Improper integrals

As we have seen so far, the definite integral ${\int }_a^{b}f(x)dx$ is defined for the (continuous) function $f(x)$ on a bounded closed interval $[a,b]$. It is not defined on semi-open intervals such as $(a,b]$ or $[a,b)$, or on unbounded intervals such as $[a,\mathrm{\infty })$ or $(-\mathrm{\infty },\mathrm{\infty })$.

Nevertheless, we may extend the definition of definite integrals to deal with such cases. For example, the function $f(x)$ on $[a,b)$ is not defined on $x=b$, but if the left limit $\underset{t\to b-0}{lim}{\int }_a^{t}f(x)dx$ exists, we may define it as ${\int }_a^{b}f(x)dx$. Such an extended notion of the integral is called the improper integral.

Integration on semi-open intervals

For the continuous function $f(x)$ on the semi-open interval $[a,b)$, if the limit

$$\underset{t\to b-0}{lim}{\int }_a^{t}f(x)dx=\underset{\epsilon \to +0}{lim}{\int }_a^{b-\epsilon }f(x)dx$$

exists, we say that the improper integral ${\int }_a^{b}f(x)dx$ converges.

Similarly, for the continuous function $f(x)$ on $[a,\mathrm{\infty })$, the limit

$$\underset{t\to \mathrm{\infty }}{lim}{\int }_a^{t}f(x)dx$$

exists, we say that the improper integral ${\int }_a^{\mathrm{\infty }}f(x)dx$ converges.

We define the improper integrals in the same manner for continuous functions on $(a,b]$ or on $(-\mathrm{\infty },b]$ as

$$\begin{array}{rcl}{\int }_a^{b}f(x)dx& =& \underset{\epsilon \to +0}{lim}{\int }_{a+\epsilon }^{b}f(x)dx,\\ {\int }_{-\mathrm{\infty }}^{b}f(x)dx& =& \underset{t\to -\mathrm{\infty }}{lim}{\int }_t^{b}f(x)dx,\end{array}$$

respectively.

Example. {eg:sqrt1x} The function $f(x)=\frac{1}{\sqrt{1-x}}$ is not defined at $x=1$ so ${\int }_0^1\frac{dx}{\sqrt{1-x}}$ is an improper integral. 

$${\int }_0^{1-\epsilon }\frac{dx}{\sqrt{1-x}}={\int }_0^{1-\epsilon }(1-x{)}^{-\frac{1}{2}}dx={[-2\sqrt{1-x}]}_0^{1-\epsilon }=-2\sqrt{\epsilon }+2.$$

Therefore,

$${\int }_0^1\frac{dx}{\sqrt{1-x}}=\underset{\epsilon \to +0}{lim}{\int }_0^{1-\epsilon }\frac{dx}{\sqrt{1-x}}=2.$$

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Example. 

$${\int }_0^{\mathrm{\infty }}{e}^{-x}dx=\underset{t\to \mathrm{\infty }}{lim}{\int }_0^t{e}^{-x}dx=\underset{t\to \mathrm{\infty }}{lim}{[-e^{-x}]}_0^t=\underset{t\to \mathrm{\infty }}{lim}(-e^{-t}+1)=1.$$

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Example. 

$${\int }_0^1\frac{dx}{x}=\underset{\epsilon \to +0}{lim}{\int }_{\epsilon }^1\frac{dx}{x}=\underset{\epsilon \to +0}{lim}{[\log x]}_{\epsilon }^1=\underset{\epsilon \to +0}{lim}(-\log \epsilon )=\mathrm{\infty }.$$□

Example. {eg:power} Let $a,b,k\in \mathbb{R}$ such that $a<b$. Then

$${\int }_a^b(b-x{)}^{k}dx={\int }_a^b(x-a{)}^{k}dx=\{\begin{array}{cc}\frac{(b-a{)}^{k+1}}{k+1}& (k>-1),\\ \text{diverges}& (k\le -1).\end{array}$$

Let's prove this.

$${\int }_a^{b-\epsilon }(b-x{)}^{k}dx=\{\begin{array}{cc}{[-\frac{(b-x{)}^{k+1}}{k+1}]}_a^{b-\epsilon }& (k\ne -1),\\ {[-\log (b-x)]}_a^{b-\epsilon }& (k=-1).\end{array}$$

If $k\ne -1$, the given equation is

$$-\frac{{\epsilon }^{k+1}}{k+1}+\frac{(b-a{)}^{k+1}}{k+1}\to \{\begin{array}{cc}\frac{(b-a{)}^{k+1}}{k+1}& (k+1>0),\\ -\mathrm{\infty }& (k+1<0)\end{array}$$

as $\epsilon \to +0$.

