Lecture 10. The Riemann Integral

Riemann Sums

Definitions (Upper and Lower Sums)

Let $-\infty < a < b < \infty$ and $f:[a,b]\to \mathbb{R}$ be a bounded function. Partition the interval $[a,b]$ by $P=\{ x_0,\dots,x_n\}$ where $$ a = x_0 < x_1 < \dots < x_n = b. $$ The partition $P$ has a norm $\| P\|$ given by its maximal interval length $$ \| P\| = \max \{ \Delta x_i=x_i - x_{i-1}\, :\, i=1,2,\dots ,n\}. $$ Define \begin{align*} m_i(f)&= \inf\{ f(x): x_{i-1} \le x < x_i \}\\ M_i(f)&= \sup\{f(x): x_{i-1} \le x < x_i \}. \end{align*} and consider the lower and upper partial sums of $f$ (with respect to the partition $P$) given by \begin{align*} L(f,P)&= \sum_{i=1}^n m_i \Delta x_i,\\ U(f,P)&= \sum_{i=1}^n M_i \Delta x_i. \end{align*} It is easy to see that $L(f,P) \le U(f,P)$.

Remark

Let $$ P_1 : a= x_0 < x_1 < \dots < x_n =b\text{ and }P_2: a=y_0 < y_1 < \dots < y_m = b, $$ and define a refinement (partition) $P$ for $P_1$ and $P_2$ by using $x_0,\dots,x_n$ and $y_0,\dots,y_m$ as end points in the appropriate natural order. Denote such a refinement by $P_1 \cup P_2$. Then $$ L(f, P_1) \leq L(f,P) \leq U(f,P) \leq U(f, P_2). $$

Definition (Riemann Integral)

Now the lower and upper integrals of $f$ on $[a,b]$ are defined by \begin{align*} L(f)&= \lim_{\| P\|\to 0} L(f,P)\text{ and}\\ U(F)&= \lim_{\| P\| \to 0} U(f,P), \end{align*} respectively. They both always exist by monotonicity. Let $f\in \operatorname{B}\bigl([a, b],\mathbb{R}\bigr)$, that is, a real valued bounded function defined on $[a,b]$. It is said to be (Riemann) integrable, or simply R-integrable iff $$ L(f) = U(f). $$

Conditions for Riemann Integrability

Theorem

Let $f\in \operatorname{B}\bigl([a, b],\mathbb{R}\bigr)$ Then $f$ is R-integrable iff, for any $\epsilon > 0$, there is a partition $P$ of $[a,b]$ such that $$ U(f,P) - L(f,P)\leq\epsilon. $$

"$\Rightarrow$": if $f$ is (R-)integrable, then $L(f) = U(f)$. Given $\epsilon>0$, there are $P_1$ and $P_2$ such that $$ -L(f,P_1) + L(f)\leq\epsilon/2\text{ and }U(f,P_2) - U(f)\leq \epsilon/2, $$ by definition of lower and upper integral. Since $L(f)=U(f)$ it follows that $$ U(f, P) - L(f,P)\leq\epsilon $$ for $P=P_1\cup P_2$.
"$\Leftarrow$": it always holds that $L(f) \ge L(f, P)$ and $U(f) \le U(f,P)$. If, for any $\epsilon>0$, there is $P$ such that $U(f,P) - L(f,P)\leq\epsilon$, then $$ U(f) - L(f)\leq \epsilon. $$ Now $U(f) - L(f) \ge 0$ is a fixed number and the inequality holds for all $\epsilon>0$ so that indeed $U(f)-L(f) = 0$ and $f$ is integrable.

