MATH 409 Lecture 22

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Lecture Slides

Improper Riemann Integrals

If a function $f:[a,b]\to \mathbb{ℝ}$ is integrable on $[a,b]$, then the function $F(x)=\underset{a}{\overset{x}{\int }}f(t)\mathrm{d}t$ is well-defined and continuous on $[a,b]$. In particular, $F(c)\to F(b)$ as $c\to b^{-}$, that is,

$\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x=\underset{c\to b+}{\lim }\underset{a}{\overset{c}{\int }}f(x)\mathrm{d}x$

Now suppose $f$ is defined on the semi-open interval $J=[a,b)$ and is integrable on any closed interval $[c,d]\subset J$ (such a function is called locally integrable on $J$). Then all integrals in the rgiht-hand side above are defined and the limit might exist even if $f$ is not integrable on $[a,b]$.

If this is the case, then $f$ is called improperly integrable on $J$, and the limit is called the (improper) integral of $f$ on $[a,b)$.

Similar definition for $(a,b]$: take limit as $c\to a^{+}$.

Precise Definition. A function $f:(a,b)\to \mathbb{ℝ}$ is called improperly integrable on the open interval $(a,b)$ if for some (and then for any) $c\in (a,b)$ it is improperly integrable on semi-open intervals $(a,c]$ and $[c,b)$. The integral of $f$ is defined by

$\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x=\underset{a}{\overset{c}{\int }}f(x)\mathrm{d}x+\underset{c}{\overset{b}{\int }}f(x)\mathrm{d}x$

Properties

Suppose a function $f$ is locally integrable on a semi-open interval $J=[a,b)$ or $J=(a,b]$. Then there are two possible obstructions for $f$ to be integrable on $[a,b]$:

1. the function $f$ is not bounded on $J$ as we approach the open endpoint,
2. The interval $J$ is not bounded: open endpoint might be infinity.

Since an improper Riemann integral is a limit of proper integrals, the properties of improper integrals are analogous to those of proper integrals (and derived using limit theorems).

Theorem. Let $f:[a,b)\to \mathbb{ℝ}$ be integrable on any closed interval $[a_1,b_1]\subset [a,b)$. Given $c\in (a,b)$, the function $f$ is improperly integrable on $[c,b)$ if and only if it is improperly integrable on $[a,b)$. In the case of integrability,

$\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x=\underset{a}{\overset{c}{\int }}f(x)\mathrm{d}x+\underset{c}{\overset{b}{\int }}f(x)\mathrm{d}x$

Sketch of proof. Choos $d\in (c,b)$. We have the following equality involving proper Riemann integrals:

$\underset{a}{\overset{d}{\int }}f(x)\mathrm{d}x=\underset{a}{\overset{c}{\int }}f(x)\mathrm{d}x+\underset{c}{\overset{d}{\int }}f(x)\mathrm{d}x$

The theorem is proved by taking the limit as $d\to b^{-}$.

quod erat demonstrandum

Theorem. Suppose that a function $f:(a,b)\to \mathbb{ℝ}$ is integrable on any closed interval $[c,d]\subset (a,b)$. Given a number $I\in \mathbb{ℝ}$, the following conditions are equivalent:

1. for some $c\in (a,b)$, the function $f$ is improperly integrable on $(a,c]$ and $[c,b)$ and $\underset{a}{\overset{c}{\int }}f(x)\mathrm{d}x+\underset{c}{\overset{b}{\int }}f(x)\mathrm{d}x=I$
2. for every $c\in (a,b)$, the function $f$ is improperly integrable on $(a,c]$ and $[c,b)$ and $\underset{a}{\overset{c}{\int }}f(x)\mathrm{d}x+\underset{c}{\overset{b}{\int }}f(x)\mathrm{d}x=I$
3. for every $c\in (a,b)$, the function $f$ is improperly integrable on $(a,c]$ and $\underset{a}{\overset{c}{\int }}f(x)\mathrm{d}x\to I$ as $c\to b^{-}$
4. for every $c\in (a,b)$, the function $f$ is improperly integrable on $[c,b)$ and $\underset{c}{\overset{b}{\int }}f(x)\mathrm{d}x\to I$ as $c\to a^{+}$

In view of the previous theorem, the integral does not depend on $c$. It can also be computed as a repeated limit:

$\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x=\underset{d\to b^{-}}{\lim }(\underset{c\to a^{+}}{\lim }\underset{c}{\overset{d}{\int }}f(x)\mathrm{d}x)=\underset{c\to a^{+}}{\lim }(\underset{d\to b^{-}}{\lim }\underset{c}{\overset{d}{\int }}f(x)\mathrm{d}x)$

Finally, the integral can be computed as a double limit (i.e., the limit of a function of two variables):

$\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x=\underset{d\to b^{-}}{\underset{c\to a^{+}}{\lim }}\underset{c}{\overset{d}{\int }}f(x)\mathrm{d}x$

If a function $f$ is integrable on a closed interval $[a,b]$ or improperly integrable on one of the semi-open intervals $[a,b)$ and $(a,b]$, then it is also improperly integrable on the open interval $(a,b)$ with the same value of the integral.

