$ , because the two ends impose contradictory conditions.
Misreading $p$ . The integrand $x^{1 / 2}$ is $1 / x^{- 1 / 2}$ , so $p = - \frac{1}{2}$ , not $\frac{1}{2}$ . Always move the power into the denominator before reading off $p$ .
Treating $p = 1$ as a near miss. It is not a rounding question: $\int_{1}^{\infty} d x / x$ and $\int_{0}^{1} d x / x$ both diverge outright, even though $\int_{1}^{\infty} d x / x^{1.001}$ and $\int_{0}^{1} d x / x^{0.999}$ both converge.

One more warning, this time about method rather than arithmetic: never apply the Fundamental Theorem of Calculus straight across an interior singularity. Writing $\int_{- 1}^{1} d x / x^{2} = [- 1 / x]_{- 1}^{1} = - 2$ produces a negative number for the integral of a positive function. The correct treatment is to split at $0$ and test each half with the Type II $p$ -test, where $p = 2 \geq 1$ shows both halves diverge.

Exercises

For which values of $p$ does each integral converge? When it converges, find its value.

$\int_{1}^{\infty} \frac{d x}{x^{p}}$
$\int_{0}^{1} \frac{d x}{x^{p}}$
$\int_{1}^{\infty} \frac{\ln x}{x^{p}} d x$

Answer

(a) converges if and only if $p > 1$ , with value $\frac{1}{p - 1}$ . (b) converges if and only if $p < 1$ , with value $\frac{1}{1 - p}$ . (c) converges if and only if $p > 1$ , with value $\frac{1}{(p - 1)^{2}}$ .

Solution

(a) and (b) are exactly the two $p$ -tests proved above: part (a) is the $p$ -test for infinite intervals, and part (b) is the $p$ -test for endpoint singularities.

(c) We treat the cases separately.

Case $p = 1$ . With $u = \ln x$ the antiderivative is $(\ln x)^{2} / 2$ , so

\begin{aligned} \int_1^\infty \frac{\ln x}{x}\,dx &= \lim_{b\to\infty}\left[\frac{(\ln x)^2}{2}\right]_1^b \\ &= \lim_{b\to\infty}\frac{(\ln b)^2}{2} \\ &= \infty. \end{aligned}

This case diverges.

Case $p \neq 1$ . Integrate by parts. Let $u = \ln x$ and $d v = x^{- p} d x$ , so that $d u = d x / x$ and $v = x^{1 - p} / (1 - p)$ :

\begin{aligned} \int_1^b \frac{\ln x}{x^p}\,dx &= \left[\frac{x^{1-p}\ln x}{1-p}\right]_1^b \\ &\quad - \frac{1}{1-p}\int_1^b x^{-p}\,dx \\ &= \frac{b^{1-p}\ln b}{1-p} - 0 \\ &\quad - \frac{1}{1-p}\cdot\frac{b^{1-p}-1}{1-p}. \end{aligned}

Sub-case $p > 1$ . Here $1 - p < 0$ , so $b^{1 - p} = b^{- (p - 1)} \to 0$ as $b \to \infty$ . Also $b^{1 - p} \ln b = (\ln b) / b^{p - 1} \to 0$ , since $\ln b$ grows far more slowly than $b^{p - 1}$ ; more precisely, L'Hopital's rule gives

\frac{\ln b}{b^{p - 1}} \sim \frac{1 / b}{(p - 1) b^{p - 2}} = \frac{1}{(p - 1) b^{p - 1}} \to 0.

Taking $b \to \infty$ :

\begin{aligned} \int_1^\infty \frac{\ln x}{x^p}\,dx &= 0 - \frac{1}{1-p}\cdot\frac{0-1}{1-p} \\ &= \frac{1}{(1-p)^2} \\ &= \frac{1}{(p-1)^2}. \end{aligned}

It converges, with value $\frac{1}{(p - 1)^{2}}$ .

Sub-case $p < 1$ . Here $1 - p > 0$ , so $b^{1 - p} \to \infty$ and $b^{1 - p} \ln b \to \infty$ . The integral diverges.

Conclusion. $\int_{1}^{\infty} (\ln x) / x^{p} d x$ converges if and only if $p > 1$ . The logarithm does not move the threshold; it only squares the answer, turning $1 / (p - 1)$ into $1 / (p - 1)^{2}$ .