Examples

(a) Continuity of $f$ implies uts uniform continuity since $[a,b]$ is compact. Thus, given $\epsilon>0$, there is $\delta>0$ such that $$ |f(x_1) - f(x_2) |\leq\frac{\epsilon}{b-a}\text{, whenever }|x_1 - x_2 |\leq\delta\text{ for }x_1, x_2 \in [a,b]\, . $$ It suffices now to choose $n$ such that $\frac{b-a}{n} \leq\delta$ and $$ P: x_0 = a < x_1< \dots < x_n =b\text{ with }x_j = x_{j-1} + \frac{b-a}{n}, $$ to see that $M_j - m_j\leq\frac{\epsilon}{b-a}$ and that $$ U(f,P) - L(f,P) \le \sum_{j=1}^n (M_j - m_j) \Delta x_j\leq\frac{\epsilon}{b-a} \sum_{i=1}^n \Delta x_i=\epsilon. $$
(c) Exercise.
(d) Let $P: a=x_0 < x_1 < \dots < x_n=b$ be any partition for $[a,b]$. Since $f$ is increasing, it follows that \begin{gather*} m_i = \inf\{f(x): x_{i-1} \le x < x_i \} = f(x_{i-1}),\\ M_i=\sup \{f(x): x_{i-1} \le x < x_i \} = f(x_i - 0)\le f(x_i). \end{gather*} Therefore \begin{equation*} U(f,P) - L(f,P) = \sum_{i=1}^n (M_i - m_i) \Delta x_i \le \| P\| \sum_{i=1}^n (M_i - m_i)\leq\big[f(b) - f(a)\big]\| P\|, \end{equation*} since $$ M_1 - m_1 + M_2 - m_2+\dots + M_n - m_n \le M_n - m_1 \le f(b) -f(a). $$ For $\epsilon>0$ and for any partition $P$ for $[a,b]$ with $\|P\|\leq\frac{\epsilon}{f(b) - f(a)}$ it follows that $$ U(f,P) - L(f,P)\leq \big[f(b) - f(a)\big]\frac{\epsilon}{f(b) - f(a)} = \epsilon $$ and $f$ is integrable.

Theorem

Let $f\in \operatorname{B}\bigl([a,b],\mathbb{R}\bigr)$. Then $$ f \text{ is integrable iff } m(D) = 0\, (\text{i.e. iff }D\text{ has measure }0). $$

Remark

Theorem

Let $f\in \operatorname{B}\bigl([a,b],\mathbb{R}\bigr)$. Then $f$ is integrable on $[a,b]$ iff $f^+$ and $f^-$ are integrable, where $$ f^+(x)=\begin{cases} f(x),&\text{if } f(x) \geq 0\\ 0,&\text{if }f(x) < 0\end{cases}\text{ and } f^-(x)=\begin{cases} f(x),&\text{if } f(x) \leq 0\\ 0,&\text{if }f(x) > 0\end{cases} $$

Corollary

If $f$ is integrable on $[a,b]$, then $|f|$ is integrable on $[a,b]$. The converse is not true in general.

Nexample

Improper Integrals

Definition

Let $f:(a,b)\to \mathbb{R}$ be a function (notice that such a function can be unbounded even if it is assume continuous). If $f\big |_{[c, d]}$ is integrable for any choice of $c,d$ with $a < c < d < b$, then it is called integrable iff $$ \int_a^b f(x) dx=:\lim_{c\to a, d\to b}\int_{c}^d f(x) dx\: \text{exists.} $$

Proposition

If $f:(a,b)\to \mathbb{R}$ is integrable on $[c_1, c_2]$ for any $a < c_1 < c_2 < b$ and $|f|$ is integrable on $(a,b)$, then $f$ is integrable on $(a,b)$. The converse is not true in general.

Nexample

For $\alpha \ne 1$ one computes $$ \int_{\epsilon}^1 f(x)\, dx=\int_{\epsilon}^1 \frac{1}{x^\alpha}\, dx=\frac{1}{1-\alpha}x^{1-\alpha} \Big | _{\epsilon}^1 = \frac{1}{\alpha-1} \epsilon^{1- \alpha} -\frac{1}{\alpha-1}, $$ and thus $$ \lim_{\epsilon \to 0^+} \int_{\epsilon}^1 \frac{1}{x^\alpha} \, dx = -\frac{1}{\alpha-1} + \lim_{\epsilon \to 0^+} \frac{1}{\alpha-1} \epsilon^{1-\alpha}= \begin{cases} < \infty\, ,&\text{if }\alpha < 1\, ,\\ +\infty\, ,&\text{if }\alpha>1\end{cases} $$ As for $\alpha=1$ it holds $$ \int_0^\epsilon \frac{1}{x}\, dx =\log(\epsilon)\to\infty\text{ as }\epsilon\to 0+. $$

Theorem (Comparison)

Assume that $f,g:(a,b)\to \mathbb{R}$ are continuous and that $|g|$ is integrable. If $$ |f(x)|\leq |g(x)|,\: x\in (a,b), $$ then $|f|$ is integrable and so is $f$.

