Type Ii Discontinuous Integrands

A Type II improper integral is one whose integrand becomes unbounded somewhere on the interval of integration: at an endpoint, or at a point hidden in the middle. You handle it by cutting the bad point out with a limit, and the integral converges only when that limit is finite.

Item	Statement
Type II improper integral	$\int_{a}^{b} f (x) d x$ in which $f$ is unbounded at some point of $[a, b]$
Singularity	a point where $\| f (x) \| \to \infty$ ; also called an infinite discontinuity
Singular right endpoint $b$	$\int_{a}^{b} f (x) d x = lim_{ϵ \to 0^{+}} \int_{a}^{b - ϵ} f (x) d x$
Singular left endpoint $a$	$\int_{a}^{b} f (x) d x = lim_{ϵ \to 0^{+}} \int_{a + ϵ}^{b} f (x) d x$
Interior singularity at $c$	$\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x$ , and both pieces must converge
Converges / diverges	converges when the limit is finite, diverges otherwise
The $p$ -test	$\int_{0}^{1} d x / x^{p}$ converges exactly when $p < 1$ , with value $1 / (1 - p)$
Benchmark	$\int_{0}^{1} d x / \sqrt{x} = 2$ (converges)
Benchmark	$\int_{0}^{1} d x / x$ diverges
Benchmark	$\int_{0}^{1} d x / x^{2}$ diverges
Benchmark	$\int_{0}^{1} \ln x d x = - 1$ (converges even though $\ln x \to - \infty$ )
Both endpoints singular	split at any convenient interior point, then treat each piece separately
Classic trap	applying the Fundamental Theorem straight across an interior singularity
Sign check	if $f \geq 0$ but your answer is negative, you missed a singularity

The Definition

The second type of improper integral occurs when the integrand is unbounded at one or both endpoints, or at a point inside the interval. We call such a point, where $| f (x) | \to \infty$ , a singularity of $f$ , or equivalently an infinite discontinuity.

The interval itself is perfectly finite here. That is the difference from Type I integrals, where the interval was infinite and the integrand was well behaved. In a Type II integral the interval is short and the function is the thing that misbehaves.

Let $a < b$ be real numbers.

(a) If $f$ is continuous on $[a, b)$ and $| f (x) | \to \infty$ as $x \to b^{-}$ , we define

\int_{a}^{b} f (x) d x = lim_{ϵ \to 0^{+}} \int_{a}^{b - ϵ} f (x) d x .

(b) If $f$ is continuous on $(a, b]$ and $| f (x) | \to \infty$ as $x \to a^{+}$ , we define

\int_{a}^{b} f (x) d x = lim_{ϵ \to 0^{+}} \int_{a + ϵ}^{b} f (x) d x .

(c) If $f$ has an infinite discontinuity at an interior point $c \in (a, b)$ and is continuous elsewhere on $[a, b]$ , we define

\int_{a}^{b} f (x) d x = \int_{a}^{c} f (x) d x + \int_{c}^{b} f (x) d x,

provided both integrals on the right converge.

In each case the integral converges if the limit is finite, and diverges otherwise.

The number $ϵ$ is a small positive amount of room we leave between the limit of integration and the singularity. Inside that shortened interval the integrand is continuous and bounded, so $\int_{a}^{b - ϵ} f$ is an ordinary Riemann integral with an ordinary value. The improper integral is then defined as what that ordinary value approaches as the gap closes.

The two endpoint rules are the ones you use constantly. When the trouble is at the left endpoint:

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int_a^b f(x)\,dx = \lim_{\epsilon\to 0^+}\int_{a+\epsilon}^{b} f(x)\,dx}

In words: start a hair to the right of the bad point, integrate normally, and then slide the starting point back toward $a$ .

When Both Endpoints Are Singular

The definition does not cover an integrand that blows up at $a$ and at $b$ , such as $1 / \sqrt{1 - x^{2}}$ on $[- 1, 1]$ . The fix is the same as in case (c): pick any convenient interior point $c$ , split the integral there, and apply rule (b) to the left piece and rule (a) to the right piece. The original integral converges exactly when both pieces converge, and its value is their sum. The choice of $c$ does not affect the answer.

A Step-by-Step Procedure

Locate the singularities. Set every denominator equal to zero, check the argument of every logarithm, and check where $\tan$ , $\sec$ , and friends blow up. Keep only the points that lie in $[a, b]$ .
Split the interval so that every singularity is an endpoint of a piece, never an interior point of a piece.
Replace each singular endpoint by a limit variable. A singular left endpoint $a$ becomes $a + ϵ$ ; a singular right endpoint $b$ becomes $b - ϵ$ .
Integrate the resulting proper integral exactly as usual, using the Fundamental Theorem, which is now legitimate because the integrand is continuous on the shortened interval.
Take the limit $ϵ \to 0^{+}$ . If it is finite, that piece converges to that number; otherwise it diverges.
Add the pieces. If even one piece diverges, the whole integral diverges. Do not let a $+ \infty$ from one piece cancel a $- \infty$ from another; that cancellation is a different object, the Cauchy principal value.

