Riemann Integration

6.1 Definition

Let $f : [a,b] \to \mathbb{R}$ be bounded. A partition of $[a,b]$ is a finite set $P = \{x_0, x_1, \ldots, x_n\}$ with $a = x_0 \lt x_1 \lt \cdots \lt x_n = b$ .

The upper sum and lower sum of $f$ with respect to $P$ are:

$U(f, P) = \sum_{i=1}^{n} M_i \Delta x_i, \quad L(f, P) = \sum_{i=1}^{n} m_i \Delta x_i$

Where $M_i = \sup\{f(x) : x \in [x_{i-1}, x_i]\}$ , $m_i = \inf\{f(x) : x \in [x_{i-1}, x_i]\}$ And $\Delta x_i = x_i - x_{i-1}$ .

The mesh of $P$ is $\|P\| = \max_{1 \leq i \leq n} \Delta x_i$ .

Definition. $f$ is Riemann integrable on $[a,b]$ if the upper and lower integrals are equal:

$\overline{\int_a^b} f(x)\, dx = \underline{\int_a^b} f(x)\, dx$

Where $\overline{\int_a^b} f = \inf\{U(f,P) : P \mathrm{\ is\ a\ partition}\}$ and $\underline{\int_a^b} f = \sup\{L(f,P) : P \mathrm{\ is\ a\ partition}\}$ .

The common value is denoted $\int_a^b f(x)\, dx$ .

6.2 Integrability Criteria

Theorem 6.1 (Riemann Integrability Criterion). A bounded function $f : [a,b] \to \mathbb{R}$ is Riemann integrable if and only if for every $\varepsilon > 0$ There exists a partition $P$ such that

$U(f,P) - L(f,P) \lt \varepsilon$

Theorem 6.2. Every continuous function on $[a,b]$ is Riemann integrable.

Proof. Let $f$ be continuous on $[a,b]$ . By the Heine-Cantor theorem, $f$ is uniformly continuous. Given $\varepsilon > 0$ Choose $\delta > 0$ such that $|x - y| \lt \delta$ implies $|f(x) - f(y)| \lt \varepsilon/(b-a)$ .

Let $P$ be any partition with $\|P\| \lt \delta$ . On each subinterval $[x_{i-1}, x_i]$ By the Extreme Value Theorem, $f$ attains its maximum $M_i$ and minimum $m_i$ . By uniform continuity: $M_i - m_i \lt \varepsilon/(b-a)$ . Therefore:

$U(f,P) - L(f,P) = \sum_{i=1}^{n}(M_i - m_i)\Delta x_i \lt \frac{\varepsilon}{b-a} \sum_{i=1}^{n} \Delta x_i = \varepsilon$

By the Riemann integrability criterion, $f$ is integrable. $\blacksquare$

Theorem 6.3. Every monotone function on $[a,b]$ is Riemann integrable.

Proof. Assume $f$ is increasing (the decreasing case is analogous). Given $\varepsilon > 0$ Let $P_n$ be the uniform partition with $n$ subintervals of length $(b-a)/n$ . On $[x_{i-1}, x_i]$ : $M_i = f(x_i)$ and $m_i = f(x_{i-1})$ . Then:

$U(f, P_n) - L(f, P_n) = \sum_{i=1}^{n} [f(x_i) - f(x_{i-1})] \cdot \frac{b-a}{n} = [f(b) - f(a)] \cdot \frac{b-a}{n}$

Choose $n$ large enough that $[f(b) - f(a)](b-a)/n \lt \varepsilon$ . $\blacksquare$

Theorem 6.4. A bounded function with finitely many discontinuities on $[a,b]$ is Riemann integrable.

Proof (sketch). Let $f$ have discontinuities at $d_1, \ldots, d_m \in [a,b]$ . Given $\varepsilon > 0$ Enclose each $d_j$ in a small interval $I_j$ of total length $\varepsilon/(2M)$ Where $M = \sup_{[a,b]} |f|$ . On the remaining set (a finite union of closed intervals), $f$ is continuous, Hence uniformly continuous. Choose a partition fine enough that the oscillation of $f$ on each Subinterval outside the $I_j$ is less than $\varepsilon/(2(b-a))$ . Then:

$U(f, P) - L(f, P) \leq \frac{\varepsilon}{2(b-a)} \cdot (b - a) + 2M \cdot \frac{\varepsilon}{2M} = \varepsilon$

$\blacksquare$

Proposition 6.4a. The set of Riemann integrable functions on $[a,b]$ forms a vector space, and If $f$ and $g$ are integrable, then so are $|f|$ , $f^2$ And $\max(f, g)$ .

