Calculus/Further Methods of Integration/Improper Integrals

The definition of a definite integral:

$\underset{a}{\overset{b}{\int }}f(x)dx$

requires the interval [a,b] be finite. The Fundamental Theorem of Calculus requires that f be continuous on [a,b]. In this section, you will be studying a method of evaluating integrals that fail these requirements—either because their limits of integration are infinite, or because a finite number of discontinuities exist on the interval [a,b]. Integrals that fail either of these requirements are improper integrals.

L'Hopital's Rule

L'Hopital's Rule is included in this section because limits involving infinity often appear in improper integration. L'Hopital's Rule describes how to evaluate limits involving infinity and or 0 if the limit evaluates to an indeterminate form.

All of the following expressions are indeterminate forms.

$\frac{\mathrm{\infty }}{\mathrm{\infty }},\frac{-\mathrm{\infty }}{-\mathrm{\infty }},\frac{0}{0},1^{\mathrm{\infty }},1\cdot \mathrm{\infty }$

These expressions are called indeterminate because you cannot determine their exact value in the indeterminate form. Depending on the situation, each indeterminate form could evaluate to a variety of values.

Definition

If $\underset{x\to c}{\lim }\frac{f(x)}{g(x)}$ is indeterminate

Then $\underset{x\to c}{\lim }\frac{f(x)}{g(x)}=\underset{x\to c}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}$

Note:

$x$ can approach a finite value c, $\mathrm{\infty }$ or $-\mathrm{\infty }$.

Example:

One might think the value $1^{\mathrm{\infty }}=1.$

Consider

$\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{x})}^x$

Plugging the value of x into the limit yields

$\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{x})}^x=1^{\mathrm{\infty }}$ (indeterminate form).

Let $k=\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{x})}^x=1^{\mathrm{\infty }}$

$\ln k$	=	$\underset{x\to \mathrm{\infty }}{\lim }\ln {(1+\frac{1}{x})}^x$
	=	$\underset{x\to \mathrm{\infty }}{\lim }x\ln (1+\frac{1}{x})$
	=	$\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln (1+\frac{1}{x})}{\frac{1}{x}}=\frac{\ln 1}{\frac{1}{x}}=\frac{0}{0}$ (indeterminate form)

We now apply L'Hopital's Rule by taking the derivative of the top and bottom with respect to x.

$\frac{d}{dx}[\ln (1+\frac{1}{x})]=\frac{x}{x+1}\cdot \frac{-1}{x^2}=\frac{-1}{x(x+1)}$

$\frac{d}{dx}\left(\frac{1}{x}\right)=\frac{-1}{x^2}$

Returning to the expression above

$\ln k$	=	$\underset{x\to \mathrm{\infty }}{\lim }\frac{-(-x^2)}{x(x+1)}$
	=	$\underset{x\to \mathrm{\infty }}{\lim }\frac{x}{x+1}=\frac{\mathrm{\infty }}{\mathrm{\infty }}$ (indeterminate form)

We apply L'Hopital's Rule once again

$\ln k=\underset{x\to \mathrm{\infty }}{\lim }\frac{1}{1}=1$

Therefore

$k=e$

And

$\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{x})}^x=e\ne 1$

Careful: this does not prove that $1^{\mathrm{\infty }}=e$ because

$\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{2}{x})}^x=1^{\mathrm{\infty }}\ne e$

Improper Integrals with Infinite Limits of Integration

Consider the integral

$\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^2}$

Assigning a finite upper bound b in place of infinity gives

$\underset{b\to \mathrm{\infty }}{\lim }\underset{1}{\overset{b}{\int }}\frac{dx}{x^2}=\underset{b\to \mathrm{\infty }}{\lim }(1-\frac{1}{b})=1$

This improper integral can be interpreted as the area of the unbounded region between f(x)=1/x^2, y=0 (the x-axis), and x=1.

Definition

1. Suppose $\underset{a}{\overset{b}{\int }}f(x)dx$ exists for all $b\ge a$. Then we define

$\underset{a}{\overset{\mathrm{\infty }}{\int }}f(x)dx$=$\underset{b\to \mathrm{\infty }}{\lim }\underset{a}{\overset{b}{\int }}f(x)dx,$ as long as this limit exists and is finite.

If it does exist we say the integral is convergent and otherwise we say it is divergent.

