# MATH 409 Lecture 22

« previous | Tuesday, November 19, 2013 | next »

Begin Exam 3 content
Lecture Slides

## Improper Riemann Integrals

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

${\displaystyle \int _{a}^{b}f(x)\,\mathrm {d} x=\lim _{c\to b+}\int _{a}^{c}f(x)\,\mathrm {d} x}$

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

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

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

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

${\displaystyle \int _{a}^{b}f(x)\,\mathrm {d} x=\int _{a}^{c}f(x)\,\mathrm {d} x+\int _{c}^{b}f(x)\,\mathrm {d} x}$

### Properties

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

1. the function ${\displaystyle f}$ is not bounded on ${\displaystyle J}$ as we approach the open endpoint,
2. The interval ${\displaystyle J}$ 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 ${\displaystyle f:\left[a,b\right)\to \mathbb {R} }$ be integrable on any closed interval ${\displaystyle [a_{1},b_{1}]\subset \left[a,b\right)}$. Given ${\displaystyle c\in (a,b)}$, the function ${\displaystyle f}$ is improperly integrable on ${\displaystyle \left[c,b\right)}$ if and only if it is improperly integrable on ${\displaystyle \left[a,b\right)}$. In the case of integrability,

${\displaystyle \int _{a}^{b}f(x)\,\mathrm {d} x=\int _{a}^{c}f(x)\,\mathrm {d} x+\int _{c}^{b}f(x)\,\mathrm {d} x}$

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

${\displaystyle \int _{a}^{d}f(x)\,\mathrm {d} x=\int _{a}^{c}f(x)\,\mathrm {d} x+\int _{c}^{d}f(x)\,\mathrm {d} x}$

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

quod erat demonstrandum

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

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

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

${\displaystyle \int _{a}^{b}f(x)\,\mathrm {d} x=\lim _{d\to b^{-}}\left(\lim _{c\to a^{+}}\int _{c}^{d}f(x)\,\mathrm {d} x\right)=\lim _{c\to a^{+}}\left(\lim _{d\to b^{-}}\int _{c}^{d}f(x)\,\mathrm {d} x\right)}$

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

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

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

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

${\displaystyle \int _{a}^{b}\left(\alpha \,f(x)+\beta \,g(x)\right)\,\mathrm {d} x=\alpha \,\int _{a}^{b}f(x)\,\mathrm {d} x+\beta \,\int _{a}^{b}g(x)\,\mathrm {d} x}$

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

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

### Comparison Theorems for Improper Integrals

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

${\displaystyle \int _{a}^{b}f(x)\,\mathrm {d} x\leq \int _{a}^{b}f(x)\,\mathrm {d} x}$

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

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 ${\displaystyle f,g,h}$ are locally integrable on ${\displaystyle (a,b)}$. If the functions ${\displaystyle g,h}$ are improperly integrable on ${\displaystyle (a,b)}$ and ${\displaystyle h(x)\leq f(x)\leq g(x)}$ for all ${\displaystyle x\in (a,b)}$, then ${\displaystyle f}$ is also improperly integrable on ${\displaystyle (a,b)}$ and

${\displaystyle \int _{a}^{b}h(x)\,\mathrm {d} x\leq \int _{a}^{b}f(x)\,\mathrm {d} x\leq \int _{a}^{b}g(x)\,\mathrm {d} x}$
Note: This theorem is derived from theorems 1 and 2. Also, ${\displaystyle h}$ and ${\displaystyle g}$ need not be identical.

### Examples

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

{\displaystyle {\begin{aligned}\int _{0}^{1}{\frac {1}{\sqrt {x}}}\,\mathrm {d} x&=\lim _{c\to 0^{+}}\int _{c}^{1}{\frac {1}{\sqrt {x}}}\,\mathrm {d} x\\&=\lim _{c\to 0^{+}}\left.2{\sqrt {x}}\right|_{x=c}^{1}\\&=\lim _{c\to 0^{+}}\left(2-2{\sqrt {c}}\right)\\&=2\end{aligned}}}

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

{\displaystyle {\begin{aligned}\int _{1}^{\infty }&=\lim _{c\to +\infty }\int _{1}^{c}x^{-2}\,\mathrm {d} x\\&=\lim _{c\to +\infty }\left.-x^{-1}\right|_{x=1}^{c}\\&=\lim _{c\to +\infty }\left(1-c^{-1}\right)\\&=1\end{aligned}}}

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

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

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

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

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

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

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

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

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

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

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

{\displaystyle {\begin{aligned}\int _{1}^{c}x^{-1}\,\cos {x}\,\mathrm {d} x&=-\int _{1}^{c}x^{-1}\,\mathrm {d} \left(\cos {x}\right)\\&=\left.-x^{-1}\,\cos {x}\right|_{x=1}^{c}+\int _{1}^{c}\cos {x}\,\,\mathrm {d} \left(x^{-1}\right)\\&=\cos {1}-c^{-1}\,\cos {c}-\int _{1}^{c}x^{-2}\,\cos {x}\,\mathrm {d} x\end{aligned}}}

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

## Absolute Integrability

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

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

${\displaystyle \left|\int _{a}^{b}f(x)\,\mathrm {d} x\right|\leq \int _{a}^{b}\left|f(x)\right|\,\mathrm {d} x}$

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

${\displaystyle -\int _{a}^{b}\left|f(x)\right|\,\mathrm {d} x\leq \int _{a}^{b}f(x)\,\mathrm {d} x\leq \int _{a}^{b}\left|f(x)\right|\,\mathrm {d} x}$
quod erat demonstrandum

### Examples

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

In particular,

• The function ${\displaystyle f_{1}(x)=x^{-2}}$ is absolutely integrable on ${\displaystyle \left[1,\infty \right)}$ and is not on ${\displaystyle \left(0,\infty \right)}$.
• The function ${\displaystyle f_{2}={\frac {1}{\sqrt {x}}}}$ is absolutely integrable on ${\displaystyle (0,1)}$.
• The function ${\displaystyle f_{3}(x)=\mathrm {e} ^{-x^{2}}}$ is absolutely integrable on ${\displaystyle \left(-\infty ,\infty \right)}$

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

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

### Counterexamples

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

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

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

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

{\displaystyle {\begin{aligned}\int _{n\,\pi }^{(n+1)\,\pi }\left|f(x)\right|\,\mathrm {d} x&\geq \int _{n\,\pi }^{(n+1)\,\pi }{\frac {\left|\sin {x}\right|}{(n+1)\,\pi }}\,\mathrm {d} x\\&={\frac {1}{(n+1)\,\pi }}\int _{0}^{\pi }\sin {x}\,\mathrm {d} x\\&={\frac {2}{(n+1)\,\pi }}&\geq {\frac {1}{n\,\pi }}&\geq {\frac {1}{\pi }}\,\int _{n\,\pi }^{(n+1)\,\pi }{\frac {1}{x}}\,\mathrm {d} x\end{aligned}}}

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

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