If functions $f,g$ are improperly integrable on $(a,b)$, then for any $\alpha ,\beta \in \mathbb{ℝ}$, the linear combination $\alpha f+\beta g$ is also improperly integrable on $(a,b)$ and

$\underset{a}{\overset{b}{\int }}(\alpha f(x)+\beta g(x))\mathrm{d}x=\alpha \underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x+\beta \underset{a}{\overset{b}{\int }}g(x)\mathrm{d}x$

Suppose a function $f:(a,b)\to \mathbb{ℝ}$ is locally integrable and has an antiderivative $F$. Then $f$ is improperly integrable on $(a,b)$ if and only if $F(x)$ has finite limits as $x\to a^{+}$ and as $x\to b^{-}$, in which case

$\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x=\underset{x\to b^{-}}{\lim }F(x)-\underset{x\to a^{+}}{\lim }F(x)$

Comparison Theorems for Improper Integrals

Theorem 1. Suppose that functions $f,g$ are improperly integrable on $(a,b)$. If $f(x)\le g(x)$ for all $x\in (a,b)$, then

$\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x\le \underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x$

Theorem 2. Suppose that functions $f,g$ are locally integrable on $(a,b)$. If the function $g$ is improperly integrable on $(a,b)$ and $0\le f(x)\le g(x)$ for all $x\in (a,b)$, then $f$ is also improperly integrable on $(a,b)$.

Note: This is the first theorem in which we can prove integrability of a function without actually computing its integral or taking an antiderivative. (not all functions have antiderivatives, but are still integrable)

Theorem 3. Suppose that functions $f,g,h$ are locally integrable on $(a,b)$. If the functions $g,h$ are improperly integrable on $(a,b)$ and $h(x)\le f(x)\le g(x)$ for all $x\in (a,b)$, then $f$ is also improperly integrable on $(a,b)$ and

$\underset{a}{\overset{b}{\int }}h(x)\mathrm{d}x\le \underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x\le \underset{a}{\overset{b}{\int }}g(x)\mathrm{d}x$

Note: This theorem is derived from theorems 1 and 2. Also, $h$ and $g$ need not be identical.

Examples

$\frac{1}{\sqrt{x}}$ is improperly integrable on $(0,1]$.

$\begin{array}{rl}\underset{0}{\overset{1}{\int }}\frac{1}{\sqrt{x}}\mathrm{d}x& =\underset{c\to 0^{+}}{\lim }\underset{c}{\overset{1}{\int }}\frac{1}{\sqrt{x}}\mathrm{d}x\\ & =\underset{c\to 0^{+}}{\lim }{2\sqrt{x}|}_{x=c}^1\\ & =\underset{c\to 0^{+}}{\lim }(2-2\sqrt{c})\\ & =2\end{array}$

$x^{-2}$ is improperly integrable on $[1,\mathrm{\infty })$.

$\begin{array}{rl}\underset{1}{\overset{\mathrm{\infty }}{\int }}& =\underset{c\to +\mathrm{\infty }}{\lim }\underset{1}{\overset{c}{\int }}{x}^{-2}\mathrm{d}x\\ & =\underset{c\to +\mathrm{\infty }}{\lim }{-x^{-1}|}_{x=1}^c\\ & =\underset{c\to +\mathrm{\infty }}{\lim }(1-c^{-1})\\ & =1\end{array}$

$f(x)=x^{-2}$ is not improperly integrable on $(0,\mathrm{\infty })$.

Indeed the antiderivative of $f$ is $F(x)=-x^{-1}$, which has a finite limit as $x\to \mathrm{\infty }$, but diverges to infinity as $x\to 0+$.

$g(x)=x^{-2}\cos x$ is improperly integrable on $[1,\mathrm{\infty })$.

This function has no nice antiderivative. Instead, we shall use the "squeeze theorem" for improper integrals. We have $-f(x)\le g(x)\le f(x)=x^{-2}$ for all $x\ge 1$ ($f$ is from the previous example). Since $f$ is improperly integrable on $[1,\mathrm{\infty })$, it follows that $-f$ is also improperly integrable on $[1,\mathrm{\infty })$. By the Comparison Theorem for improper integrals, the function $g$ is improperly integrable on $[1,\mathrm{\infty })$ as well.

$f(x)={\mathrm{e}}^{-x}$ is improperly integrable on $[0,\mathrm{\infty })$.

Indeed, the antiderivative of the function is $-{\mathrm{e}}^{-x}$, which has a finite limit as $x\to +\mathrm{\infty }$.

$g(x)={\mathrm{e}}^{-x^2}$ is improperly integrable on $(-\mathrm{\infty },\mathrm{\infty })$.

Similar to above, there is no nice antiderivative of this function, but this formula shows up all the time in statistics.

We have $0\le g(x)\le f(x)={\mathrm{e}}^{-x}$ for all $x\ge 1$. Since $f$ is improperly integrable on $[0,\mathrm{\infty })$, it follows that $g$ is improperly integrable on $[1,\mathrm{\infty })$. Since $g$ is even, $g(-x)=g(x)$, it follows that $g$ is also improperly integrable on $(-\mathrm{\infty },-1]$. Finally, $g$ is properly integrable on $[-1,1]$.