The Three Cases, Geometrically

In each picture below, the shaded area is the honest, finite area of the shortened region. The improper integral asks what happens to that area as the gap around the singularity closes: the integral converges if and only if the total area stays finite.

Case (a): Singularity at the Right Endpoint

The curve runs off to $+ \infty$ at the right endpoint $b$ , so we stop short at $b - ϵ$ and shade the area we can actually compute. Then we let $ϵ \to 0^{+}$ and watch whether that area settles on a number.

Case (b): Singularity at the Left Endpoint

Here the curve escapes at the left endpoint $a$ , so we start at $a + ϵ$ instead. The shaded area grows as $ϵ$ shrinks, and the question is whether it grows without bound or approaches a limit.

Case (c): Singularity at an Interior Point

An interior singularity at $c$ splits the problem into two independent Type II integrals, one shaded blue and one shaded orange. Both must converge on their own; a single divergent half kills the whole integral.

Worked Examples

Evaluate $\int_{0}^{1} \frac{d x}{\sqrt{x}}$ .

Solution

The integrand blows up as $x \to 0^{+}$ . Here $a = 0$ , so we integrate from $a + ϵ = ϵ$ :

\begin{aligned} \int_0^1 \frac{dx}{\sqrt{x}} &= \lim_{\epsilon\to 0^+}\int_\epsilon^1 x^{-1/2}\,dx \\ &= \lim_{\epsilon\to 0^+}\bigl[2\sqrt{x}\bigr]_\epsilon^1 \\ &= \lim_{\epsilon\to 0^+}(2 - 2\sqrt{\epsilon}) \\ &= 2. \end{aligned}

The region under $y = 1 / \sqrt{x}$ over $(0, 1]$ is infinitely tall, yet it has finite area $2$ . The curve rises to infinity slowly enough that the extra area near $x = 0$ adds up to something finite.

Evaluate $\int_{0}^{1} \ln x d x$ .

Solution

The singularity is at $a = 0$ , so we integrate from $a + ϵ = ϵ$ up to $1$ and let $ϵ \to 0^{+}$ . Integrate by parts with $u = \ln x$ and $d v = d x$ , so $d u = d x / x$ and $v = x$ :

\begin{aligned} \int_\epsilon^1 \ln x\,dx &= \bigl[x\ln x\bigr]_\epsilon^1 - \int_\epsilon^1 dx \\ &= (0 - \epsilon\ln \epsilon) - (1-\epsilon). \end{aligned}

Now $lim_{ϵ \to 0^{+}} ϵ \ln ϵ = 0$ . (By L'Hôpital's rule, $ϵ \ln ϵ = \frac{\ln ϵ}{1 / ϵ} \to \frac{1 / ϵ}{- 1 / ϵ^{2}} = - ϵ \to 0$ : the factor $ϵ$ shrinks faster than $\ln ϵ$ grows.) Therefore:

\int_{0}^{1} \ln x d x = lim_{ϵ \to 0^{+}} (- ϵ \ln ϵ - 1 + ϵ) = - 1.

The answer is negative because $\ln x < 0$ on $(0, 1)$ , so the region lies below the $x$ -axis and contributes negative signed area.

The shaded region is unbounded downward as $x \to 0^{+}$ , and yet its total signed area is exactly $- 1$ . Being unbounded does not automatically mean divergent.

A Warning: Hidden Discontinuities

The most dangerous situation is an integrand with an infinite discontinuity at a point inside the interval. Such discontinuities are easy to miss, and applying the Fundamental Theorem of Calculus without noticing them can lead to a completely wrong answer.

The reason is that the Fundamental Theorem has a hypothesis, and the hypothesis is exactly what fails. To write $\int_{a}^{b} f (x) d x = F (b) - F (a)$ you need $f$ to be continuous on all of $[a, b]$ and $F$ to be an antiderivative there. If $f$ has a pole partway through, no single $F$ is an antiderivative across the gap, and the subtraction $F (b) - F (a)$ is meaningless.

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

Solution

The integrand $1 / x^{2}$ has an infinite discontinuity at $x = 0$ , which lies inside $[- 1, 1]$ . We must split there:

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

For the right half, the singularity is at the left endpoint $x = 0$ , so we integrate from $0 + ϵ = ϵ$ :

\begin{aligned} \int_\epsilon^1 \frac{dx}{x^2} &= \left[-\frac{1}{x}\right]_\epsilon^1 \\ &= -1 + \frac{1}{\epsilon}. \end{aligned}

As $ϵ \to 0^{+}$ , the term $1 / ϵ \to \infty$ , so $\int_{0}^{1} d x / x^{2}$ diverges. One divergent half is enough: the full integral diverges.