2. Similarly if $\underset{a}{\overset{b}{\int }}f(x)dx$ exists for all $a\le b$ we define

$\underset{-\mathrm{\infty }}{\overset{b}{\int }}f(x)dx$=$\underset{a\to -\mathrm{\infty }}{\lim }\underset{a}{\overset{b}{\int }}f(x)dx.$

3. Finally suppose c is a fixed real number and that $\underset{-\mathrm{\infty }}{\overset{c}{\int }}f(x)dx$ and $\underset{c}{\overset{\mathrm{\infty }}{\int }}f(x)dx$ are both convergent. Then we define

$\underset{-\mathrm{\infty }}{\overset{\mathrm{\infty }}{\int }}f(x)dx$=$\underset{-\mathrm{\infty }}{\overset{c}{\int }}f(x)dx+\underset{c}{\overset{\mathrm{\infty }}{\int }}f(x)dx.$

Example: Convergent Improper Integral

We claim that

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-x}dx=1$

To do this we calculate

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-x}dx$	=	$\underset{b\to \mathrm{\infty }}{\lim }\underset{0}{\overset{b}{\int }}{e}^{-x}dx$
	=	$\underset{b\to \mathrm{\infty }}{\lim }{[-e^{-x}]}_0^b$
	=	$\underset{b\to \mathrm{\infty }}{\lim }(-e^{-b}+1)$
	=	$1$

Example: Divergent Improper Integral

We claim that the integral

$\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x}$ diverges.

This follows as

$\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x}$	=	$\underset{b\to \mathrm{\infty }}{\lim }\underset{1}{\overset{b}{\int }}\frac{dx}{x}$
	=	$\underset{b\to \mathrm{\infty }}{\lim }{[\ln x]}_1^b$
	=	$\underset{b\to \mathrm{\infty }}{\lim }(\ln b-0)$
	=	$\mathrm{\infty }$

Therefore

$\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x}$ diverges.

Example: Improper Integral

Find $\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{-x}dx.$

To calculate the integral use integration by parts twice to get

$\begin{array}{ccc}\underset{0}{\overset{b}{\int }}{x}^2{e}^{-x}dx& =& [-x^2{e}^{-x}{]}_0^b+2\underset{2}{\overset{b}{\int }}x{e}^{-x}dx\\ & =& -b^2{e}^{-b}-[2x{e}^{-x}{]}_0^b+\underset{0}{\overset{b}{\int }}{e}^{-x}dx\\ & =& -b^2{e}^{-b}-2b{e}^{-b}-[e^{-x}{]}_0^b\\ & =& -b^2{e}^{-b}-2b{e}^{-b}-e^{-b}+1.\end{array}$

Now $\underset{b\to \mathrm{\infty }}{\lim }{e}^{-b}=0$ and because exponentials overpower polynomials, we see that $\underset{b\to \mathrm{\infty }}{\lim }{b}^2{e}^{-b}=0$ and $\underset{b\to \mathrm{\infty }}{\lim }b{e}^{-b}=0$ as well. Hence,

$\underset{0}{\overset{\mathrm{\infty }}{\int }}{x}^2{e}^{-x}dx=\underset{b\to \mathrm{\infty }}{\lim }\underset{0}{\overset{b}{\int }}{x}^2{e}^{-x}dx=0+0+1=1.$

Example: Powers

Show $\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^p}=\{\begin{array}{cc}\frac{1}{p-1},& \text{if }p>1\\ \text{diverges},& \text{if }p\le 1\end{array}$}}

If $p\ne 1$ then

$\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{dx}{x^p}$	=	$\underset{b\to \mathrm{\infty }}{\lim }\underset{1}{\overset{b}{\int }}{x}^{-p}dx$
	=	$\underset{b\to \mathrm{\infty }}{\lim }{\left[\frac{1}{-p+1}{x}^{-p+1}\right]}_1^b$
	=	$\frac{1}{1-p}\underset{b\to \mathrm{\infty }}{\lim }(b^{-p+1}-1)$
	=	$\{\begin{array}{cc}\frac{1}{p-1},& \text{if }p>1\\ \text{diverges},& \text{if }p<1.\end{array}$

Notice that we had to assume that $p\ne 1$ do avoid dividing by zero. However the $p=1$ case was done in a previous example.

Improper Integrals with a Finite Number Discontinuities

First we give a definition for the integral of functions which have a discontinuity at one point.

Definition of improper integrals with a single discontinuity

If f is continuous on the interval [a,b) and is discontinuous at b, we define :$\underset{a}{\overset{b}{\int }}f(x)dx$=$\underset{c\to b^{-}}{\lim }\underset{a}{\overset{c}{\int }}f(x)dx.$

If the limit in question exists we say the integral converges and otherwise we say it diverges.

