Integrate an infinite expansion, integrals from easy to advanced lesson

The integrand $(e^x - 1)/x$ on $[0, 1]$ reads $0/0$ at the left endpoint, and no antiderivative built from elementary functions exists, so the fundamental theorem has no formula to work with. Expanding $e^x$ as a power series turns the quotient into a sum of plain powers of $x$ , and each power is easy to integrate. This page shows why the endpoint is harmless, when an infinite sum may be integrated one term at a time, and how a bound on the omitted terms certifies the value $1.317902$ to six decimal places.

Key terms

Removable singularity

Definition Let $f$ be defined at every point of an interval $D$ containing $c$ , except possibly at $c$ itself. The point $c$ is a removable singularity of $f$ if the limit $L = \lim_{x \to c} f(x)$ , taken over the points $x \ne c$ of $D$ , exists as a finite number, but $f$ is undefined at $c$ or $f(c) \ne L$ . Setting $f(c) = L$ gives the continuous extension of $f$ , which is continuous at $c$ . When $c$ is an endpoint of $D$ the limit is one-sided. For integrals this has a direct consequence. If $f$ is continuous on $(a, b]$ and $\lim_{x \to a^+} f(x) = L$ is finite, the extension is continuous on $[a, b]$ , its integral over $[a, b]$ is an ordinary integral, and that integral equals

\lim_{\varepsilon \to 0^+} \int_{a + \varepsilon}^{b} f(x)\,dx .

In plain words The formula fails at one point, or gives the wrong value there, but near that point the graph has only a hole at a finite height. Filling the hole with the limiting value makes the function continuous on the whole closed interval. Changing a value at a single point does not change an integral, so filling the hole adds no area. What it does show is that the integrand is continuous, and therefore bounded, on the closed interval, so the integral is an ordinary one.

Example Because $e^0 = 1$ , the quotient $(e^x - 1)/x$ is the difference quotient of $e^x$ at $0$ . Its limit is the derivative of $e^x$ at $0$ , so

\lim_{x \to 0} \frac{e^x - 1}{x} = e^0 = 1 .

The continuous extension $f$ has $f(0) = 1$ and is continuous on $[0, 1]$ . By contrast, $1/x$ has no finite limit as $x \to 0^+$ , so $0$ is not a removable singularity of $1/x$ . For $0 < \varepsilon < 1$ ,

\int_{\varepsilon}^{1} \frac{dx}{x} = -\ln \varepsilon ,

which grows without bound as $\varepsilon \to 0^+$ , so the integral of $1/x$ over $[0, 1]$ diverges.

Power series

Definition A power series centered at $x_0$ is a series

\sum_{n=0}^{\infty} c_n (x - x_0)^n

with real coefficients $c_n$ , read as a function of $x$ , with the convention $(x - x_0)^0 = 1$ even at $x = x_0$ . Every power series has a radius of convergence $R$ , which is $0$ , a positive number or $\infty$ , such that the series converges absolutely when $\lvert x - x_0 \rvert < R$ and diverges when $\lvert x - x_0 \rvert > R$ . At the endpoints $x = x_0 \pm R$ of a finite positive radius the series may converge or diverge, and each endpoint needs its own check. When $R > 0$ , the sum is a continuous function, and even a differentiable one, on the open interval $(x_0 - R, x_0 + R)$ .

In plain words A power series is like a polynomial with infinitely many terms. Inside its interval of convergence it defines a well-behaved function. Outside that interval the series diverges and defines no value at all, even if some closed formula for its sum still makes sense there.

Example The geometric series has center $0$ , every coefficient equal to $1$ and radius $R = 1$ . For $\lvert x \rvert < 1$ its sum is

\sum_{n=0}^{\infty} x^n = \frac{1}{1 - x} .

At $x = 1$ and at $x = -1$ its terms do not tend to $0$ , so it diverges at both endpoints. The formula $1/(1 - x)$ gives $-1$ at $x = 2$ , but the series $1 + 2 + 4 + \cdots$ diverges there, so the identity holds only for $\lvert x \rvert < 1$ .

Exponential series

Definition For a whole number $n \ge 1$ the factorial $n! = 1 \cdot 2 \cdots n$ is the product of the positive integers up to $n$ , and $0! = 1$ . The exponential series is the power series

e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} ,

valid for every real $x$ . Its radius of convergence is infinite, because for $x \ne 0$ the ratio of the absolute values of consecutive terms is $\lvert x \rvert / (n + 1)$ , which tends to $0$ . Taylor’s theorem shows that the sum is $e^x$ .

In plain words Every power of $x$ appears, each divided by a factorial. The factorials grow so fast that for every fixed $x$ the terms tend to zero quickly enough for the series to converge, on the whole real line.

