Integration of Real Functions

Riemann Integral

Definition: Riemann Sum

Let $f:\mathcal{D}\subseteq \mathbb{R}\to \mathbb{R}$ be a real function, let $I=[a;b]\subseteq \mathcal{D}$ be a closed interval and let $x_0<x_1<\cdots <x_n\in I$.

A Riemann sum of $f$ over $I$ is any sum of the form

$$\sum _{i=1}^{n}f(x_i^{\ast })\Delta x_i,$$

where $\Delta x_i=(x_i-x_{i-1})$ and $x_i^{\ast }\in [x_{i-1};x_i]$.

Note: Choice of $x_i^{\ast }$

Different choices for $x_i^{\ast }$ yield different Riemann sums.

Definition: Left Riemann Sum

A left Riemann sum has $x_i^{\ast }=x_{i-1}$ for all $i$.

Definition: Right Riemann Sum

A right Riemann sum has $x_i^{\ast }=x_i$ for all $i$.

Definition: Left Riemann Sum

A middle Riemann sum has $x_i^{\ast }=\frac{x_{i-1}+x_i}{2}$ for all $i$.

Definition: Riemann Integral

Let $f:\mathcal{D}\subseteq \mathbb{R}\to \mathbb{R}$ be a real function and let $I=[a;b]\subseteq \mathcal{D}$ be a closed interval.

We say that $f$ is Riemann-integrable on $I$ if all of its Riemann sums have the same limit for $\Delta x_i\to 0$:

$$\underset{\Delta x_i\to 0}{\lim }\sum _{i=1}^{n}f(x_i^{\ast })\Delta x_i=S\in \mathbb{R}$$

In this case, the value of this limit is known as $f$‘s Riemann integral or definite integral over $I$.

NOTATION

The most common notation for the definite integral is

$${\int }_a^{b}f(x)\mathrm{dx}$$

The upside of this notation is that it clearly shows where the integrand, i.e. the thing being integrated, begins and where it ends. This is not very useful when we are referring to $f$ by its name, but it helps to remove ambiguity when we substitute $f$ with an expression such as ${\int }_a^b{x}^3+\ln x\mathrm{dx}$.

Its main downside is that it forces us to assign a symbol to the function’s argument which is redundant and can even be confusing in some contexts where we refer to $f$ by its name. In particular, it is irrelevant whether we write ${\int }_a^{b}f(x)\mathrm{dx}$ or ${\int }_a^{b}f(y)\mathrm{dy}$ or ${\int }_a^{b}f(\omega )\mathrm{d\omega }$, hence we can shorten the notation to just

$${\int }_a^{b}f$$

The main downside of this notation is that it implies that the order of $a$ and $b$ matters, which is not the case - what matters is the actual closed interval which $[a;b]$ represents. To emphasise this, we can use the following notations:

$${\int }_{[a;b]}f{\int }_{I}f$$

In the latter case, we can also add $\mathrm{d}I$ to clarify where the integrand begins and ends, such as

$${\int }_{I}f\mathrm{dI}$$

All of these notations are useful in specific contexts and less so in others.

Notation: Definite Integrals with Special Bounds

A common convention is to define the notations

$${\int }_b^{a}f=-{\int }_a^{b}f$$

and

$${\int }_c^{c}f=0$$

for each $c\in [a;b]$. This is merely notation which makes the formulation of many theorems easier and more natural.

Note: Definite Integrals over Non-Interval Domains

The definition of the Riemann integral can be naturally extended as a sum when $I$ is not a closed interval but can be expressed as the union $I_1\cup \cdots \cup I_n$ of finitely many closed intervals such that each two overlap at most at a single point:

$${\int }_{I}f\mathrm{dI}\stackrel{\text{def}}{=}{\int }_{I_1}f\mathrm{d}{I}_1+\cdots +{\int }_{I_n}f\mathrm{d}{I}_n$$

Theorem: Linearity of the Definite Integral

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ and $g:{\mathcal{D}}_g\subseteq \mathbb{R}\to \mathbb{R}$ be real functions and let $I\subseteq {\mathcal{D}}_f\cap {\mathcal{D}}_g$ be a closed interval.

If $f$ and $g$ are Riemann-integrable on $I$, then for all $\alpha ,\beta \in \mathbb{R}$ we have

$${\int }_I\alpha f(x)+\beta g(x)\mathrm{dx}=\alpha {\int }_{I}f(x)\mathrm{dx}+\beta {\int }_{I}g(x)\mathrm{dx}$$

PROOF

TODO

Theorem: Integration by Parts

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ and $g:{\mathcal{D}}_g\subseteq \mathbb{R}\to \mathbb{R}$ be real functions and let $[a;b]\subseteq {\mathcal{D}}_f\cap {\mathcal{D}}_g$ be a closed interval.

If $f$ and $g$ are continuously differentiable on the open interval $(a;b)$, then they are Riemann-integrable on $[a;b]$ with

$${\int }_{[a;b]}f(x)g^{\prime }(x)\mathrm{dx}=f(b)g(b)-f(a)g(a)-{\int }_a^b{f}^{\prime }(x)g(x)\mathrm{dx}$$

PROOF

TODO

Theorem: Substitution

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ and $g:{\mathcal{D}}_g\subseteq \mathbb{R}\to \mathbb{R}$ be real functions such that the image of $g$ is contained in the domain of $f$ and let $[a;b]\subseteq {\mathcal{D}}_g$ be a closed interval.

