Chapter 7 · Integrals

Exercise 7.5 · Q21: $\int\frac{dx}{e^x - 1}$

Exercise 7.5Substitution and partial fractions · Section 7.5

The question

Integrate the function

\frac{1}{e^x - 1}

[Hint: Put $e^x = t$ ⁠.]

Exam answer

Let $I = \displaystyle\int\frac{dx}{e^x - 1}$ ⁠.

Put $e^x = t$ ⁠. Then $e^x\,dx = dt$ ⁠, so $dx = \frac{dt}{t}$ ⁠.

I = \int\frac{dt}{t(t - 1)}

Let $\frac{1}{t(t - 1)} = \frac{A}{t} + \frac{B}{t - 1}$ ⁠. Then $1 = A(t - 1) + Bt$ ⁠.

Putting $t = 0$ ⁠: $A = -1$ ⁠. Putting $t = 1$ ⁠: $B = 1$ ⁠.

\begin{aligned} \therefore\ I &= \int\left(\frac{1}{t - 1} - \frac1t\right)dt \\ &= \log|t - 1| - \log|t| + C \\ &= \log\left|\frac{t - 1}{t}\right| + C \end{aligned}

$\therefore\ I = \log\left|\dfrac{e^x - 1}{e^x}\right| + C$ ⁠.

The answer key prints $\log\left(\frac{e^x - 1}{e^x}\right) + C$ ⁠, without the modulus. This agrees for $x > 0$ ⁠; for $x < 0$ the bracket is negative, so the modulus is needed.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

With $t = e^x$ we have $dt = e^x\,dx = t\,dx$ ⁠, so $dx = \frac{dt}{t}$ ⁠. The integral becomes $\int\frac{dt}{t(t - 1)}$ ⁠, a rational function in $t$ with two different linear factors, which splits by Table 7.2 (row 1).

Given, to find, and the plan

Given: The integrand $\frac{1}{e^x - 1}$ ⁠, defined for $x \ne 0$ (since $e^x = 1$ only at $x = 0$ ⁠).
To find: $\int\frac{dx}{e^x - 1}$ ⁠.
Plan: (1) Put $t = e^x$ ⁠. (2) Split $\frac{1}{t(t - 1)}$ ⁠. (3) Integrate and return to $x$ ⁠. (4) Look at the sign of $e^x - 1$ on each side of 0 and compare with the textbook's form. (5) Differentiate back.

Basics you need

1The exponential

$\frac{d}{dx}e^x = e^x$ (standard formula (xi), Section 7.2); $e^x > 0$ for all $x$ ⁠; $e^x > 1$ exactly when $x > 0$ ⁠; $\log e^x = x$ ⁠. Basics: logarithms and 1/x

2Substitution

If $t = e^x$ then $dt = e^x\,dx$ ⁠. Basics: substitution

3Table 7.2, row 1

$\frac{1}{(t - a)(t - b)} = \frac{A}{t - a} + \frac{B}{t - b}$ ⁠, $a \ne b$ ⁠. Lesson: partial fractions

Solution

Substitute $t = e^x$

Then $dt = e^x\,dx = t\,dx$ ⁠, so $dx = \frac{dt}{t}$ (allowed since $t = e^x > 0$ ⁠). Hence
$\int\frac{dx}{e^x - 1} = \int\frac{dt}{t(t - 1)},$
with $t > 0$ ⁠, $t \ne 1$ ⁠.

There is no $e^x$ in the numerator to absorb $dt$ ⁠, so we solve $dt = t\,dx$ for $dx$ ⁠. This brings an extra factor $t$ into the denominator and gives a rational function of $t$ ⁠.

Integration by substitution, Section 7.3.1. Lesson: substitution
Split into partial fractions
$\begin{gathered} \frac{1}{t(t - 1)} = \frac{A}{t} + \frac{B}{t - 1}, \\ 1 = A(t - 1) + Bt. \end{gathered}$
Put $t = 0$ ⁠: $A = -1$ ⁠. Put $t = 1$ ⁠: $B = 1$ ⁠. So
$\frac{1}{t(t - 1)} = \frac{1}{t - 1} - \frac1t.$
Two different linear factors: row 1 of Table 7.2. Check at $t = 2$ ⁠: $\frac12 = 1 - \frac12$ ⁠. ✓

Table 7.2, row 1. Basics: splitting fractions
Integrate and return to $x$
$\begin{aligned} \int\Big(\frac{1}{t - 1} - \frac1t\Big)dt &= \log|t - 1| - \log|t| + C \\ &= \log\left|\frac{t - 1}{t}\right| + C \\ &= \log\left|\frac{e^x - 1}{e^x}\right| + C. \end{aligned}$
Dividing top and bottom by $e^x$ ⁠, this is also $\log|1 - e^{-x}| + C$ ⁠.

Each simple fraction gives a logarithm (standard formula (xii)); combine with $\log|p| - \log|q| = \log\left|\frac pq\right|$ ⁠; then $t = e^x$ ⁠.