Example Subtract $1$ to remove the constant term, then divide by $x \ne 0$ . Each remaining power drops by one, so

\begin{aligned} \frac{e^x - 1}{x} &= \sum_{n=1}^{\infty} \frac{x^{n-1}}{n!} \\[0.4em] &= 1 + \frac{x}{2} + \frac{x^2}{6} + \cdots \end{aligned}

The right side is again a power series with infinite radius. At $x = 0$ only its first term survives, and its value $1$ matches the continuous extension, so the series equals the extended integrand $f$ at every real $x$ . Since a power series is continuous inside its interval of convergence, this gives a second proof that the limit at $0$ is $1$ . All the coefficients are positive, so $f$ increases on $[0, 1]$ , from $f(0) = 1$ to $f(1) = e - 1 \approx 1.718$ .

Uniform convergence

Definition A series $\sum_{n \ge 1} g_n$ of functions on a set $E$ , with partial sums $S_N = g_1 + \cdots + g_N$ , converges uniformly on $E$ to a function $S$ if

\sup_{x \in E} \, \lvert S(x) - S_N(x) \rvert \to 0

as $N \to \infty$ . Equivalently, for every $\varepsilon > 0$ there is one $N_0$ such that $\lvert S(x) - S_N(x) \rvert < \varepsilon$ for every $N \ge N_0$ and every $x$ in $E$ at once. The Weierstrass M-test gives a practical way to prove it. If $\lvert g_n(x) \rvert \le M_n$ for every $n$ and every $x$ in $E$ , and the series of numbers $\sum M_n$ converges, then $\sum g_n$ converges uniformly and absolutely on $E$ , because at every $x$ the error after $N$ terms is at most $\sum_{n > N} M_n$ . For a power series with radius $R > 0$ and any $r$ with $0 < r < R$ , the series converges absolutely at $x = x_0 + r$ , so the M-test with $M_n = \lvert c_n \rvert r^n$ gives uniform convergence on $[x_0 - r, x_0 + r]$ .

In plain words Pointwise convergence allows every $x$ its own number of terms. Uniform convergence asks for one number of terms that pushes the error below a chosen tolerance everywhere on the set at the same time. Draw a band of height $\varepsilon$ above and below the graph of $S$ . From some $N$ on, the graph of every partial sum lies inside that band.

Example On $[0, 1]$ the terms of the series for $(e^x - 1)/x$ satisfy

\left\lvert \frac{x^{n-1}}{n!} \right\rvert \le \frac{1}{n!} ,

because $\lvert x^{n-1} \rvert \le 1$ there. The series $\sum_{n \ge 1} 1/n!$ converges to $e - 1$ , so the M-test gives uniform convergence on $[0, 1]$ . After four terms the error is below $0.01$ at every point of the interval. The geometric series shows that uniform convergence can fail on the whole open interval of convergence. After the terms through $x^N$ its error is

\begin{aligned} &\frac{1}{1 - x} - \sum_{n=0}^{N} x^n \\[0.4em] &\quad = \frac{x^{N+1}}{1 - x} , \end{aligned}

which grows without bound as $x \to 1^-$ , whatever $N$ is. On $[-r, r]$ with $0 < r < 1$ the error is at most $r^{N+1}/(1 - r)$ , which tends to $0$ , so there the convergence is uniform.

Term-by-term integration

Definition Let $g_1, g_2, \dots$ be integrable on $[a, b]$ , and suppose $\sum g_n$ converges uniformly on $[a, b]$ to $S$ . Then $S$ is integrable on $[a, b]$ and

\begin{aligned} &\int_a^b S(x)\,dx \\[0.4em] &\quad = \sum_{n=1}^{\infty} \int_a^b g_n(x)\,dx . \end{aligned}

The reason is that the integral of the remainder is at most a width times a height in absolute value,

\begin{aligned} &\left\lvert \int_a^b \bigl(S - S_N\bigr)\,dx \right\rvert \\[0.4em] &\quad \le (b - a) \sup_{[a, b]} \lvert S - S_N \rvert , \end{aligned}

and uniform convergence sends the height to $0$ . A power series with radius $R > 0$ converges uniformly on every closed interval inside $(x_0 - R, x_0 + R)$ , so it may be integrated term by term over any such interval.

In plain words Swapping the sum and the integral exchanges two limits, so it needs a reason. Uniform convergence supplies one, since the leftover part of the sum is uniformly small, and a uniformly small function over an interval of finite length has a small integral. Convergence at each point alone is not enough. The functions equal to $N$ on $(0, 1/N)$ and to $0$ elsewhere tend to $0$ at every point of $[0, 1]$ , yet each of them has integral $1$ over $[0, 1]$ .