$f(x)=x^{-1}\sin x$ is improperly integrable on $[1,\mathrm{\infty })$.

In the $\cos x$ example above, we used the comparison theorem. This will not work here because the antiderivative of $x^{-1}$ is $\ln x$, which has no limit as $x\to \mathrm{\infty }$. To show improper integrability, we integrate by parts:

$\begin{array}{rl}\underset{1}{\overset{c}{\int }}{x}^{-1}\cos x\mathrm{d}x& =-\underset{1}{\overset{c}{\int }}{x}^{-1}\mathrm{d}(\cos x)\\ & ={-x^{-1}\cos x|}_{x=1}^c+\underset{1}{\overset{c}{\int }}\cos x\mathrm{d}\left(x^{-1}\right)\\ & =\cos 1-c^{-1}\cos c-\underset{1}{\overset{c}{\int }}{x}^{-2}\cos x\mathrm{d}x\end{array}$

Since the function $g(x)=x^{-2}\cos x$ is improperly integrable on $[1,\mathrm{\infty })$ and $c^{-1}\cos c\to 0$ as $c\to \mathrm{\infty }$, it follows that $f$ is improperly integrable on $[1,\mathrm{\infty })$.

Absolute Integrability

Definition. A funciton $f:(a,b)\to \mathbb{ℝ}$ is called absolutely integrable on $(a,b)$ if $f$ is locally integrable on $(a,b)$ and $\left|f\right|$ is improperly integrable on $(a,b)$.

Theorem. If a function $f$ is absolutely integrable on $(a,b)$, then it is also improperly integrable on $(a,b)$ and

$|\underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x|\le \underset{a}{\overset{b}{\int }}|f(x)|\mathrm{d}x$

Proof. Since $\left|f\right|$ is improperly integrable on $(a,b)$, so is $-\left|f\right|$. Clearly, $-|f(x)|\le f(x)\le |f(x)|$ for all $x\in (a,b)$. By the Comparison Theorems for improper integrals, the function $f$ is improperly integrable on $(a,b)$ and

$-\underset{a}{\overset{b}{\int }}|f(x)|\mathrm{d}x\le \underset{a}{\overset{b}{\int }}f(x)\mathrm{d}x\le \underset{a}{\overset{b}{\int }}|f(x)|\mathrm{d}x$

quod erat demonstrandum

Examples

For any nonnegataive function, the absolute integrability is equivalent to improper integrability.

In particular,

• The function $f_1(x)=x^{-2}$ is absolutely integrable on $[1,\mathrm{\infty })$ and is not on $(0,\mathrm{\infty })$.
• The function $f_2=\frac{1}{\sqrt{x}}$ is absolutely integrable on $(0,1)$.
• The function $f_3(x)={\mathrm{e}}^{-x^2}$ is absolutely integrable on $(-\mathrm{\infty },\mathrm{\infty })$

The function $f(x)={\mathrm{e}}^{-x^2}\sin x$ is absolutely integrable on $(-\mathrm{\infty },\mathrm{\infty })$

Indeed, the function $f$ is locally integrable on $(-\mathrm{\infty },\mathrm{\infty })$, a function $g(x)={\mathrm{e}}^{-x^2}$ is improperly integrable on $(-\mathrm{\infty },\mathrm{\infty })$, and $|f(x)|\le g(x)$ for all $x\in \mathbb{ℝ}$

Counterexamples

The function $f(x)=\{\begin{array}{ll}1& x\in \mathbb{\mathbb{Q}}\\ -1& x\in \mathbb{ℝ}\setminus \mathbb{\mathbb{Q}}\end{array}$ is not absolutely integrable on $(0,1)$:

Indeed, the function $f$ is not locally integrable on $(0,1)$. At the same time, the function $\left|f\right|$ is constant and hence (properly) integrable on $(0,1)$.

$f(x)=x^{-1}\sin x$ is not absolutely integrable on $[1,\mathrm{\infty })$.

For any $n\in \mathbb{\mathbb{N}}$,

$\begin{array}{rl}\underset{n\pi }{\overset{(n+1)\pi }{\int }}|f(x)|\mathrm{d}x& \ge \underset{n\pi }{\overset{(n+1)\pi }{\int }}\frac{|\sin x|}{(n+1)\pi }\mathrm{d}x\\ & =\frac{1}{(n+1)\pi }\underset{0}{\overset{\pi }{\int }}\sin x\mathrm{d}x\\ & =\frac{2}{(n+1)\pi }& \ge \frac{1}{n\pi }& \ge \frac{1}{\pi }\underset{n\pi }{\overset{(n+1)\pi }{\int }}\frac{1}{x}\mathrm{d}x\end{array}$

It remains to notice that $g(x)=\frac{1}{x}$ is not improperly integrable on $[\pi ,\mathrm{\infty })$.

Main idea: The function $f$ behaves similarly to $x^{-1}$, so the improper integral on $[1,\mathrm{\infty })$ does not exist.