If $k=-1$, the given equation is

$$-\log (\epsilon )+\log (b-a)\to \mathrm{\infty }(\epsilon \to +0).$$

The case for ${\int }_a^b(x-a{)}^{k}dx$ is similarly proved. □

Integration on open intervals

Suppose $f(x)$ is a continuous function on $(a,b)$. Pick an arbitrary $c\in (a,b)$. If both of the improper integrals ${\int }_a^{c}f(x)dx$ on $(a,c]$ and ${\int }_c^{b}f(x)dx$ on $[c,b)$ exist, we can define the improper integral on $(a,b)$ as

$${\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx.$$

Similarly, we can define the integral of the continuous function $f(x)$ on $(-\mathrm{\infty },\mathrm{\infty })$ as

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)dx={\int }_{-\mathrm{\infty }}^{c}f(x)dx+{\int }_c^{\mathrm{\infty }}f(x)dx.$$

Example. The function $f(x)=\frac{1}{\sqrt{1-x^2}}$ is undefined at $x=\pm 1$ so ${\int }_{-1}^1\frac{dx}{\sqrt{1-x^2}}$ is an improper integral.

$$\begin{array}{rcl}{\int }_{-1}^1\frac{dx}{\sqrt{1-x^2}}& =& \underset{\epsilon \to +0}{lim}{\int }_{-1+\epsilon }^0\frac{dx}{\sqrt{1-x^2}}+\underset{{\epsilon }^{\prime }\to +0}{lim}{\int }_0^{1-{\epsilon }^{\prime }}\frac{dx}{\sqrt{1-x^2}}\\ & =& {[\arcsin x]}_{-1+\epsilon }^0+{[\arcsin x]}_0^{1-{\epsilon }^{\prime }}\\ & =& -\arcsin (-1+\epsilon )+\arcsin (1-{\epsilon }^{\prime })\\ & \to & -(-\frac{\pi }{2})+\frac{\pi }{2}=\pi .\end{array}$$

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Remark. Computing improper integrals as

$$\begin{array}{rcl}{\int }_a^{b}f(x)dx& =& \underset{\epsilon \to +0}{lim}{\int }_{a+\epsilon }^{b-\epsilon }f(x)dx,\\ {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(x)dx& =& \underset{t\to \mathrm{\infty }}{lim}{\int }_{-t}^{t}f(x)dx,\end{array}$$

is wrong! 

Consider ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}xdx$.

$$\begin{array}{rcl}{\int }_{-\mathrm{\infty }}^{0}xdx& =& {\left[\frac{1}{2}{x}^2\right]}_{-\mathrm{\infty }}^0=-\mathrm{\infty },\\ {\int }_0^{\mathrm{\infty }}xdx& =& {\left[\frac{1}{2}{x}^2\right]}_0^{\mathrm{\infty }}=\mathrm{\infty }\end{array}$$

so the improper integral does not converge. However

$$\underset{t\to \mathrm{\infty }}{lim}{\int }_{-t}^{t}xdx=\underset{t\to \mathrm{\infty }}{lim}{\left[\frac{1}{2}{x}^2\right]}_{-t}^t=0.$$

Note that $\mathrm{\infty }-\mathrm{\infty }$ is undefined and hence, is not equal to 0. □

A function that is continuous except finitely many points on an interval

Suppose that $f(x)$ is a function on $[a,b]$ that is continuous except finitely many points in $[a,b]$. Let $a_1,a_2,\cdots ,a_n$ be those discontinuous points such that $a<a_1<a_2<\cdots <a_n<b$. If the $n+1$ improper integrals 

$${\int }_a^{a_1}f(x)dx,{\int }_{a_1}^{a_2}f(x)dx,\cdots ,{\int }_{a_n}^{b}f(x)dx$$

all converge, then we say that the improper integral ${\int }_a^{b}f(x)dx$ converges and define

$${\int }_a^{b}f(x)dx={\int }_a^{a_1}f(x)dx+{\int }_{a_1}^{a_2}f(x)dx+\cdots +{\int }_{a_n}^{b}f(x)dx.$$