Warning. Applying the Fundamental Theorem naively:

\int_{- 1}^{1} \frac{d x}{x^{2}} \overset{?}{=} {[- \frac{1}{x}]}_{- 1}^{1} = - 1 - 1 = - 2.

This is nonsensical: an integral of the nonnegative function $1 / x^{2}$ cannot be negative. The Fundamental Theorem requires the integrand to be continuous on the entire interval.

Both shaded areas, to the left and to the right of $x = 0$ , grow without bound. The integral $\int_{- 1}^{1} d x / x^{2}$ is improper and diverges, even though the naive antiderivative calculation returns the tidy looking number $- 2$ .

The splitting rule is the one to commit to memory:

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int_a^b f(x)\,dx = \int_a^c f(x)\,dx + \int_c^b f(x)\,dx, \quad \text{both halves must converge}}

Requiring both halves separately is not fussiness. It is what keeps a $+ \infty$ on one side from silently cancelling a $- \infty$ on the other. When such a cancellation does happen and you want to record it, the correct name for the resulting number is the Cauchy principal value, and it is a weaker statement than convergence.

A Checklist Before You Integrate

Checklist before evaluating $\int_{a}^{b} f (x) d x$ . Check that $f$ is continuous at every point of $[a, b]$ , including all interior points. If $f$ has an infinite discontinuity anywhere on $[a, b]$ , the integral is improper and must be handled with limits. Skipping this check can give not just a wrong answer, but an impossible one.

Where are the zeros of every denominator? Any that land in $[a, b]$ are singularities.
Where is the argument of a logarithm equal to zero? There $\ln \to - \infty$ .
Where does a negative power, a root, $\tan$ , $\cot$ , $\sec$ , or $\csc$ blow up inside $[a, b]$ ?
Split at every singularity so that each bad point is an endpoint, never an interior point, of a piece.
Evaluate each piece as a limit, and declare divergence as soon as one piece diverges.
Sanity check the sign: a nonnegative integrand can never produce a negative answer.

Exercises

Evaluate $\int_{0}^{1} \frac{d x}{x}$ , or show it diverges.

Answer

Diverges.

Solution

The integrand $1 / x$ has an infinite discontinuity at $x = 0$ , the left endpoint. Here $a = 0$ , so we integrate from $a + ϵ = ϵ$ . By definition:

\begin{aligned} \int_0^1 \frac{dx}{x} &= \lim_{\epsilon\to 0^+}\int_\epsilon^1 \frac{dx}{x} \\ &= \lim_{\epsilon\to 0^+}\bigl[\ln x\bigr]_\epsilon^1 \\ &= \lim_{\epsilon\to 0^+}(0 - \ln \epsilon) \\ &= \lim_{\epsilon\to 0^+}(-\ln \epsilon). \end{aligned}

As $ϵ \to 0^{+}$ , $\ln ϵ \to - \infty$ , so $- \ln ϵ \to + \infty$ . The integral diverges.

Evaluate the following integral, or show it diverges.

\int_{- 1}^{1} \frac{d x}{\sqrt{1 - x^{2}}}

Answer

$π$

Solution

The integrand $1 / \sqrt{1 - x^{2}}$ is undefined at both endpoints $x = \pm 1$ . We split at $x = 0$ :

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

For the right piece, the singularity is at $b = 1$ , so we integrate up to $b - ϵ = 1 - ϵ$ :

\begin{aligned} \int_0^1 \frac{dx}{\sqrt{1-x^2}} &= \lim_{\epsilon\to 0^+}\int_0^{1-\epsilon} \frac{dx}{\sqrt{1-x^2}} \\ &= \lim_{\epsilon\to 0^+}\bigl[\arcsin x\bigr]_0^{1-\epsilon} \\ &= \lim_{\epsilon\to 0^+}\bigl(\arcsin(1-\epsilon) - 0\bigr) \\ &= \arcsin 1 = \frac{\pi}{2}. \end{aligned}

For the left piece, the singularity is at $a = - 1$ , so we integrate from $a + ϵ = - 1 + ϵ$ :

\begin{aligned} \int_{-1}^0 \frac{dx}{\sqrt{1-x^2}} &= \lim_{\epsilon\to 0^+}\int_{-1+\epsilon}^0 \frac{dx}{\sqrt{1-x^2}} \\ &= \lim_{\epsilon\to 0^+}\bigl[\arcsin x\bigr]_{-1+\epsilon}^0 \\ &= \lim_{\epsilon\to 0^+}\bigl(0 - \arcsin(-1+\epsilon)\bigr) \\ &= -\arcsin(-1) = -\!\left(-\frac{\pi}{2}\right) = \frac{\pi}{2}. \end{aligned}

Both pieces converge, so

\int_{- 1}^{1} \frac{d x}{\sqrt{1 - x^{2}}} = \frac{π}{2} + \frac{π}{2} = π .

Evaluate the following integral, or show it diverges.

\int_{0}^{3} \frac{d x}{x - 1}

Answer

Diverges.

