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 $F(x)=\int _{a}^{x}f(t)\,\mathrm {d} t$ is well-defined and continuous on $[a,b]$. In particular, $F(c)\to F(b)$ as $c\to b^{-}$, that is,

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

Now suppose $f$ is defined on the semi-open interval $J=\left[a,b\right)$ 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 $\left[a,b\right)$.

Similar definition for $\left(a,b\right]$: 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 $\left(a,c\right]$ and $\left[c,b\right)$. The integral of $f$ is defined by

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

Properties

Suppose a function $f$ is locally integrable on a semi-open interval $J=\left[a,b\right)$ or $J=\left(a,b\right]$. 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 $f:\left[a,b\right)\to \mathbb {R}$ be integrable on any closed interval $[a_{1},b_{1}]\subset \left[a,b\right)$. Given $c\in (a,b)$, the function $f$ is improperly integrable on $\left[c,b\right)$ if and only if it is improperly integrable on $\left[a,b\right)$. In the case of integrability,

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

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

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

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

quod erat demonstrandum

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

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

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

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

If a function $f$ is integrable on a closed interval $[a,b]$ or improperly integrable on one of the semi-open intervals $\left[a,b\right)$ and $\left(a,b\right]$, 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

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

Suppose a function $f:(a,b)\to \mathbb {R}$ 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

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

Comparison Theorems for Improper Integrals

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

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

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

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

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

Examples

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $x\to \infty$. To show improper integrability, we integrate by parts:

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

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

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

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

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

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

quod erat demonstrandum

Examples

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

In particular,

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

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

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

Counterexamples

The function $f(x)={\begin{cases}1&x\in \mathbb {Q} \\-1&x\in \mathbb {R} \setminus \mathbb {Q} \end{cases}}$ 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 $\left[1,\infty \right)$.

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

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

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

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