Decide whether each integral converges, and evaluate the convergent ones.

$\int_{1}^{\infty} \frac{d x}{x^{7 / 3}}$
$\int_{1}^{\infty} \frac{d x}{\sqrt[3]{x}}$
$\int_{1}^{\infty} \frac{d x}{x^{1.01}}$
$\int_{1}^{\infty} \frac{d x}{x^{π}}$

Answer

(a) converges, $\frac{3}{4}$ . (b) diverges. (c) converges, $100$ . (d) converges, $\frac{1}{π - 1} \approx 0.4669$ .

Solution

Every part is a Type I $p$ -integral on $[1, \infty)$ , so the test is: converges exactly when $p > 1$ , with value $1 / (p - 1)$ .

$p = \frac{7}{3} > 1$ , so it converges, and
$\frac{1}{p - 1} = \frac{1}{\frac{7}{3} - 1} = \frac{1}{\frac{4}{3}} = \frac{3}{4} .$
$\sqrt[3]{x} = x^{1 / 3}$ , so $p = \frac{1}{3}$ . Since $p \leq 1$ , the integral diverges.
$p = 1.01 > 1$ , so it converges, and
$\frac{1}{p - 1} = \frac{1}{0.01} = 100.$
The exponent beats $1$ by only one part in a hundred, and the area is correspondingly large.
$p = π \approx 3.1416 > 1$ , so it converges, and its value is
$\frac{1}{p - 1} = \frac{1}{π - 1} \approx 0.4669 .$
Nothing requires $p$ to be rational; the test uses only the inequality $p > 1$ .

Decide whether each integral converges, and evaluate the convergent ones.

$\int_{0}^{1} \frac{d x}{\sqrt{x}}$
$\int_{0}^{1} \frac{d x}{x^{5 / 4}}$
$\int_{0}^{1} \frac{d x}{x^{2 / 5}}$
$\int_{0}^{1} \frac{d x}{x^{- 2}}$

Answer

(a) converges, $2$ . (b) diverges. (c) converges, $\frac{5}{3}$ . (d) converges, $\frac{1}{3}$ (and it is not even improper).

Solution

Every part is a Type II $p$ -integral with the trouble at $x = 0$ , so the test is: converges exactly when $p < 1$ , with value $1 / (1 - p)$ .

$\sqrt{x} = x^{1 / 2}$ , so $p = \frac{1}{2} < 1$ : it converges, and
$\frac{1}{1 - p} = \frac{1}{1 - \frac{1}{2}} = 2.$
$p = \frac{5}{4} \geq 1$ , so it diverges.
$p = \frac{2}{5} < 1$ , so it converges, and
$\frac{1}{1 - p} = \frac{1}{1 - \frac{2}{5}} = \frac{1}{\frac{3}{5}} = \frac{5}{3} .$
Here $\frac{1}{x^{- 2}} = x^{2}$ , so $p = - 2 < 1$ and the test predicts convergence with value
$\frac{1}{1 - p} = \frac{1}{1 - (- 2)} = \frac{1}{3} .$
This is a good consistency check: the integrand $x^{2}$ is continuous on $[0, 1]$ , so the integral is an ordinary proper integral, and indeed $\int_{0}^{1} x^{2} d x = {[x^{3} / 3]}_{0}^{1} = \frac{1}{3}$ . Whenever $p \leq 0$ there is no singularity at all, and the $p$ -test simply reproduces the elementary answer.

One of the two integrals

\int_{0}^{1} \frac{d x}{x^{0.999}} and \int_{0}^{1} \frac{d x}{x^{1.001}}

converges and one diverges. Decide which is which, evaluate the convergent one, and explain why the two behave so differently despite the exponents differing by only $0.002$ .

Answer

$\int_{0}^{1} x^{- 0.999} d x$ converges, to $1000$ . $\int_{0}^{1} x^{- 1.001} d x$ diverges.

Solution

Both are Type II $p$ -integrals with the singularity at $x = 0$ , so we compute the truncated integral on $[ϵ, 1]$ and let $ϵ \to 0^{+}$ . In general

\int_{ϵ}^{1} x^{- p} d x = \frac{1 - ϵ^{1 - p}}{1 - p} .