Theorem 6.4b (Lebesgue”s Criterion for Riemann Integrability). A bounded function $f : [a,b] \to \mathbb{R}$ Is Riemann integrable if and only if the set of its discontinuities has (Lebesgue) measure zero.

Remark. A set has measure zero if it can be covered by countably many intervals of arbitrarily Small total length. In particular, every countable set has measure zero. This means:

Every continuous function is integrable (empty set of discontinuities).
Every function with countably many discontinuities is integrable (Theorem 6.4 is a special case).
The Dirichlet function $f(x) = 1$ for $x \in \mathbb{Q}$ and $f(x) = 0$ for $x \notin \mathbb{Q}$ is discontinuous everywhere (set of discontinuities = $[a,b]$ Measure $> 0$ ), hence not integrable.
Thomae’s function $f(x) = 1/q$ if $x = p/q$ in lowest terms, and $f(x) = 0$ if $x$ is irrational, is continuous at every irrational and discontinuous at every rational. Since $\mathbb{Q}$ is countable (measure zero), Thomae’s function is Riemann integrable, with $\int_0^1 f = 0$ .

6.3 Properties of the Integral

Theorem 6.5 (Linearity). If $f$ and $g$ are integrable on $[a,b]$ and $\alpha, \beta \in \mathbb{R}$ :

$\int_a^b (\alpha f + \beta g) = \alpha \int_a^b f + \beta \int_a^b g$

Theorem 6.6 (Monotonicity). If $f(x) \leq g(x)$ for all $x \in [a,b]$ Then $\int_a^b f \leq \int_a^b g$ .

Theorem 6.7 (Triangle Inequality). $\left|\int_a^b f\right| \leq \int_a^b |f|$ .

6.4 The Fundamental Theorem of Calculus

Theorem 6.8 (FTC Part 1). If $f$ is continuous on $[a,b]$ Then the function

$F(x) = \int_a^x f(t)\, dt$

Is differentiable on $(a,b)$ and $F'(x) = f(x)$ .

Proof. Let $h > 0$ (the case $h \lt 0$ is similar). By the Mean Value Theorem for Integrals (which follows from the EVT), there exists $\xi \in [x, x+h]$ such that

$\frac{F(x+h) - F(x)}{h} = \frac{1}{h}\int_x^{x+h} f(t)\, dt = f(\xi)$

As $h \to 0^+$ We have $\xi \to x^+$ (since $\xi \in [x, x+h]$ ). By continuity of $f$ $f(\xi) \to f(x)$ . Hence $F'_+(x) = f(x)$ . A similar argument gives $F'_-(x) = f(x)$ . $\blacksquare$

Theorem 6.9 (FTC Part 2). If $F$ is differentiable on $[a,b]$ with $F' = f$ (and $f$ is integrable), Then

$\int_a^b f(x)\, dx = F(b) - F(a)$

Proof. Let $P = \{x_0, \ldots, x_n\}$ be any partition of $[a,b]$ . By the Mean Value Theorem, For each $i$ there exists $\xi_i \in [x_{i-1}, x_i]$ with $F(x_i) - F(x_{i-1}) = f(\xi_i)\Delta x_i$ . Summing:

$F(b) - F(a) = \sum_{i=1}^{n} [F(x_i) - F(x_{i-1})] = \sum_{i=1}^{n} f(\xi_i) \Delta x_i$

The right-hand side is a Riemann sum for $\int_a^b f$ . As $\|P\| \to 0$ This converges to the Integral. Hence $F(b) - F(a) = \int_a^b f(x)\, dx$ . $\blacksquare$

6.5 Worked Examples

Problem. Compute $\int_0^1 x^2\, dx$ from the definition.

Solution. Let $P_n = \{0, 1/n, 2/n, \ldots, 1\}$ . On $[x_{i-1}, x_i] = [(i-1)/n, i/n]$ , $f(x) = x^2$ Has $M_i = (i/n)^2$ and $m_i = ((i-1)/n)^2$ .

$U(f, P_n) = \sum_{i=1}^{n} \frac{i^2}{n^2} \cdot \frac{1}{n} = \frac{1}{n^3} \sum_{i=1}^{n} i^2 = \frac{1}{n^3} \cdot \frac{n(n+1)(2n+1)}{6}$

As $n \to \infty$ : $\lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \frac{2}{6} = \frac{1}{3}$ .