Example The series for the extended integrand converges uniformly on $[0, 1]$ and its terms are continuous, so

\begin{aligned} I &= \int_0^1 \frac{e^x - 1}{x}\,dx \\[0.4em] &= \sum_{n=1}^{\infty} \frac{1}{n!} \int_0^1 x^{n-1}\,dx . \end{aligned}

Here the integrand takes the value $1$ at $x = 0$ , and the improper reading as a limit of integrals over $[\varepsilon, 1]$ gives the same number. Since the integral of $x^{n-1}$ over $[0, 1]$ is $1/n$ for every $n \ge 1$ ,

\begin{aligned} I &= \sum_{n=1}^{\infty} \frac{1}{n\,n!} \\[0.4em] &= 1 + \frac{1}{4} + \frac{1}{18} + \cdots \end{aligned}

The next terms are $1/96$ and $1/600$ . The constant term of the integrand contributes area $1$ and every other term adds positive area, so $I > 1$ , and since the integrand rises from $1$ to $e - 1$ , also $I < e - 1$ . The series is itself an exact answer. In terms of special functions it equals $\operatorname{Ei}(1) - \gamma$ , where $\operatorname{Ei}$ is the exponential integral and $\gamma \approx 0.5772$ is Euler’s constant.

Geometric tail bound

Definition Let $\sum_{n \ge 1} a_n$ be a series of positive terms, fix a whole number $N \ge 0$ , and write $S_N = a_1 + \cdots + a_N$ for the partial sum and $T_N = \sum_{n > N} a_n$ for the tail. Suppose there is a number $q$ with $0 < q < 1$ such that $a_{n+1} \le q\,a_n$ for every $n \ge N + 1$ . Then $a_{N+1+k} \le q^k a_{N+1}$ for every $k \ge 0$ , and comparison with a geometric series shows that the series converges and

0 < T_N \le \frac{a_{N+1}}{1 - q} .

So the sum $S$ satisfies $S_N < S \le S_N + a_{N+1}/(1 - q)$ , and the upper inequality is strict when $a_{n+1} < q\,a_n$ for every $n \ge N + 1$ .

In plain words Once every term is at most a fixed fraction $q$ of the one before it, the omitted terms add up to no more than a geometric series that starts with the first omitted term. That geometric series has a simple sum, so the whole infinite remainder is bounded by a number you can compute. With positive terms every partial sum is an underestimate, and the bound says by how much at most.

Example For $a_n = 1/(n\,n!)$ the ratio of consecutive terms is

\frac{a_{n+1}}{a_n} = \frac{n}{(n+1)^2} < \frac{1}{n + 1} .

For $n \ge N + 1$ this is less than $1/(N + 2)$ , so $q = 1/(N + 2)$ works, with strict inequality, and

0 < I - S_N < \frac{N + 2}{N + 1}\, a_{N+1} .

$N$	$S_N$ , rounded	Bound on $I - S_N$ , rounded up
$4$	$1.315972222$	$0.002$
$6$	$1.317870370$	$3.3 \times 10^{-5}$
$8$	$1.317901815$	$3.5 \times 10^{-7}$
$10$	$1.317902149$	$2.5 \times 10^{-9}$

With $N = 8$ the bound gives $1.3179018 < I < 1.3179022$ , and every number in that interval rounds to $1.317902$ , so $I \approx 1.317902$ to six decimal places. With $N = 7$ the bound is about $3.5 \times 10^{-6}$ , more than one unit in the sixth decimal, so the interval it gives cannot fix that digit. Ten terms give $1.317902148 < I < 1.317902152$ , which fixes eight decimals, $I \approx 1.31790215$ .

Check yourself

Use the geometric series to write $\ln 2$ as a series, starting from the identity

\ln 2 = \int_0^{1/2} \frac{dx}{1 - x} .

Why is term-by-term integration allowed on this interval?

Show the answer

The starting identity holds because $-\ln(1 - x)$ is an antiderivative of $1/(1 - x)$ for $x < 1$ , and $-\ln \tfrac12 + \ln 1 = \ln 2$ . For $\lvert x \rvert < 1$ the integrand equals $\sum_{n \ge 0} x^n$ . The interval $[0, \tfrac12]$ lies inside the interval of convergence $(-1, 1)$ . On it $\lvert x^n \rvert \le (\tfrac12)^n$ , and $\sum_{n \ge 0} (\tfrac12)^n = 2$ converges, so the M-test gives uniform convergence of a series of continuous terms. Integrating term by term,

\int_0^{1/2} x^n\,dx = \frac{1}{(n + 1)\,2^{n+1}} ,

and with $k = n + 1$ ,

\begin{aligned} \ln 2 &= \sum_{k=1}^{\infty} \frac{1}{k\,2^k} \\[0.4em] &= \frac{1}{2} + \frac{1}{8} + \frac{1}{24} + \cdots \end{aligned}

The first four terms add up to about $0.6823$ , a little below $\ln 2 \approx 0.6931$ , as expected for a series of positive terms.