First integral, $p = 0.999$ . Here $1 - p = 0.001 > 0$ , so $ϵ^{0.001} \to 0$ and

\int_{0}^{1} \frac{d x}{x^{0.999}} = lim_{ϵ \to 0^{+}} \frac{1 - ϵ^{0.001}}{0.001} = \frac{1}{0.001} = 1000.

This agrees with $1 / (1 - p)$ from the $p$ -test.

Second integral, $p = 1.001$ . Here $1 - p = - 0.001 < 0$ , so

\int_{ϵ}^{1} \frac{d x}{x^{1.001}} = \frac{1 - ϵ^{- 0.001}}{- 0.001} = \frac{ϵ^{- 0.001} - 1}{0.001},

and $ϵ^{- 0.001} \to \infty$ as $ϵ \to 0^{+}$ . The integral diverges.

Why the difference. Convergence is not decided by the size of $ϵ^{1 - p}$ at any particular $ϵ$ ; it is decided by the sign of the exponent $1 - p$ , and a sign is not a matter of degree. A positive exponent, however tiny, forces $ϵ^{1 - p} \to 0$ ; a negative exponent, however tiny, forces $ϵ^{1 - p} \to \infty$ .

The approach is spectacularly slow in both directions, which is what makes the example instructive. For the convergent integral, $ϵ^{0.001}$ is still $0.1$ when $ϵ = 10^{- 1000}$ , so the running area has reached only about $900$ out of its eventual $1000$ . For the divergent integral, $ϵ^{- 0.001} = 10$ at $ϵ = 10^{- 1000}$ , so the partial areas have crawled up to about $9000$ and are still climbing. Numerical evidence would be worthless here; only the exponent test settles it.

Show that

\int_{0}^{\infty} \frac{d x}{x^{p}}

diverges for every real number $p$ .

Answer

It diverges for every $p$ : no exponent can satisfy the requirements of both ends at once.

Solution

The integral is improper at both ends: the upper limit is infinite, and (whenever $p > 0$ ) the integrand blows up at $x = 0$ . By definition it is assigned a value only when it is split at some interior point and both pieces converge. Split at $x = 1$ :

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

Now apply the two $p$ -tests:

The left piece $\int_{0}^{1} d x / x^{p}$ converges if and only if $p < 1$ .
The right piece $\int_{1}^{\infty} d x / x^{p}$ converges if and only if $p > 1$ .

Convergence of the whole integral would require $p < 1$ and $p > 1$ simultaneously, and no real number satisfies both. Case by case:

If $p < 1$ , the right piece diverges.
If $p > 1$ , the left piece diverges.
If $p = 1$ , both pieces diverge.

In every case at least one piece diverges, so $\int_{0}^{\infty} d x / x^{p}$ diverges for every real $p$ .

Two remarks. First, the splitting point $1$ is not special: any $d > 0$ works, since $\int_{0}^{d} d x / x^{p}$ and $\int_{d}^{\infty} d x / x^{p}$ obey the same two conditions. Second, the argument is really a statement about the shape of the family: a single power law cannot be steep at one end and shallow at the other, because it has the same exponent everywhere.

Evaluate $\int_{3}^{\infty} \frac{d x}{(x - 2)^{5 / 2}}$ or show that it diverges.

Answer

Converges, to $\frac{2}{3}$ .

Solution

The integrand is continuous on $[3, \infty)$ , since $x - 2 \geq 1 > 0$ there, so the only impropriety is the infinite upper limit. This is a shifted tail with $c = 2$ , $a = 3$ and $p = \frac{5}{2}$ .

Substitute $u = x - 2$ , so $d u = d x$ , and the limits $x = 3$ , $x \to \infty$ become $u = 1$ , $u \to \infty$ :

\begin{aligned} \int_3^\infty \frac{dx}{(x-2)^{5/2}} &= \int_1^\infty u^{-5/2}\,du \\ &= \lim_{b\to\infty}\left[\frac{u^{-3/2}}{-3/2}\right]_1^b \\ &= \lim_{b\to\infty}\left(-\frac{2}{3}b^{-3/2} + \frac{2}{3}\right) \\ &= \frac{2}{3}. \end{aligned}

The substitution turned it into the standard Type I $p$ -integral with $p = \frac{5}{2} > 1$ , whose value is $1 / (p - 1) = 1 / \frac{3}{2} = \frac{2}{3}$ . The general formula $\frac{(a - c)^{1 - p}}{p - 1} = \frac{1^{- 3 / 2}}{3 / 2} = \frac{2}{3}$ gives the same thing.