Solution

The integrand $1 / (x - 1)$ has an infinite discontinuity at $x = 1$ , which lies inside $(0, 3)$ . We split there:

\int_{0}^{3} \frac{d x}{x - 1} = \int_{0}^{1} \frac{d x}{x - 1} + \int_{1}^{3} \frac{d x}{x - 1} .

For the left piece, the singularity is at the right endpoint $x = 1$ , so we integrate up to $1 - ϵ$ :

\begin{aligned} \int_0^{1-\epsilon} \frac{dx}{x-1} &= \bigl[\ln\lvert x-1\rvert\bigr]_0^{1-\epsilon} \\ &= \ln\lvert -\epsilon\rvert - \ln\lvert -1\rvert \\ &= \ln\epsilon - 0 \\ &= \ln\epsilon. \end{aligned}

As $ϵ \to 0^{+}$ , $\ln ϵ \to - \infty$ . The left piece diverges, so the entire integral diverges.

Evaluate the following integral, or show it diverges.

\int_{0}^{4} \frac{d x}{\sqrt{4 - x}}

Answer

$4$

Solution

The integrand $(4 - x)^{- 1 / 2}$ is unbounded at $x = 4$ , the right endpoint, so we integrate up to $b - ϵ = 4 - ϵ$ . Let $w = 4 - x$ , so $d w = - d x$ . When $x = 0$ , $w = 4$ ; when $x = 4 - ϵ$ , $w = ϵ$ :

\begin{aligned} \int_0^{4-\epsilon} \frac{dx}{\sqrt{4-x}} &= \int_4^{\epsilon} w^{-1/2}(-dw) \\ &= \int_{\epsilon}^4 w^{-1/2}\,dw \\ &= \bigl[2\sqrt{w}\bigr]_{\epsilon}^4 \\ &= 2\sqrt{4} - 2\sqrt{\epsilon} \\ &= 4 - 2\sqrt{\epsilon}. \end{aligned}

Taking the limit as $ϵ \to 0^{+}$ :

\int_{0}^{4} \frac{d x}{\sqrt{4 - x}} = lim_{ϵ \to 0^{+}} (4 - 2 \sqrt{ϵ}) = 4 - 0 = 4.

Evaluate $\int_{0}^{1} x \ln x d x$ .

Hint: Integrate by parts.

Answer

$- \frac{1}{4}$

Solution

The integrand $x \ln x$ is not defined at $x = 0$ , so we start from $a + ϵ = ϵ$ (with $a = 0$ ) and let $ϵ \to 0^{+}$ .

Integrate by parts. Let $u = \ln x$ and $d v = x d x$ , so $d u = d x / x$ and $v = x^{2} / 2$ :

\begin{aligned} \int_\epsilon^1 x\ln x\,dx &= \left[\frac{x^2\ln x}{2}\right]_\epsilon^1 - \int_\epsilon^1 \frac{x^2}{2}\cdot\frac{dx}{x} \\ &= \left(\frac{1^2\cdot\ln 1}{2} - \frac{\epsilon^2\ln \epsilon}{2}\right) - \frac{1}{2}\int_\epsilon^1 x\,dx \\ &= \left(0 - \frac{\epsilon^2\ln \epsilon}{2}\right) - \frac{1}{2}\left[\frac{x^2}{2}\right]_\epsilon^1 \\ &= -\frac{\epsilon^2\ln \epsilon}{2} - \frac{1}{2}\left(\frac{1}{2}-\frac{\epsilon^2}{2}\right) \\ &= -\frac{\epsilon^2\ln \epsilon}{2} - \frac{1}{4} + \frac{\epsilon^2}{4}. \end{aligned}

Now we take the limit as $ϵ \to 0^{+}$ . The term $ϵ^{2} / 4 \to 0$ and the constant $- 1 / 4$ stays. For the remaining term, apply L'Hôpital's rule:

lim_{ϵ \to 0^{+}} ϵ^{2} \ln ϵ = lim_{ϵ \to 0^{+}} \frac{\ln ϵ}{1 / ϵ^{2}} = lim_{ϵ \to 0^{+}} \frac{1 / ϵ}{- 2 / ϵ^{3}} = lim_{ϵ \to 0^{+}} \frac{- ϵ^{2}}{2} = 0.

Therefore:

\int_{0}^{1} x \ln x d x = lim_{ϵ \to 0^{+}} (- \frac{ϵ^{2} \ln ϵ}{2} - \frac{1}{4} + \frac{ϵ^{2}}{4}) = 0 - \frac{1}{4} + 0 = - \frac{1}{4} .

Determine whether $\int_{0}^{1} \frac{d x}{\sqrt{x} (1 + x)}$ converges. If it does, find its value.

Hint: Let $u = \sqrt{x}$ .

Answer

$\frac{π}{2}$

Solution

The integrand has a singularity at $x = 0$ , so this is a Type II improper integral.