Similarly, $L(f, P_n) \to 1/3$ . So $\int_0^1 x^2\, dx = 1/3$ . $\blacksquare$

Worked Example: Compute $\int_0^1 \sqrt{x}\, dx$ from the definition

Solution. Let $P_n = \{0, 1/n, 2/n, \ldots, 1\}$ . On $[(i-1)/n, i/n]$ , $f(x) = \sqrt{x}$ has $M_i = \sqrt{i/n}$ and $m_i = \sqrt{(i-1)/n}$ .

$U(f, P_n) = \sum_{i=1}^{n} \sqrt{\frac{i}{n}} \cdot \frac{1}{n} = \frac{1}{n^{3/2}} \sum_{i=1}^{n} \sqrt{i}$

Using $\sum_{i=1}^{n} \sqrt{i} = \frac{2}{3} n^{3/2} + O(n^{1/2})$ (obtained from comparing with $\int_0^n \sqrt{x}\, dx$ ):

$\lim_{n \to \infty} U(f, P_n) = \lim_{n \to \infty} \frac{1}{n^{3/2}} \cdot \frac{2}{3}n^{3/2} = \frac{2}{3}$

Similarly $L(f, P_n) \to 2/3$ Confirming $\int_0^1 \sqrt{x}\, dx = 2/3$ . $\blacksquare$

6.6 Improper Integrals

Definition. An improper integral is a Riemann integral where either the interval of integration Is unbounded or the integrand is unbounded.

Type I (Infinite Intervals). If $f$ is Riemann integrable on $[a, b]$ for every $b > a$ Define:

$\int_a^{\infty} f(x)\, dx = \lim_{b \to \infty} \int_a^b f(x)\, dx$

The integral converges if this limit exists as a finite number; otherwise it diverges.

Type II (Unbounded Integrands). If $f$ is unbounded near $a$ but integrable on $[c, b]$ for every $c \in (a, b]$ :

$\int_a^b f(x)\, dx = \lim_{c \to a^+} \int_c^b f(x)\, dx$

Theorem 6.10 (Comparison Test for Improper Integrals). If $0 \leq f(x) \leq g(x)$ for $x \geq a$ :

If $\int_a^{\infty} g$ converges, then $\int_a^{\infty} f$ converges.
If $\int_a^{\infty} f$ diverges, then $\int_a^{\infty} g$ diverges.

Theorem 6.11 (Absolute Convergence). If $\int_a^{\infty} |f(x)|\, dx$ converges, then $\int_a^{\infty} f(x)\, dx$ converges.

Theorem 6.12 ( $p$ -Test for Improper Integrals).

Type I: $\int_1^{\infty} \frac{1}{x^p}\, dx$ converges if and only if $p > 1$ .
Type II: $\int_0^1 \frac{1}{x^p}\, dx$ converges if and only if $p < 1$ .

Proof. For Type I with $p \neq 1$ :

$\int_1^{\infty} x^{-p}\, dx = \lim_{b \to \infty} \left[\frac{x^{1-p}}{1-p}\right]_1^b = \lim_{b \to \infty} \frac{b^{1-p} - 1}{1-p}$

This converges when $1 - p < 0$ I.e., $p > 1$ . For $p = 1$ : $\int_1^{\infty} 1/x\, dx = \lim_{b \to \infty} \ln b = \infty$ .

For Type II: $\int_0^1 x^{-p}\, dx = \lim_{c \to 0^+} \frac{1 - c^{1-p}}{1-p}$ . This converges when $1 - p > 0$ I.e., $p < 1$ . $\blacksquare$

Remark. The $p$ -test for Type I integrals mirrors the $p$ -series test: $\sum 1/n^p$ converges Iff $p > 1$ . This is not a coincidence --- the integral test establishes the connection.

Worked Example: Evaluate $\int_0^{\infty} e^{-x}\, dx$

Solution. This is a Type I improper integral:

$\int_0^{\infty} e^{-x}\, dx = \lim_{b \to \infty} \int_0^b e^{-x}\, dx = \lim_{b \to \infty} \left[-e^{-x}\right]_0^b = \lim_{b \to \infty} (-e^{-b} + 1) = 1$

So the integral converges to $1$ . $\blacksquare$

Worked Example: Does $\int_1^{\infty} \frac{\sin x}{x}\, dx$ converge?