Examples

(a) Observe that $\frac{1}{\sqrt{x}} \sin(\frac{1}{x})$ is continuous on $(0,1]$ and that $$ \Big |\frac{1}{\sqrt{x}} \sin(\frac{1}{x})\Big |\leq\frac{1}{\sqrt{x}},\: x \in (0,1]. $$ Since it is known that $\int_0^1 \frac{1}{\sqrt{x}} < + \infty$ the theorem yields convergence.
(b) Observe that $\sin(x) \leq x$ for $x \in[0, \pi/2]$ so that $$ 0 \le \frac{\sin(x)}{x^{3/2}} \le \frac{x}{x^{3/2}} = \frac{1}{x^{1/2}} $$ Comparison again yields convergence.
(c) Since $$ \frac{\cos x}{x^{3/2}} \geq\frac{1-x^2/2}{x^{3/2}}, \: x\in [0,1], $$ it follows that $$ \int_\epsilon^1\frac{\cos x}{x^{3/2}} dx\geq\int _\epsilon^1\frac{1-x^2/2}{x^{3/2}}=\int _\epsilon^1\frac{1}{x^{3/2}}-\int_0^1 \frac{1}{x^{1/2}}\, dx\to\infty\text{ as }\epsilon\to 0+, $$

Definition

Let $f:\mathbb{R}\to \mathbb{R}$ be such that $f\big |_{[a,b]}$ is integrable for any $a,b\in \mathbb{R}$ with $a < b$. Then
(i) $\int_a^\infty f(x)\, dx = \lim_{b \to +\infty} \int_a^b f(x)\, dx$ if the limit exists.
(ii) $\int_{-\infty}^b f(x) \, dx = \lim_{a \to -\infty} \int_a^b f(x) \, dx$ if the limit exists.
(iii) $\int_{\infty}^\infty f(x) \, dx = \lim_{a \to -\infty, b \to \infty} \int_a^b f(x) \, dx$ if the limit exists.

Examples

(b) First observe that \begin{equation*} \int_0^\infty \frac{\sin(x)}{\ln(x+2)} \, dx= -\frac{\cos(x)}{\ln (x+2)}\Big | _{x=0}^\infty+ \int_0^\infty \bigl( \frac{1}{\ln(x+2)}\bigr)' \cos(x) \, dx =\frac{1}{\ln 2} -\int_0^\infty \frac{\cos x}{(x+2)\ln ^2(x+2)}\, dx. \end{equation*} Next for $p>1$ one has that \begin{equation*} \int_2^\infty \frac{1}{x}\frac{1}{(\ln x)^p} \, dx =\int_2^\infty \frac{1}{(\ln x)^p} d \ln x=-\frac{1}{p-1} (\ln x)^{1-p} \Big | _{x=2}^\infty=\frac{1}{p-2}(\ln 2)^{1-p}<\infty. \end{equation*} Combining this with \begin{equation*} \int_0^\infty \Big |\frac{-1}{(\ln (x+2)^2} \frac{1}{x+2} \cos(x)\Big |\, dx\le\int_0^\infty\frac{1}{(x+2)\ln ^2(x+2)} \, dx =-\ln(x+2)^{-1} \Big |_{x=0}^\infty =(\ln 2)^{-1}<+\infty \end{equation*} and the fact that the integrand is continuous on $(0, \infty)$, and using the comparison test yields the convergence of $\int_0^\infty \frac{\cos(x)}{(x+2)\ln ^2(x+2)}\, dx$ and therefore that of $\int_0^\infty \frac{\sin (x) }{\ln(x+2)}$.

Theorem (Newton-Leibniz)

If $f:[a,b]\to \mathbb{R}$ is integrable and $$ F(x):=\int_a^x f(t)\, dt\, ,\: x\in [a,b]\, , $$ then $F'(x_0) = f(x_0)$ whenever $f$ is continuous at $x_0$.