Let $u = \sqrt{x}$ , so $x = u^{2}$ and $d x = 2 u d u$ . When $x = 0$ , $u = 0$ ; when $x = 1$ , $u = 1$ . Also $\sqrt{x} = u$ , so:

\begin{aligned} \int_0^1 \frac{dx}{\sqrt{x}\,(1+x)} &= \int_0^1 \frac{2u\,du}{u\,(1+u^2)} \\ &= 2\int_0^1 \frac{du}{1+u^2}. \end{aligned}

The integrand $1 / (1 + u^{2})$ is continuous on $[0, 1]$ , so this is now a proper Riemann integral. Evaluating:

\begin{aligned} 2\int_0^1 \frac{du}{1+u^2} &= 2\bigl[\arctan u\bigr]_0^1 \\ &= 2\left(\arctan 1 - \arctan 0\right) \\ &= 2\left(\frac{\pi}{4} - 0\right) \\ &= \frac{\pi}{2}. \end{aligned}

The integral converges to $π / 2$ .

Determine whether each integral converges or diverges. For those that converge, identify the type (Type I, Type II, or both) and find the value.

$\int_{0}^{2} \frac{d x}{x^{2} - 1}$
$\int_{- \infty}^{\infty} \frac{d x}{x^{2} + 2 x + 2}$

Answer

(a) Diverges. (b) Converges; value $π$ ; Type I only.

Solution

(a) Write $x^{2} - 1 = (x - 1) (x + 1)$ . The denominator is zero at $x = 1$ , which lies inside $(0, 2)$ . We split there:

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

For the left piece, use partial fractions:

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

The singularity is at the right endpoint $x = 1$ , so we integrate from $0$ to $1 - ϵ$ :

\begin{aligned} \int_0^{1-\epsilon} \frac{dx}{x^2-1} &= \frac{1}{2}\bigl[\ln\lvert x-1\rvert\bigr]_0^{1-\epsilon} - \frac{1}{2}\bigl[\ln\lvert x+1\rvert\bigr]_0^{1-\epsilon} \\ &= \frac{1}{2}(\ln\epsilon - \ln 1) - \frac{1}{2}(\ln(2-\epsilon) - \ln 1) \\ &= \frac{1}{2}\ln\epsilon - \frac{1}{2}\ln(2-\epsilon). \end{aligned}

As $ϵ \to 0^{+}$ , $\ln ϵ \to - \infty$ . Since $\frac{1}{2} \ln (2 - ϵ) \to \frac{1}{2} \ln 2$ (a finite number), the left piece diverges. Therefore the full integral diverges.

(b) Complete the square: $x^{2} + 2 x + 2 = (x + 1)^{2} + 1$ . The denominator $(x + 1)^{2} + 1 \geq 1 > 0$ for all $x$ , so there are no discontinuities. This is a Type I integral only.

Let $u = x + 1$ , $d u = d x$ ; the limits are still $- \infty$ to $\infty$ :

\int_{- \infty}^{\infty} \frac{d x}{(x + 1)^{2} + 1} = \int_{- \infty}^{\infty} \frac{d u}{u^{2} + 1} .

This standard integral is evaluated in Type I: Infinite Limits of Integration, where splitting at $u = 0$ gives $\frac{π}{2} + \frac{π}{2}$ . The answer is $π$ .

One of these converges and the other diverges. Decide which is which, and evaluate the convergent one.

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

Answer

(a) Converges to $3$ . (b) Diverges.

Solution

Both integrands blow up at the left endpoint $x = 0$ , so both are Type II improper integrals and both are handled by integrating from $ϵ$ and letting $ϵ \to 0^{+}$ .

(a) Here the exponent is $p = 2 / 3$ . Since $- 2 / 3 + 1 = 1 / 3$ , the antiderivative of $x^{- 2 / 3}$ is $3 x^{1 / 3}$ :

\begin{aligned} \int_\epsilon^1 x^{-2/3}\,dx &= \bigl[3x^{1/3}\bigr]_\epsilon^1 \\ &= 3 - 3\epsilon^{1/3}. \end{aligned}

As $ϵ \to 0^{+}$ we have $ϵ^{1 / 3} \to 0$ , so the limit is $3$ and the integral converges:

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

(b) Here $p = 3 / 2$ . Since $- 3 / 2 + 1 = - 1 / 2$ , the antiderivative of $x^{- 3 / 2}$ is $- 2 x^{- 1 / 2}$ :

\begin{aligned} \int_\epsilon^1 x^{-3/2}\,dx &= \bigl[-2x^{-1/2}\bigr]_\epsilon^1 \\ &= -2 + \frac{2}{\sqrt{\epsilon}}. \end{aligned}

As $ϵ \to 0^{+}$ , $2 / \sqrt{ϵ} \to + \infty$ , so the integral diverges.