Evaluate $\int_{2}^{5} \frac{d x}{(x - 2)^{1 / 3}}$ or show that it diverges.

Answer

Converges, to $\frac{3 \sqrt[3]{9}}{2} \approx 3.1201$ .

Solution

The interval $[2, 5]$ is finite, so this is not a Type I integral. The integrand blows up at the left endpoint $x = 2$ , where $x - 2 = 0$ , so it is a Type II integral with a shifted singularity: $a = 2$ , $b = 5$ and $p = \frac{1}{3}$ .

Substitute $u = x - 2$ , so $d u = d x$ , and the limits $x = 2$ , $x = 5$ become $u = 0$ , $u = 3$ :

\begin{aligned} \int_2^5 \frac{dx}{(x-2)^{1/3}} &= \int_0^3 u^{-1/3}\,du \\ &= \lim_{\epsilon\to 0^+}\left[\frac{u^{2/3}}{2/3}\right]_\epsilon^3 \\ &= \lim_{\epsilon\to 0^+}\frac{3}{2}\left(3^{2/3}-\epsilon^{2/3}\right) \\ &= \frac{3}{2}\cdot 3^{2/3}. \end{aligned}

Since $p = \frac{1}{3} < 1$ , the $p$ -test predicted convergence, and the general formula gives the same value: $\frac{(b - a)^{1 - p}}{1 - p} = \frac{3^{2 / 3}}{2 / 3} = \frac{3 \cdot 3^{2 / 3}}{2}$ . Writing $3^{2 / 3} = \sqrt[3]{9}$ , the answer is $\frac{3 \sqrt[3]{9}}{2} \approx 3.1201$ .

Evaluate $\int_{0}^{4} \frac{d x}{\sqrt{4 - x}}$ or show that it diverges.

Answer

Converges, to $4$ .

Solution

Here the integrand blows up at the upper limit $x = 4$ , so this is the right-endpoint form of the shifted Type II result, with $a = 0$ , $b = 4$ and $p = \frac{1}{2}$ . Since $p = \frac{1}{2} < 1$ , it converges.

Substitute $u = 4 - x$ , so $d u = - d x$ ; the limits $x = 0$ , $x = 4$ become $u = 4$ , $u = 0$ , and reversing the order of the limits cancels the minus sign:

\begin{aligned} \int_0^4 \frac{dx}{\sqrt{4-x}} &= \int_0^4 u^{-1/2}\,du \\ &= \lim_{\epsilon\to 0^+}\left[2u^{1/2}\right]_\epsilon^4 \\ &= \lim_{\epsilon\to 0^+}\left(4 - 2\sqrt{\epsilon}\right) \\ &= 4. \end{aligned}

The general formula agrees: $\frac{(b - a)^{1 - p}}{1 - p} = \frac{4^{1 / 2}}{1 / 2} = 4$ . The moral is that a singularity at the right endpoint obeys exactly the same condition $p < 1$ as one at the left endpoint; only the substitution changes.

For which values of $p$ and $q$ does

\int_{0}^{1} \frac{d x}{x^{p} (1 - x)^{q}}

converge?

Answer

It converges if and only if $p < 1$ and $q < 1$ .

Solution

The integrand can be singular at both endpoints: at $x = 0$ because of $x^{- p}$ , and at $x = 1$ because of $(1 - x)^{- q}$ . It is continuous everywhere in between. Split at the midpoint:

\int_{0}^{1} \frac{d x}{x^{p} (1 - x)^{q}} = \int_{0}^{1 / 2} \frac{d x}{x^{p} (1 - x)^{q}} + \int_{1 / 2}^{1} \frac{d x}{x^{p} (1 - x)^{q}},

and the original integral converges exactly when both pieces do.

One elementary fact is used twice below, and it is worth stating on its own. Suppose $0 \leq u (x) \leq v (x)$ on $(0, c]$ and both are continuous there. As $ϵ$ decreases, $\int_{ϵ}^{c} u$ increases, because the integrand is nonnegative, and it never exceeds $\int_{0}^{c} v$ . An increasing quantity that is bounded above has a limit, so if $\int_{0}^{c} v$ converges then so does $\int_{0}^{c} u$ . Reading the same statement backwards: if $\int_{0}^{c} u$ diverges, so does $\int_{0}^{c} v$ .