Example

It holds that \begin{equation*} F(x)=\int_0^{x^2 - \sin(x)} e^{t^2} (t+2)\, dt + \int_{x- \cos(x)}^0 e^{t^2} (t+2)\, dt, \end{equation*} and we thus have that \begin{align*} F'(x)&=\Big[ \int_0^{x^2 - \sin(x)} e^{t^2} (t+2) dt \Big] ' -\Big[\int_0^{x-\cos(x)} e^{t^2} (t+2) dt \Big]'\\ &=e^{[x^2 - \sin (x)]^2} \big[x^2 - \sin (x) + 2\big](x^2-\sin x)' - e^{(x- \cos(x))^2} (x- \cos x +2) (x - \cos x)' \\ &=e^{[x^2- \sin (x)]^2} \big[x^2 - \sin (x) + 2\big](2x-\cos x)- e^{(x- \cos(x))^2} (x- \cos x +2) (1 +\sin x). \end{align*}

Important inequalities

Theorem

Let $f,g\in\operatorname{B}\bigl([a,b],\mathbb{R}\bigr)$ be integrable. Then
(i) $f\cdot g$ is integrable on $[a,b]$
(ii) (Cauchy-Schwarz inequality) $$ \bigl( \int_a^b f(x)g(x) \, dx \bigr)^2 \leq \int_a^b f^2(x)\, dx\int_a^b g^2(x)\, dx. $$

(i) Exercise.
(ii) It follows from \begin{equation*} 0\leq\int_a^b\big[f(x) + \lambda g(x)\big]^2\, dx=\int_a^b f^2(x) \,dx +2\lambda\int _a^b f(x) g(x) \, dx+ \lambda^2\int_a^bg^2(x)\, dx, \end{equation*} for any $\lambda\in\mathbb{R}$ that $$ \big[2 \int _a^b f(x) g(x) \, dx\big]^2-4\int_a^b g^2(x)\, dx\int_a^b f^2(x)\, dx\le 0, $$ which is equivalent to the claim.

Theorem (Hölder Inequality)

Let $f,g\in\operatorname{B}\bigl([a,b],\mathbb{R}\bigr)$ be integrable and assume that $1\leq p,q < \infty$ satisfy $\frac{1}{p}+\frac{1}{q} =1$. Then $f\cdot g$ is integrable and $$ \Big |\int_a^b f(x) g(x)\, dx\Big |\leq\Bigl(\int_a^b |f(x) |^p \, dx\Bigr)^{1/p}\Bigl(\int_a^b |g(x)|^q\, dx\Bigr)^{1/q}. $$

W.l.o.g we can assume that $f,g\geq 0$. We consider two cases. First assume that $\int_a^b f^p(x)\, dx=1$ and $\int_a^b g^q(x)\, dx =1$. In this case the claim amounts to proving that $$ \int_a^b f(x)g(x) \, dx \le 1\, . $$ This readily follows from \begin{equation*} \int_a^b f(x) g(x) \, dx \le\int_a^b \Big(\frac{1}{p} f(x) ^p + \frac{1}{q} g(x)^q \Big)\, dx=\frac{1}{p} \int_a^b f^p(x)\, dx + \frac{1}{q} \int_a^b g^q(x) \, dx=\frac{1}{p}+\frac{1}{q}=1. \end{equation*} In general, define $$ F(x) = \frac{f(x)}{\bigl(\int_a^b f(x)^p \, dx\bigr)^{1/p}},\quad G(x) = \frac{g(x)}{\bigl(\int_a^b g(x)^q \, dx\bigr)^{1/q}} $$ so that $$ \int_a^b F^p(x)\, dx=\frac{1}{\int_a^b f^p(x)\, dx}\int_a^b f^p(x)\, dx=1\text{ and }\int_a^b G^q(x)\, dx =1. $$ Thus it follows from the previous case that $\int_a^b F(x) G(x) \, dx \le 1$ and therefore that $$ \int_a^b f(x) g(x) \, dx \le \Big(\int_a^b f(x)^p \, dx\Big)^{1/p} \Big(\int_a^bg(x)^q \, dx\Big)^{1/q}. $$ as claimed.

Reminder