Standard formula (xii), Section 7.2. Basics: logarithms and 1/x
The modulus matters for $x < 0$

For $x > 0$ ⁠: $e^x > 1$ ⁠, so $\frac{e^x - 1}{e^x} > 0$ and the answer may be written $\log\frac{e^x - 1}{e^x} + C$ ⁠, the textbook's form.

For $x < 0$ ⁠: $0 < e^x < 1$ ⁠, so $\frac{e^x - 1}{e^x} < 0$ ⁠. Its logarithm is not defined, and the answer must be written with the modulus: $\log\frac{1 - e^x}{e^x} + C = \log(e^{-x} - 1) + C$ ⁠.

The integrand is defined for $x < 0$ too (for example, at $x = -1$ it is $\frac{1}{e^{-1} - 1} \approx -1.582$ ⁠), so an answer valid only for $x > 0$ is incomplete. The textbook's answer omits the modulus, so it covers only $x > 0$ ⁠; with $|\cdot|$ the same expression covers every $x \ne 0$ ⁠.

Logarithms are defined only for positive numbers. Basics: logarithms and 1/x
Check: differentiate the answer

Write the answer as $\log|e^x - 1| - \log e^x = \log|e^x - 1| - x$ ⁠. Then
$\begin{aligned} \frac{d}{dx}\big(\log|e^x - 1| - x\big) &= \frac{e^x}{e^x - 1} - 1 \\ &= \frac{e^x - (e^x - 1)}{e^x - 1} = \frac{1}{e^x - 1}. \ \checkmark \end{aligned}$
Valid for every $x \ne 0$ ⁠, on both sides of 0.

Chain rule on $\log|e^x - 1|$ (inner derivative $e^x$ ⁠); $\log e^x = x$ ⁠. Over the common denominator the $e^x$ terms cancel.

Property (I), Section 7.2.1. Basics: integration undoes differentiation

Result

\int\frac{dx}{e^x - 1} = \log\left|\frac{e^x - 1}{e^x}\right| + C = \log|1 - e^{-x}| + C,

for $x \ne 0$ ⁠. (Textbook answer: $\log\left(\frac{e^x - 1}{e^x}\right) + C$ ⁠. This agrees with ours for $x > 0$ ⁠; for $x < 0$ the bracket is negative, so the modulus is needed, as shown in Step 4.)

Check it with numbers

From 1 to 2 (continuous there): $F(2) - F(1) = \log\frac{1 - e^{-2}}{1 - e^{-1}}$ $= \log(1 + e^{-1}) \approx 0.3132617$ ⁠. A numerical integration over $[1, 2]$ gives $0.3132617$ too. ✓

From $-2$ to $-1$ ⁠, where the modulus is needed: $F(-1) - F(-2)$ $= \log(e - 1) - \log(e^2 - 1)$ $= -\log(1 + e) \approx -1.3132617$ ⁠. A numerical integration over $[-2, -1]$ gives $-1.3132617$ too. ✓

Common mistakes

Writing $dt = dx$ ⁠. With $t = e^x$ ⁠, $dt = e^x dx$ ⁠, so $dx = \frac{dt}{t}$ ⁠.
Answer $\log|e^x - 1|$ ⁠. The numerator is 1, not $e^x$ (the derivative of the denominator), so this is wrong: its derivative is $\frac{e^x}{e^x - 1}$ ⁠.
Dropping the modulus everywhere. $e^x - 1 < 0$ for $x < 0$ ⁠.

What to take away

For integrands in $e^x$ ⁠, $t = e^x$ with $dx = \frac{dt}{t}$ gives a rational function of $t$ ⁠.
A second route: multiply by $\frac{e^{-x}}{e^{-x}}$ to get $\frac{e^{-x}}{1 - e^{-x}}$ ⁠, whose numerator is the derivative of $1 - e^{-x}$ ⁠; this gives $\log|1 - e^{-x}|$ at once.
Keep the modulus unless the quantity is known to be positive on the whole domain.

Every fact used, and where it comes from

Fact	Used in	Source
$\frac{d}{dx}e^x = e^x$ ⁠; substitution $t = e^x$	Step 1	Standard formula (xi), Section 7.2; Section 7.3.1
$\frac{A}{t - a} + \frac{B}{t - b}$	Step 2	Table 7.2, row 1, Section 7.5
$\int\frac{dt}{t} = \log\|t\| + C$	Step 3	Standard formula (xii), Section 7.2
$e^x > 1 \iff x > 0$ ⁠; $\log$ needs a positive argument	Step 4	Exponential and logarithmic functions (Basics: logarithms)
$\log e^x = x$	Step 5	Laws of logarithms
Derivative of an antiderivative is the integrand	Step 5	Property (I), Section 7.2.1

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Exercise 7.5 · Q21: ∫dxex−1\int\frac{dx}{e^x - 1}∫ex−1dx​

Basics you need

Solution

Substitute t=ext = e^xt=ex

Split into partial fractions

Integrate and return to xxx

The modulus matters for x<0x < 0x<0

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.5 · Q21: $\int\frac{dx}{e^x - 1}$

Substitute $t = e^x$

Integrate and return to $x$

The modulus matters for $x < 0$