Left piece. For $x \in (0, \frac{1}{2}]$ we have $1 - x \in [\frac{1}{2}, 1)$ , so the factor $(1 - x)^{- q}$ stays between the two positive constants $m = min (1, 2^{q})$ and $M = max (1, 2^{q})$ . Therefore

m \int_{0}^{1 / 2} \frac{d x}{x^{p}} \leq \int_{0}^{1 / 2} \frac{d x}{x^{p} (1 - x)^{q}} \leq M \int_{0}^{1 / 2} \frac{d x}{x^{p}},

and the fact just stated, applied in each direction, shows that the left piece converges exactly when $\int_{0}^{1 / 2} d x / x^{p}$ does, that is (by the Type II $p$ -test) exactly when $p < 1$ . The bounded factor cannot create or destroy convergence; only the singular factor matters.

Right piece. Symmetrically, for $x \in [\frac{1}{2}, 1)$ the factor $x^{- p}$ lies between two positive constants, so the right piece converges exactly when $\int_{1 / 2}^{1} d x / (1 - x)^{q}$ does. By the right-endpoint shifted rule that happens exactly when $q < 1$ .

Conclusion. The integral converges if and only if $p < 1$ and $q < 1$ . Both conditions point the same way, because both ends are singularities of a finite interval; neither end is a tail.

A check with familiar numbers: taking $p = q = \frac{1}{2}$ gives

\int_{0}^{1} \frac{d x}{\sqrt{x (1 - x)}} = [2 \arcsin \sqrt{x}]_{0}^{1} = π,

which is finite, as predicted. Taking $p = 1$ instead makes the left piece behave like $\int_{0}^{1 / 2} d x / x$ , which diverges. (For the record, the general value of this integral is the Beta function $B (1 - p, 1 - q)$ , defined precisely on the region $p < 1$ , $q < 1$ found here.)

For which values of $p$ does

\int_{2}^{\infty} \frac{d x}{x (\ln x)^{p}}

converge, and what is its value then? Explain also why the lower limit cannot be replaced by $1$ .

Answer

Converges if and only if $p > 1$ , with value $\frac{(\ln 2)^{1 - p}}{p - 1}$ . With lower limit $1$ the integral diverges for every $p$ .

Solution

Substitute $u = \ln x$ , so that $d u = d x / x$ . The limits $x = 2$ and $x \to \infty$ become $u = \ln 2 > 0$ and $u \to \infty$ , and

\int_{2}^{\infty} \frac{d x}{x (\ln x)^{p}} = \int_{\ln 2}^{\infty} \frac{d u}{u^{p}} .

This is a Type I $p$ -integral starting at $\ln 2$ instead of $1$ , which is the shifted tail result with $c = 0$ and $a = \ln 2$ . It converges if and only if $p > 1$ , and then

\int_{\ln 2}^{\infty} u^{- p} d u = {[\frac{u^{1 - p}}{1 - p}]}_{\ln 2}^{\infty} = \frac{(\ln 2)^{1 - p}}{p - 1} .

For example, $p = 2$ gives $1 / \ln 2 \approx 1.4427$ and $p = \frac{3}{2}$ gives $2 / \sqrt{\ln 2} \approx 2.4022$ .

So $1 / (x (\ln x)^{p})$ is the logarithmic analogue of $1 / x^{p}$ : the threshold is again $p = 1$ , but the whole family now decays only a hair faster than $1 / x$ , which is why these integrals are the standard tie-breakers when the ordinary $p$ -test is inconclusive.

Why the lower limit matters. If the lower limit were $1$ , the same substitution would give

\int_{1}^{\infty} \frac{d x}{x (\ln x)^{p}} = \int_{0}^{\infty} \frac{d u}{u^{p}},

because $\ln 1 = 0$ . That is precisely the integral of Exercise 5, which diverges for every real $p$ : the new lower endpoint $u = 0$ is itself a singularity, and it demands $p < 1$ while the tail demands $p > 1$ . Concretely, $\ln x \to 0$ as $x \to 1^{+}$ , so the integrand blows up at $x = 1$ . Any lower limit strictly greater than $1$ (such as $2$ , or $e$ ) avoids this and leaves only the tail condition $p > 1$ .