If $g$ is continuously differentiable on the open interval $(a;b)$ and $f$ is continuous on the closed interval $[g(a);g(b)]$, then the function $(f\circ g)g^{\prime }$ is Riemann-integrable on $[a;b]$ with

$${\int }_a^b(f\circ g)g^{\prime }={\int }_{g(a)}^{g(b)}f$$

PROOF

TODO

Mean Value Theorem for Definite Integrals

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ and $g:{\mathcal{D}}_g\subseteq \mathbb{R}\to \mathbb{R}$ be real functions and let $[a;b]\subseteq {\mathcal{D}}_f\cap {\mathcal{D}}_g$ be a closed interval.

If $f$ and $g$ are continuous on $[a;b]$ and $g(x)\ne 0$ for all $x\in [a;b]$, then their product $fg$ is Riemann-integrable on $[a;b]$ and there exists some $\xi \in (a;b)$ such that

$${\int }_{[a;b]}fg=f(\xi ){\int }_{[a;b]}g$$

PROOF

TODO

Tip: $g(x)=1$

In the case of $g(x)=1$ for all $x$, the mean value theorem simplifies to

$${\int }_{[a;b]}f=f(\xi )(b-a)$$

Theorem: Integrability of the Absolute Value

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ be a real function and let $I\subseteq {\mathcal{D}}_f$ be a closed interval.

If $f$ is Riemann-integrable on $I$, then so is its absolute value $|f|$ with

$$\left|{\int }_{I}f\right|\le {\int }_I|f|$$

PROOF

TODO

The Fundamental Theorem of Real Analysis (Part I)

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ be a real function and let $I=[a;b]\subseteq {\mathcal{D}}_f$ be a closed interval.

If $f$ is Riemann-integrable on $I$, then the function $F:I\to \mathbb{R}$ defined as

$$F(x)\stackrel{\text{def}}{=}{\int }_a^{x}f$$

is continuous. Moreover, if $f$ is continuous at some $c\in I$, then $F$ is an antiderivative of $f$ at $c$, i.e. $F^{\prime }(c)=f(c)$.

PROOF

According to the definition of the derivative we need to prove

$$\underset{\Delta x\to 0}{\lim }\frac{F(x+\Delta x)-F(x)}{\Delta x}=f(x)$$

For all $c,x_0,\Delta x\in [a;b]$, where $x_0+\Delta x\in [a;b]$, it holds that

$$\begin{array}{rl}F(x_0+\Delta x)-F(x_0)& ={\int }_c^{x_0+\Delta x}f(t)\mathrm{dt}-{\int }_c^{x_0}f(t)\mathrm{dt}\\ & ={\int }_c^{x_0+\Delta x}f(t)\mathrm{dt}+{\int }_{x_0}^{c}f(t)\mathrm{dt}\\ & ={\int }_{x_0}^{c}f(t)\mathrm{dt}+{\int }_c^{x_0+\Delta x}f(t)\mathrm{dt}\\ & ={\int }_{x_0}^{x_0+\Delta x}f(t)\mathrm{dt}\end{array}$$

The mean value theorem for integrals says that there is at least one $\xi \in [x_0;x_0+\Delta x]$ such that

$${\int }_{x_0}^{x_0+\Delta x}f(t)\mathrm{dt}=f(\xi )(x_0+\Delta x-x_0)=f(\xi )\cdot \Delta x$$ $$\frac{F(x_0+\Delta x)-F(x_0)}{\Delta x}=f(\xi )$$

We now take the limit as $\Delta x\to 0$.

$$\underset{\Delta x\to 0}{\lim }\frac{F(x_0+\Delta x)-F(x_0)}{\Delta x}=\underset{\Delta x\to 0}{\lim }f(\xi )$$

The left-hand side is just the derivative of $F$ at $x_0$. Since $\xi$ is between $x_0$ and $x_0+\Delta x$, it must approach $x_0$ for $\Delta x\to 0$. This means that

$$\underset{\Delta x\to 0}{\lim }f(\xi )=f(x_0)$$

and so we have

$$\underset{\Delta x\to 0}{\lim }\frac{F(x_0+\Delta x)-F(x_0)}{\Delta x}=f(x_0)$$

We have thus proven

$$F^{\prime }(x_0)=f(x_0)$$

The Fundamental Theorem of Real Analysis (Part II)

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ be a real function, let $[a;b]\subseteq {\mathcal{D}}_f$ be a closed interval.

If $f$ is Riemann-integrable on $[a;b]$, then

$${\int }_a^{b}f(x)\mathrm{dx}=F(b)-F(a)$$

for any real function $F$ which is continuous on $[a;b]$ and is an antiderivative of $f$ on $(a;b)$.

NOTATION

The expression $F(b)-F(a)$ is usually shortened to just $F(x){|}_a^b$.