A classmate evaluates the integral of $e^x/x$ over $[0, 1]$ by writing $e^x/x = \sum_{n \ge 0} x^{n-1}/n!$ and integrating term by term. What goes wrong, and does the integral have a value?

Show the answer

The term with $n = 0$ is $x^{-1}$ , so the right side is not a power series, and the singularity at $0$ is not removable, since $e^x/x \to \infty$ as $x \to 0^+$ . Split the integrand as $1/x + (e^x - 1)/x$ for $x \ne 0$ . For $0 < \varepsilon < 1$ ,

\begin{aligned} \int_{\varepsilon}^{1} \frac{e^x}{x}\,dx &= -\ln \varepsilon \\[0.4em] &\quad + \int_{\varepsilon}^{1} \frac{e^x - 1}{x}\,dx . \end{aligned}

As $\varepsilon \to 0^+$ the last integral tends to $I \approx 1.3179$ , while $-\ln \varepsilon$ grows without bound. So the improper integral diverges and has no value. Term-by-term integration would start with the integral of $1/x$ over $[0, 1]$ , which already diverges. The method works only after the $1/x$ part is removed, and subtracting $1$ in $e^x - 1$ does exactly that.

Find the value of the integral below to four decimal places with a certified error bound, and then check it against $I$ with integration by parts.

J = \int_0^1 \frac{e^x - 1 - x}{x^2}\,dx

Show the answer

At $x = 0$ the quotient reads $0/0$ . For $x \ne 0$ , subtracting $1 + x$ from the exponential series and dividing by $x^2$ gives

\begin{aligned} \frac{e^x - 1 - x}{x^2} &= \sum_{n=2}^{\infty} \frac{x^{n-2}}{n!} \\[0.4em] &= \frac{1}{2} + \frac{x}{6} + \frac{x^2}{24} + \cdots \end{aligned}

This is a power series with infinite radius, so its sum is continuous and the limit at $0$ is its value there, $\tfrac12$ . The singularity is removable, and the integral uses the extension with value $\tfrac12$ at $0$ . On $[0, 1]$ each term is at most $1/n!$ in absolute value and $\sum_{n \ge 2} 1/n! = e - 2$ converges, so the convergence is uniform. Since the integral of $x^{n-2}$ over $[0, 1]$ is $1/(n - 1)$ for $n \ge 2$ ,

\begin{aligned} J &= \sum_{n=2}^{\infty} \frac{1}{(n - 1)\,n!} \\[0.4em] &= \frac{1}{2} + \frac{1}{12} + \frac{1}{72} + \cdots \end{aligned}

With $b_n = 1/((n - 1)\,n!)$ the ratio of consecutive terms is

\frac{b_{n+1}}{b_n} = \frac{n - 1}{n(n + 1)} < \frac{1}{n + 1} .

Keep the terms $n = 2$ to $6$ , whose sum is $\tfrac12 + \tfrac{1}{12} + \tfrac{1}{72} + \tfrac{1}{480} + \tfrac{1}{3600} = \tfrac{1439}{2400} \approx 0.5995833$ . For $n \ge 7$ the ratio is below $\tfrac18$ , so the tail is less than $b_7/(1 - \tfrac18) = \tfrac87 \cdot \tfrac{1}{30240} = \tfrac{1}{26460} \approx 0.0000378$ . Hence $0.5995833 < J < 0.5996212$ , and both ends round to $0.5996$ , so $J \approx 0.5996$ to four decimal places.

For the check, integrate by parts on $[\varepsilon, 1]$ with $u = e^x - 1 - x$ and $dv = dx/x^2$ , so $du = (e^x - 1)\,dx$ and $v = -1/x$ . The boundary term is $-(e^x - 1 - x)/x$ evaluated from $\varepsilon$ to $1$ , so

\begin{aligned} &\int_{\varepsilon}^{1} \frac{e^x - 1 - x}{x^2}\,dx \\[0.4em] &\quad = -(e - 2) + \frac{e^{\varepsilon} - 1 - \varepsilon}{\varepsilon} \\[0.4em] &\qquad + \int_{\varepsilon}^{1} \frac{e^x - 1}{x}\,dx . \end{aligned}

The middle term equals $\varepsilon$ times a quotient that tends to $\tfrac12$ , so it tends to $0$ as $\varepsilon \to 0^+$ . Therefore $J = I - (e - 2)$ . With $I \approx 1.31790215$ and $e - 2 \approx 0.71828183$ this gives $J \approx 0.59962032$ , which lies inside the certified interval.