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

Answer

Diverges.

Solution

The integrand is continuous on $[e, \infty)$ , since $\ln x \geq 1 > 0$ there, so the only impropriety is the infinite limit. Substitute $u = \ln x$ , $d u = d x / x$ ; the limits $x = e$ and $x \to \infty$ become $u = 1$ and $u \to \infty$ :

\begin{aligned} \int_e^\infty \frac{dx}{x\ln x} &= \int_1^\infty \frac{du}{u} \\ &= \lim_{b\to\infty}\bigl[\ln u\bigr]_1^b \\ &= \lim_{b\to\infty}\ln b \\ &= \infty. \end{aligned}

The integral diverges. This is the $p = 1$ case of the logarithmic $p$ -test in the previous exercise, and it fails for exactly the same reason $\int_{1}^{\infty} d x / x$ fails: the antiderivative is a logarithm, which is unbounded. Written out in the original variable, the antiderivative is $\ln (\ln x)$ , which grows to infinity so slowly that the divergence is invisible numerically. At $x = 10^{100}$ the partial integral is still only $\ln (\ln 10^{100}) \approx 5.4$ .

For which values of $p$ does $\int_{0}^{1} x^{p} \ln x d x$ converge, and what is its value then?

Answer

Converges if and only if $p > - 1$ , with value $- \frac{1}{(p + 1)^{2}}$ .

Solution

The integrand is continuous on $(0, 1]$ but $\ln x \to - \infty$ as $x \to 0^{+}$ , so this is a Type II integral with the trouble at $0$ . Note the value will be negative, since $\ln x < 0$ throughout $(0, 1)$ .

Case $p \neq - 1$ : integrate by parts. Take $u = \ln x$ and $d v = x^{p} d x$ , so $d u = d x / x$ and $v = x^{p + 1} / (p + 1)$ . On $[ϵ, 1]$ ,

\begin{aligned} \int_\epsilon^1 x^p\ln x\,dx &= \left[\frac{x^{p+1}\ln x}{p+1}\right]_\epsilon^1 \\ &\quad - \frac{1}{p+1}\int_\epsilon^1 x^p\,dx \\ &= 0 - \frac{\epsilon^{p+1}\ln\epsilon}{p+1} \\ &\quad - \frac{1}{p+1}\cdot\frac{1-\epsilon^{p+1}}{p+1}, \end{aligned}

using $\ln 1 = 0$ for the upper endpoint.

Sub-case $p > - 1$ . Then $p + 1 > 0$ , so $ϵ^{p + 1} \to 0$ , and also $ϵ^{p + 1} \ln ϵ \to 0$ , because a positive power of $ϵ$ beats the logarithm (by L'Hopital's rule, $\frac{\ln ϵ}{ϵ^{- (p + 1)}} \sim \frac{1 / ϵ}{- (p + 1) ϵ^{- (p + 2)}} = - \frac{ϵ^{p + 1}}{p + 1} \to 0$ ). Hence

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

Sub-case $p < - 1$ . The same formula settles this case; it only needs regrouping. Write $q = p + 1$ , so $q < 0$ , and collect the two terms containing $ϵ^{q}$ :

\begin{aligned} \int_\epsilon^1 x^p\ln x\,dx &= \epsilon^{q}\left(\frac{1}{q^{2}}-\frac{\ln\epsilon}{q}\right) - \frac{1}{q^{2}}. \end{aligned}

Now let $ϵ \to 0^{+}$ . Since $q < 0$ , the factor $ϵ^{q} = ϵ^{- | q |}$ tends to $+ \infty$ . Inside the bracket, $\ln ϵ \to - \infty$ and $q < 0$ , so $\ln ϵ / q \to + \infty$ and the whole bracket tends to $- \infty$ . A quantity tending to $+ \infty$ times one tending to $- \infty$ tends to $- \infty$ , and subtracting the constant $1 / q^{2}$ leaves that unchanged. The integral diverges to $- \infty$ .

A numerical check with $p = - 2$ , where $q = - 1$ : the formula gives $ϵ^{- 1} (1 + \ln ϵ) - 1$ , which is about $- 362$ at $ϵ = 10^{- 2}$ and about $- 5909$ at $ϵ = 10^{- 3}$ , sinking steadily.