What decided the outcome. The two parts differ only in the exponent, and the antiderivative shows why that is enough. Writing the integrand as $x^{- p}$ with $p \neq 1$ , an antiderivative is $\frac{x^{1 - p}}{1 - p}$ , so

\begin{aligned} \int_\epsilon^1 x^{-p}\,dx &= \frac{1-\epsilon^{1-p}}{1-p}. \end{aligned}

Everything now rides on the sign of the exponent $1 - p$ . In part (a), $p = 2 / 3$ gives $1 - p = 1 / 3 > 0$ , so $ϵ^{1 / 3} \to 0$ and the limit is the finite number $1 / (1 - p) = 3$ . In part (b), $p = 3 / 2$ gives $1 - p = - 1 / 2 < 0$ , so $ϵ^{- 1 / 2} \to + \infty$ and there is no limit. The dividing line is $p = 1$ : a singularity of the form $x^{- p}$ at an endpoint is integrable exactly when $p < 1$ , and when it is, the value $1 / (1 - p)$ can be read straight off the antiderivative.

Evaluate $\int_{0}^{2} \frac{d x}{\sqrt{2 - x}}$ , or show it diverges.

Answer

$2 \sqrt{2}$

Solution

The denominator $\sqrt{2 - x}$ vanishes at $x = 2$ , the right endpoint, and the integrand runs off to $+ \infty$ there. The integrand is continuous on $[0, 2)$ , so this is case (a): integrate up to $b - ϵ = 2 - ϵ$ .

An antiderivative of $(2 - x)^{- 1 / 2}$ is $- 2 \sqrt{2 - x}$ , because differentiating $- 2 (2 - x)^{1 / 2}$ gives $- 2 \cdot \frac{1}{2} (2 - x)^{- 1 / 2} \cdot (- 1) = (2 - x)^{- 1 / 2}$ . Hence:

\begin{aligned} \int_0^{2-\epsilon} \frac{dx}{\sqrt{2-x}} &= \bigl[-2\sqrt{2-x}\bigr]_0^{2-\epsilon} \\ &= -2\sqrt{2-(2-\epsilon)} + 2\sqrt{2-0} \\ &= -2\sqrt{\epsilon} + 2\sqrt{2}. \end{aligned}

Now let $ϵ \to 0^{+}$ . The term $2 \sqrt{ϵ} \to 0$ , so:

\int_{0}^{2} \frac{d x}{\sqrt{2 - x}} = lim_{ϵ \to 0^{+}} (2 \sqrt{2} - 2 \sqrt{ϵ}) = 2 \sqrt{2} \approx 2.828 .

The integral converges. Notice the shape of the integrand: it is an inverse power of the distance to the singular point, namely $(2 - x)^{- 1 / 2}$ , with exponent $1 / 2$ . Because that exponent is smaller than $1$ , the antiderivative $- 2 (2 - x)^{1 / 2}$ still approaches a finite value as $x$ climbs to $2$ , and that is precisely what makes the integral converge.

Determine whether $\int_{- 1}^{1} \frac{d x}{x^{1 / 3}}$ converges, and if so find its value. Treat each half separately; do not appeal to symmetry until you have checked both halves.

Answer

Converges to $0$ . Each half converges: the right half to $\frac{3}{2}$ and the left half to $- \frac{3}{2}$ .

Solution

The integrand $x^{- 1 / 3}$ has an infinite discontinuity at $x = 0$ , an interior point of $[- 1, 1]$ : it tends to $+ \infty$ as $x \to 0^{+}$ and to $- \infty$ as $x \to 0^{-}$ . Here $x^{1 / 3}$ means the real cube root, which is defined and negative for $x < 0$ . By case (c) we must split at $0$ and check both halves:

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

An antiderivative of $x^{- 1 / 3}$ is $\frac{3}{2} x^{2 / 3}$ , and $x^{2 / 3} = {(x^{1 / 3})}^{2} \geq 0$ is defined for negative $x$ as well, so the same antiderivative works on both sides.

Right half. The singularity is at the left endpoint, so integrate from $ϵ$ :

\begin{aligned} \int_\epsilon^1 x^{-1/3}\,dx &= \left[\tfrac{3}{2}x^{2/3}\right]_\epsilon^1 \\ &= \tfrac{3}{2} - \tfrac{3}{2}\epsilon^{2/3} \;\longrightarrow\; \tfrac{3}{2}. \end{aligned}

Left half. The singularity is now at the right endpoint, so integrate up to $- ϵ$ :

\begin{aligned} \int_{-1}^{-\epsilon} x^{-1/3}\,dx &= \left[\tfrac{3}{2}x^{2/3}\right]_{-1}^{-\epsilon} \\ &= \tfrac{3}{2}\epsilon^{2/3} - \tfrac{3}{2} \;\longrightarrow\; -\tfrac{3}{2}. \end{aligned}

Both halves converge, so the original integral converges and

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

Why the caution matters. The integrand is odd and the answer is $0$ , but oddness alone proves nothing. Compare $\int_{- 1}^{1} d x / x$ , whose integrand is also odd about $x = 0$ . There each half diverges, so the whole integral diverges, and no appeal to symmetry can rescue it: the definition requires each half to be finite on its own before the two are added, and $+ \infty$ and $- \infty$ are not allowed to cancel. The difference between the two problems is the exponent. With exponent $1 / 3$ , each half of $\int x^{- 1 / 3} d x$ genuinely converges, so adding them is legitimate and the symmetry is real. With exponent $1$ , neither half of $\int d x / x$ converges, so there is nothing to add.