Solution. The integral $\int_1^{\infty} \left|\frac{\sin x}{x}\right|\, dx$ diverges (compare with $\int_1^{\infty} \frac{|\sin x|}{x}\, dx \geq \sum_{k=1}^{\infty} \int_{k\pi}^{(k+1)\pi} \frac{|\sin x|}{x}\, dx \geq \sum_{k=1}^{\infty} \frac{2}{(k+1)\pi}$ , which diverges by comparison with the harmonic series).

However, $\int_1^{\infty} \frac{\sin x}{x}\, dx$ converges by Dirichlet’s test for integrals. Let $F(b) = \int_1^b \sin x\, dx = \cos 1 - \cos b$ Which is bounded by $|\cos 1 - \cos b| \leq 2$ . Since $1/x$ decreases to $0$ By integration by parts:

$\int_1^b \frac{\sin x}{x}\, dx = \frac{-\cos x}{x}\bigg|_1^b - \int_1^b \frac{\cos x}{x^2}\, dx$

As $b \to \infty$ The boundary term $\cos b / b \to 0$ and $\int_1^{\infty} \frac{|\cos x|}{x^2}\, dx \leq \int_1^{\infty} \frac{1}{x^2}\, dx = 1$ , so the improper integral converges (conditionally). $\blacksquare$

Worked Example: Evaluate $\int_0^1 \frac{1}{\sqrt{x}}\, dx$ (Type II improper integral)

Solution. The integrand $f(x) = 1/\sqrt{x}$ is unbounded as $x \to 0^+$ . Compute:

$\int_0^1 \frac{1}{\sqrt{x}}\, dx = \lim_{c \to 0^+} \int_c^1 x^{-1/2}\, dx = \lim_{c \to 0^+} \left[2\sqrt{x}\right]_c^1 = \lim_{c \to 0^+} (2 - 2\sqrt{c}) = 2$

The improper integral converges to $2$ . Note that $\int_0^1 x^{-p}\, dx$ converges for $p \lt 1$ and Diverges for $p \geq 1$ . $\blacksquare$

Worked Example: Determine convergence of $\int_1^{\infty} \frac{1}{x^p}\, dx$ for various $p$

Solution. By the $p$ -test (Theorem 6.12): $\int_1^{\infty} x^{-p}\, dx$ converges iff $p > 1$ .

Specifically:

$p = 2$ : $\int_1^{\infty} 1/x^2\, dx = \lim_{b \to \infty} [-1/x]_1^b = 0 - (-1) = 1$ . Converges.
$p = 1$ : $\int_1^{\infty} 1/x\, dx = \lim_{b \to \infty} \ln b = \infty$ . Diverges.
$p = 1/2$ : $\int_1^{\infty} 1/\sqrt{x}\, dx = \lim_{b \to \infty} [2\sqrt{x}]_1^b = \infty$ . Diverges.

This mirrors the $p$ -series test: $\sum 1/n^p$ converges iff $p > 1$ . $\blacksquare$

Worked Example: Evaluate $\int_0^{\infty} x e^{-x}\, dx$

Solution. This integral requires both a Type I and Type II limit:

$\int_0^{\infty} x e^{-x}\, dx = \lim_{a \to 0^+} \lim_{b \to \infty} \int_a^b x e^{-x}\, dx$

Integrate by parts with $u = x$ , $dv = e^{-x}\, dx$ So $du = dx$ , $v = -e^{-x}$ :

$\int x e^{-x}\, dx = -xe^{-x} + \int e^{-x}\, dx = -xe^{-x} - e^{-x} = -(x+1)e^{-x}$

Evaluating: $\lim_{b \to \infty} [-(b+1)e^{-b}] - \lim_{a \to 0^+} [-(a+1)e^{-a}] = 0 - (-1) = 1$ .

So $\int_0^{\infty} x e^{-x}\, dx = 1$ . This equals $\Gamma(2) = 1! = 1$ . $\blacksquare$

:::caution Common Pitfall The Riemann integral is defined for bounded functions on closed, bounded intervals. For unbounded Functions or infinite intervals, one must use the improper Riemann integral. A common error is Applying the FTC directly to improper integrals without taking the limit. Also, conditional Convergence of improper integrals behaves differently from absolute convergence: rearranging the “terms” (subintervals) of a conditionally convergent improper integral can change its value.

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