PROOF

TODO

Improper Integrals

The notion of Riemann integrals can be extended to open and semi-open intervals using limits.

Definition: Improper Integrals

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ be a real function and let $I\subseteq {\mathcal{D}}_f$ be an open or semi-open interval.

An improper integral is defined via limits of Riemann integrals in one of the following ways:

• If $I=[a;b)$ with $a,b\in \mathbb{R}$, then

$${\int }_a^{b}f(x)\mathrm{dx}\stackrel{\text{def}}{=}\underset{\beta \to b^{-}}{\lim }{\int }_a^{\beta }f(x)\mathrm{dx}$$

• If $I=(a;b]$ with $a,b\in \mathbb{R}$, then

$${\int }_a^{b}f(x)\mathrm{dx}\stackrel{\text{def}}{=}\underset{\alpha \to a^{+}}{\lim }{\int }_{\alpha }^{b}f(x)\mathrm{dx}$$

• If $I=[a;+\infty )$ with $a\in \mathbb{R}$, then

$${\int }_a^{\infty }f(x)\mathrm{dx}\stackrel{\text{def}}{=}\underset{\beta \to \infty }{\lim }{\int }_a^{\beta }f(x)\mathrm{dx}$$

• If $D=(-\infty ;b]$ with $b\in \mathbb{R}$, then

$${\int }_{-\infty }^{b}f(x)\mathrm{dx}\stackrel{\text{def}}{=}\underset{\alpha \to -\infty }{\lim }{\int }_{\alpha }^{b}f(x)\mathrm{dx}$$

• If $D=(-\infty ;\infty )$, then

$${\int }_{-\infty }^{\infty }f(x)\mathrm{dx}\stackrel{\text{def}}{=}{\int }_{-\infty }^{c}f(x)\mathrm{dx}+{\int }_c^{\infty }f(x)\mathrm{dx},$$

where $c\in \mathbb{R}$ is such that the improper integrals ${\int }_{-\infty }^{c}f(x)\mathrm{dx}$ and ${\int }_c^{\infty }f(x)\mathrm{dx}$ exist.

We say that $f$ is improperly Riemann-integrable on $I$ if the corresponding improper integral is finite. In this case, the improper integral is said to converge or simply exist. Otherwise, it diverges or does not exist.

Lebesgue Integrals

Definition: Lebesgue Integral

Let $(\mathbb{R},\Sigma ,\mu )$ be the measure space formed using the Lebesgue measure on $\mathbb{R}$.

Let $f:\mathcal{D}\subseteq \mathbb{R}\to \mathbb{R}$ be a measurable real function on a Lebesgue-measurable subset $\mathcal{D}$ and let $S\subseteq \mathbb{R}$ also be Lebesgue-measurable.

The Lebesgue integral of $f$ over $S$ is the Lebesgue integral of $f$ over $S$ with respect to the Lebesgue measure $\mu$:

$${\int }_{S}f\mathrm{d\mu }$$

Definition: Lebesgue-Integrability

We say that $f$ is Lebesgue-integrable on $S$ if its Lebesgue integral is finite.

Theorem: Riemann-Integrability $⟹$ Lebesgue-Integrability

Let $f:I\subset \mathbb{R}\to \mathbb{R}$ be a real function on a closed interval $I=[a;b]$.

If $f$ is Riemann-integrable on $I$, then $f$ is also Lebesgue-integrable on $I$ with

$${\int }_a^{b}f(x)\mathrm{dx}={\int }_{I}f\mathrm{d\mu }$$

PROOF

TODO

Theorem: Equality of Lebesgue Integrals

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ and $g:{\mathcal{D}}_g\subseteq \mathbb{R}\to \mathbb{R}$ be measurable real functions on Lebesgue-measurable subsets ${\mathcal{D}}_f$ and ${\mathcal{D}}_g$ and let $S\subseteq \mathbb{R}$ also be Lebesgue-measurable.

If $f$ and $g$ are non-negative on $S$ and the set $\{x\in S\mid f(x)\ne g(x)\}$ is a subset of a null set, then the Lebesgue integrals of $f$ and $g$ over $S$ are equal.

$${\int }_{S}f\mathrm{d\mu }={\int }_{S}g\mathrm{d\mu }$$

PROOF

TODO

Theorem: Linearity of Lebesgue Integrals

Let $f:{\mathcal{D}}_f\subseteq \mathbb{R}\to \mathbb{R}$ and $g:{\mathcal{D}}_g\subseteq \mathbb{R}\to \mathbb{R}$ be measurable real functions on Lebesgue-measurable subsets ${\mathcal{D}}_f$ and ${\mathcal{D}}_g$ and let $S\subseteq \mathbb{R}$ also be Lebesgue-measurable.

If $f$ and $g$ are Lebesgue-integrable on $S$, then so is the function $\alpha f+\beta g$ for all $\alpha ,\beta \in \mathbb{R}$. Furthermore,

$${\int }_S\alpha f+\beta g\mathrm{d\mu }=\alpha {\int }_{S}f\mathrm{d\mu }+\beta \int g\mathrm{d\mu }$$

PROOF

TODO