Evaluate $\int_{0}^{π / 2} \tan x d x$ , or show it diverges.

Answer

Diverges.

Solution

Write $\tan x = \sin x / \cos x$ . On $[0, π / 2)$ the denominator $\cos x$ is positive and continuous, but $\cos x \to 0^{+}$ as $x \to (π / 2)^{-}$ , so $\tan x \to + \infty$ . The singularity is at the right endpoint, so we integrate up to $\frac{π}{2} - ϵ$ .

An antiderivative of $\tan x$ is $- \ln (\cos x)$ , valid wherever $\cos x > 0$ :

\begin{aligned} \int_0^{\pi/2-\epsilon}\tan x\,dx &= \bigl[-\ln(\cos x)\bigr]_0^{\pi/2-\epsilon} \\ &= -\ln\!\left(\cos\!\left(\tfrac{\pi}{2}-\epsilon\right)\right) + \ln(\cos 0) \\ &= -\ln(\sin\epsilon) + \ln 1 \\ &= -\ln(\sin\epsilon). \end{aligned}

Here we used the identity $\cos (\frac{π}{2} - ϵ) = \sin ϵ$ .

As $ϵ \to 0^{+}$ we have $\sin ϵ \to 0^{+}$ , so $\ln (\sin ϵ) \to - \infty$ and therefore $- \ln (\sin ϵ) \to + \infty$ . The limit is infinite, so the integral diverges.

This is unsurprising once you see the shape of the blow-up: near $x = π / 2$ we have $\cos x \approx \frac{π}{2} - x$ , so $\tan x$ behaves like $\frac{1}{\frac{π}{2} - x}$ , the reciprocal of the distance to the bad point. That is the same shape as $1 / x$ near the origin, which the first exercise of this set showed to be divergent.

Show that $\int_{0}^{1} \frac{\ln x}{\sqrt{x}} d x$ converges and find its value.

Answer

$- 4$

Solution

Near $x = 0$ both factors misbehave: $\ln x \to - \infty$ and $1 / \sqrt{x} \to + \infty$ , so the integrand tends to $- \infty$ . The singularity is at the left endpoint $a = 0$ , so we integrate from $ϵ$ to $1$ and let $ϵ \to 0^{+}$ .

Integrate by parts with $u = \ln x$ and $d v = x^{- 1 / 2} d x$ , so $d u = d x / x$ and $v = 2 \sqrt{x}$ :

\begin{aligned} \int_\epsilon^1 \frac{\ln x}{\sqrt{x}}\,dx &= \bigl[2\sqrt{x}\,\ln x\bigr]_\epsilon^1 - \int_\epsilon^1 2\sqrt{x}\cdot\frac{dx}{x} \\ &= \left(2\ln 1 - 2\sqrt{\epsilon}\,\ln\epsilon\right) - 2\int_\epsilon^1 x^{-1/2}\,dx \\ &= -2\sqrt{\epsilon}\,\ln\epsilon - 2\bigl[2\sqrt{x}\bigr]_\epsilon^1 \\ &= -2\sqrt{\epsilon}\,\ln\epsilon - 4 + 4\sqrt{\epsilon}. \end{aligned}

Two of the three terms are easy: $4 \sqrt{ϵ} \to 0$ and the constant $- 4$ stays. For the first term, apply L'Hôpital's rule to the indeterminate product:

lim_{ϵ \to 0^{+}} \sqrt{ϵ} \ln ϵ = lim_{ϵ \to 0^{+}} \frac{\ln ϵ}{ϵ^{- 1 / 2}} = lim_{ϵ \to 0^{+}} \frac{1 / ϵ}{- \frac{1}{2} ϵ^{- 3 / 2}} = lim_{ϵ \to 0^{+}} (- 2 \sqrt{ϵ}) = 0.

Therefore:

\int_{0}^{1} \frac{\ln x}{\sqrt{x}} d x = lim_{ϵ \to 0^{+}} (- 2 \sqrt{ϵ} \ln ϵ - 4 + 4 \sqrt{ϵ}) = 0 - 4 + 0 = - 4.

The value is negative because $\ln x < 0$ throughout $(0, 1)$ , so the whole region lies below the $x$ -axis. The factor $1 / \sqrt{x}$ makes the region deeper than in the earlier example $\int_{0}^{1} \ln x d x = - 1$ , but not so deep that the area becomes infinite.