Case $p = - 1$ . Now the antiderivative is $(\ln x)^{2} / 2$ :

\int_{ϵ}^{1} \frac{\ln x}{x} d x = {[\frac{(\ln x)^{2}}{2}]}_{ϵ}^{1} = - \frac{(\ln ϵ)^{2}}{2} ⟶ - \infty .

It diverges.

Conclusion. The integral converges exactly when $p > - 1$ , with value $- \frac{1}{(p + 1)^{2}}$ . Spot checks: $p = 0$ gives $\int_{0}^{1} \ln x d x = - 1$ ; $p = 1$ gives $- \frac{1}{4}$ ; $p = 2$ gives $- \frac{1}{9}$ ; $p = - \frac{1}{2}$ gives $- 4$ .

Compare this with the threshold $p < 1$ of the plain Type II test. Writing $x^{p} = 1 / x^{- p}$ , the condition $p > - 1$ is the condition $- p < 1$ , so the logarithm has again left the threshold alone: it is too weak a factor to change which powers are integrable.

Explain why $p = 1$ is the exact borderline for both $p$ -tests, and why it lies on the divergent side in both. Your explanation should account for the fact that the two convergence conditions, $p > 1$ and $p < 1$ , point in opposite directions.

Answer

Because the antiderivative of $x^{- p}$ is the power $x^{1 - p} / (1 - p)$ when $p \neq 1$ , and the exponent $1 - p$ changes sign exactly at $p = 1$ ; at $p = 1$ the antiderivative is instead $\ln x$ , which is unbounded at $0$ and at $\infty$ , so it fails both tests.

Solution

1. One antiderivative controls both tests. For $p \neq 1$ ,

\int x^{- p} d x = \frac{x^{1 - p}}{1 - p},

and each $p$ -test is nothing more than the question of whether this power stays bounded as $x$ approaches the bad end of the interval.

2. The bad ends are opposite, so the requirements are opposite. For $\int_{1}^{\infty}$ the bad end is $x \to \infty$ , and $x^{1 - p}$ stays bounded there exactly when the exponent is negative:

1 - p < 0 ⟺ p > 1.

For $\int_{0}^{1}$ the bad end is $x \to 0^{+}$ , and $x^{1 - p}$ stays bounded there exactly when the exponent is positive:

1 - p > 0 ⟺ p < 1.

The same quantity $1 - p$ governs both, but a power of $x$ that dies at infinity necessarily blows up at $0$ , and vice versa. That is the source of the reversal, and it also explains why no single exponent can serve both ends, as Exercise 5 shows.

3. The sign change happens at $p = 1$ and nowhere else. The expression $1 - p$ is zero only at $p = 1$ , so that is the only value where the direction of the inequality can flip. Both tests therefore have the same threshold even though they use it in opposite ways.

4. The threshold itself fails, on both sides. At $p = 1$ the exponent $1 - p$ is zero, the formula $x^{1 - p} / (1 - p)$ is meaningless, and the antiderivative is

\int \frac{d x}{x} = \ln x,

which is unbounded at both ends: $\ln b \to \infty$ as $b \to \infty$ , and $\ln ϵ \to - \infty$ as $ϵ \to 0^{+}$ . So $p = 1$ does not merely sit on the fence, it fails both tests outright. This is the precise sense in which $1 / x$ is the universal borderline: it decays too slowly to have a finite tail and blows up too strongly to have a finite spike.

5. The logarithm is the limit of the family, not an outsider. Setting $t = 1 - p$ ,

\frac{x^{1 - p} - 1}{1 - p} = \frac{x^{t} - 1}{t} ⟶ \ln x as t \to 0,

so $\ln x$ is exactly what the antiderivatives converge to as $p \to 1$ . The logarithm grows more slowly than every positive power of $x$ and more quickly than every negative power, which is another way of saying it sits precisely at the threshold and belongs to neither side.

The Type I $p$ -Test

The Type II p-Test

Why the Two Conditions Are Reversed

The Role of p=1

How to Apply the p-Test

Worked Examples

Shifted p-Integrals

Common Pitfalls

Exercises

Quick Links

Social Media

The P Integrals

The Type I p -Test

The Type II p-Test

Why the Two Conditions Are Reversed

The Role of p=1

How to Apply the p-Test

Worked Examples

Shifted p-Integrals

Common Pitfalls

Exercises

Quick Links

Social Media

The Type I $p$ -Test