Example. The function  $f(x)=\frac{1}{\sqrt{|x|}}$ is not defined at $x=0$ so, for example, ${\int }_{-1}^1\frac{dx}{\sqrt{|x|}}$ is an improper integral (otherwise $f(x)$ is continuous).  As $\epsilon ,{\epsilon }^{\prime }\to +0$,

$$\begin{array}{rcl}{\int }_{-1}^{-\epsilon }\frac{dx}{\sqrt{|x|}}+{\int }_{{\epsilon }^{\prime }}^1\frac{dx}{\sqrt{|x|}}& =& {\int }_{-1}^{-\epsilon }\frac{dx}{\sqrt{-x}}+{\int }_{{\epsilon }^{\prime }}^1\frac{dx}{\sqrt{x}}\\ & =& {[-2\sqrt{-x}]}_{-1}^{-\epsilon }+{\left[2\sqrt{x}\right]}_{{\epsilon }^{\prime }}^1\\ & =& -2\sqrt{\epsilon }+2+2-2\sqrt{{\epsilon }^{\prime }}\\ & \to & 4.\end{array}$$

Hence this improper integral converges and ${\int }_{-1}^1\frac{dx}{\sqrt{|x|}}=4$. □

Convergence criteria of improper integrals

In many cases, computing the value of improper integrals is very difficult, if not impossible. Yet, we may be able to decide whether an improper integral converges or not.

Theorem

Let $f(x)$ and $g(x)$ be continuous functions on the interval $(a,b]$ such that the following two conditions are satisfied:

1. For all $x\in (a,b]$, $|f(x)|\le g(x)$.
2. The improper integral ${\int }_a^{b}g(x)dx$ converges.

Then the improper integral ${\int }_a^{b}f(x)dx$ converges.

Remark. The function $g(x)$ in this theorem is called a dominating function of $f(x)$. □

Proof. Let $F(t)={\int }_t^{b}f(x)dx$ and $G(t)={\int }_t^{b}g(x)dx$. By the condition 2, the limit $\underset{t\to a+0}{lim}G(t)$ exists. By the Cauchy criterion, for any $\epsilon >0$, there exists a $\delta >0$ such that if $t,s\in (a,a+\delta ]$ then $|G(t)-G(s)|<\epsilon$. Then

$$\begin{array}{rcl}|F(t)-F(s)|& =& |{\int }_t^{s}f(x)dx|\\ & \le & {\int }_t^s|f(x)|dx\\ & \le & {\int }_t^{s}g(x)dx\\ & =& G(t)-G(s)<\epsilon .\end{array}$$

Thus, again by the Cauchy criterion, the limit $\underset{t\to a+0}{lim}F(t)$ exists.  In other words, the improper integral ${\int }_a^{b}f(x)dx$ converges. ■

See also: For the Cauchy criterion of convergence of a function, see More on the limit of functions. □

Example. Let $f(x)=\frac{\sin x}{\sqrt{1-x}}$ and $g(x)=\frac{1}{\sqrt{1-x}}$ defined on $[0,1)$. We can see (why?)

$$\left|\frac{\sin x}{\sqrt{1-x}}\right|\le \frac{1}{\sqrt{1-x}}$$

for all $x\in [0,1)$. Therefore $g(x)$ is a dominating function of $f(x)$. Furthermore, we have seen in a previous example (Example {eg:sqrt1x}) that

$${\int }_0^1\frac{dx}{\sqrt{1-x}}=2.$$

Therefore the improper integral ${\int }_0^1\frac{\sin x}{\sqrt{1-x}}dx$ converges. □

Corollary

Let $f(x)$ and $g(x)$ be continuous functions on $(a,b]$. The improper integral ${\int }_a^{b}f(x)dx$ converges if the following three conditions are satisfied.

1. For all $x\in (a,b]$, $g(x)>0$.
2. $\frac{f(x)}{g(x)}$ is bounded on $(a,b]$.
3. The improper integral ${\int }_a^{b}g(x)dx$ converges.

Proof. Since $\frac{f(x)}{g(x)}$ is bounded on $(a,b]$, there exists some $M\in \mathbb{R}$ such that, for all $x\in (a,b]$,

$$\left|\frac{f(x)}{g(x)}\right|=\frac{|f(x)|}{g(x)}\le M.$$

Hence $|f(x)|\le Mg(x)$ holds and $Mg(x)$ is a dominating function of $f(x)$. Therefore ${\int }_a^{b}f(x)dx$ converges. ■

Example. Let $f(x)$ be a continuous function on $(a,b]$. If there exists a real number $k<1$ such that $f(x)(x-a{)}^k$ is bounded on $(a,b]$, then the improper integral ${\int }_a^{b}f(x)dx$ converges. In fact, if we define $g(x)=(x-a{)}^{-k}$, since $-k>-1$, the improper integral ${\int }_a^{b}g(x)dx$ converges (c.f., Example {eg:power}). Therefore, by the above corollary, ${\int }_a^{b}f(x)dx$ converges. □