Evaluate $\int_{1}^{2} \frac{d x}{x \ln x}$ , or show it diverges.

Answer

Diverges.

Solution

The trap here is that nothing in the expression looks undefined until you notice that $\ln 1 = 0$ . So the denominator $x \ln x$ vanishes at $x = 1$ , the left endpoint, and the integrand tends to $+ \infty$ as $x \to 1^{+}$ . This is a Type II integral with a singular left endpoint, so we integrate from $1 + ϵ$ .

Substitute $u = \ln x$ , so $d u = d x / x$ . When $x = 1 + ϵ$ , $u = \ln (1 + ϵ)$ ; when $x = 2$ , $u = \ln 2$ :

\begin{aligned} \int_{1+\epsilon}^2 \frac{dx}{x\ln x} &= \int_{\ln(1+\epsilon)}^{\ln 2} \frac{du}{u} \\ &= \bigl[\ln u\bigr]_{\ln(1+\epsilon)}^{\ln 2} \\ &= \ln(\ln 2) - \ln\!\bigl(\ln(1+\epsilon)\bigr). \end{aligned}

The first term is a finite constant, $\ln (\ln 2) \approx - 0.367$ . For the second, as $ϵ \to 0^{+}$ we have $\ln (1 + ϵ) \to 0^{+}$ , so $\ln (\ln (1 + ϵ)) \to - \infty$ and its negative tends to $+ \infty$ . Hence:

lim_{ϵ \to 0^{+}} \int_{1 + ϵ}^{2} \frac{d x}{x \ln x} = + \infty,

and the integral diverges. The substitution makes the reason plain: in the variable $u$ the integral is $\int_{0}^{\ln 2} d u / u$ , which is the divergent integral of the first exercise of this set with a different upper limit. The substitution moved the singularity to the origin without changing its strength.

A student evaluates

\int_{- 1}^{2} \frac{d x}{x^{4}} \overset{?}{=} {[- \frac{1}{3 x^{3}}]}_{- 1}^{2} = - \frac{1}{24} - \frac{1}{3} = - \frac{3}{8} .

Explain why this computation is invalid, identify the clue that should have raised the alarm, and determine what the integral really does.

Answer

The computation is invalid because $1 / x^{4}$ has an infinite discontinuity at $x = 0$ , which is interior to $[- 1, 2]$ . The clue is the sign: $1 / x^{4} > 0$ , so no correct answer can be negative. The integral diverges.

Solution

Why the computation is invalid. The Fundamental Theorem of Calculus states that $\int_{a}^{b} f (x) d x = F (b) - F (a)$ when $f$ is continuous on $[a, b]$ and F'=f there. The integrand $f (x) = x^{- 4}$ is not even defined at $x = 0$ , and $x = 0$ lies inside $[- 1, 2]$ . So $F (x) = - 1 / (3 x^{3})$ is not an antiderivative on the whole interval, and the subtraction $F (2) - F (- 1)$ has no meaning as an integral.

The clue. Since $x^{4} > 0$ for every $x \neq 0$ , the integrand is strictly positive everywhere it is defined. A positive integrand cannot produce a negative answer, so the value $- 3 / 8$ is impossible on its face. Any time a nonnegative integrand yields a negative number, look for a singularity you stepped over.

What really happens. Split at the singularity:

\int_{- 1}^{2} \frac{d x}{x^{4}} = \int_{- 1}^{0} \frac{d x}{x^{4}} + \int_{0}^{2} \frac{d x}{x^{4}} .

For the right piece the singularity is at the left endpoint, so integrate from $ϵ$ :

\begin{aligned} \int_\epsilon^{2} x^{-4}\,dx &= \left[-\frac{1}{3x^{3}}\right]_\epsilon^{2} \\ &= -\frac{1}{24} + \frac{1}{3\epsilon^{3}}. \end{aligned}

As $ϵ \to 0^{+}$ , the term $1 / (3 ϵ^{3}) \to + \infty$ , so this piece diverges. That alone settles the question: the whole integral diverges. (The left piece diverges too, since $\int_{- 1}^{- ϵ} x^{- 4} d x = \frac{1}{3 ϵ^{3}} - \frac{1}{3} \to + \infty$ .)

The exponent alone predicts the outcome. Integrating $x^{- 4}$ up from $ϵ$ produces a term in $ϵ^{- 3}$ , and any power $x^{- p}$ with $p > 1$ produces a term in $ϵ^{1 - p}$ with a negative exponent, which blows up as $ϵ \to 0^{+}$ . Here $p = 4$ , so divergence was certain before a single antiderivative was written down.

The Definition

When Both Endpoints Are Singular

A Step-by-Step Procedure

The Three Cases, Geometrically

Case (a): Singularity at the Right Endpoint

Case (b): Singularity at the Left Endpoint

Case (c): Singularity at an Interior Point

Worked Examples

A Warning: Hidden Discontinuities

A Checklist Before You Integrate

Exercises

Quick Links

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