Similarly if f is continuous on the interval(a,b] and is discontinuous at a, we define

$\underset{a}{\overset{b}{\int }}f(x)dx$=$\underset{c\to a^{+}}{\lim }\underset{c}{\overset{b}{\int }}f(x)dx.$

Finally suppose f has an discontinuity at a point c in (a,b) and is continuous at all other points in [a,b]. If $\underset{a}{\overset{c}{\int }}f(x)dx$ and $\underset{c}{\overset{b}{\int }}f(x)dx$ converge we define

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

Example 1

Show $\underset{0}{\overset{1}{\int }}\frac{dx}{x^p}=\{\begin{array}{cc}\frac{1}{1-p},& \text{if }p<1\\ \text{diverges},& \text{if }p\ge 1\end{array}$}}

If $p\ne 1$ then

$\underset{0}{\overset{1}{\int }}\frac{dx}{x^p}$	=	$\underset{a\to 0^{+}}{\lim }\underset{a}{\overset{1}{\int }}{x}^{-p}dx$
	=	$\underset{a\to 0^{+}}{\lim }{\left[\frac{1}{-p+1}{x}^{-p+1}\right]}_a^1$
	=	$\frac{1}{1-p}\underset{a\to 0^{+}}{\lim }(1-a^{-p+1})$
	=	$\{\begin{array}{cc}\frac{1}{1-p},& \text{if }p<1\\ \text{diverges},& \text{if }p>1.\end{array}$

Notice that we had to assume that $p\ne 1$ do avoid dividing by zero. So instead we do the $p=1$ case separately,

$\underset{0}{\overset{1}{\int }}\frac{1}{x}dx=\underset{a\to 0^{+}}{\lim }[\ln |x|{]}_a^0=\underset{a\to 0^{+}}{\lim }\ln a$

which diverges.

We can also give a definition of the integral of a function with a finite number of discontinuities

Definition: Improper integrals with finite number of discontinuities

Suppose f is continuous on [a,b] except at points $c_1<c_2<\dots <c_n$ in [a,b]. We define $\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{c_1}{\int }}f(x)dx+\underset{c_1}{\overset{c_2}{\int }}f(x)dx+\underset{c_2}{\overset{c_3}{\int }}f(x)dx+\cdots +\underset{c_{n-1}}{\overset{c_n}{\int }}f(x)dx+\underset{c_n}{\overset{b}{\int }}f(x)dx$ as long as each integral on the right converges.

Notice that by combining this definition with the definition for improper integrals with infinite endpoints, we can define the integral of a function with a finite number of discontinuities with one or more infinite endpoints.

Example 2

The integral $\underset{-1}{\overset{3}{\int }}\frac{1}{x-2}dx$ is improper because the integrand is not continuous at x=2. However had we not notice that we might have been tempted to apply the fundamental theorem of calculus and conclude that it equals

$[\ln |x-2|{]}_{-1}^3=\ln (5)-\ln (3)=\ln (5/3),$

which is not correct. In fact the integral diverges.

Comparison Test

There are integrals which cannot easily be evaluated. However it may still be possible to show they are convergent by comparing them to an integral we already know converges.

Theorem (Comparison Test) Let f and g be continuous functions defined for all $x\ge a$.

1. Suppose $g(x)\ge f(x)\ge 0$ for all $x\ge a$. Then if $\underset{a}{\overset{\mathrm{\infty }}{\int }}g(x)dx$ converges so does $\underset{a}{\overset{\mathrm{\infty }}{\int }}f(x)dx.$
2. Suppose $f(x)\ge h(x)\ge 0$ for all $x\ge a$. Then if $\underset{a}{\overset{\mathrm{\infty }}{\int }}h(x)dx$ diverges so does $\underset{a}{\overset{\mathrm{\infty }}{\int }}f(x)dx.$

A similar theorem holds for improper integrals of the form $\underset{-\mathrm{\infty }}{\overset{b}{\int }}f(x)dx$ and for improper integrals with discontinuities.

Example: Use of comparsion test to show convergence

Show that $\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x+2}{x^2}dx$ converges.

For all x we know that $-1\le \sin x\le 1$ so $1\le \sin x+2\le 3$. This implies that

$0\le \frac{\sin x+2}{x^2}\le \frac{3}{x^2}$.

We have seen that $\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{3}{x^2}dx=3\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{1}{x^2}dx$ converges. So putting $f(x)=\frac{\sin x+2}{x^2}$ and $g(x)=\frac{3}{x^2}$ into the comparison test we get that the integral $\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x+2}{x^2}dx$ converges as well.

Example: Use of Comparsion Test to show divergence

Show that $\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x+2}{x}dx$ diverges.

Just as in the previous example we know that $\sin x+2\ge 1$ for all x. Thus

$\frac{\sin x+2}{x}\ge \frac{1}{x}\ge 0.$

We have seen that $\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{1}{x}dx$ diverges. So putting $f(x)=\frac{\sin x+2}{x}$ and $g(x)=\frac{1}{x}$ into the comparison test we get that $\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{\sin x+2}{x}dx$ diverges as well.

An extension of the comparison theorem

To apply the comparison theorem you do not really need $g(x)\ge f(x)\ge 0$ for all $x\ge a$. What we actually need is this inequality holds for sufficiently large x (i.e. there is a number c such that $g(x)\ge f(x)$ for all $x\ge c$). For then

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

so the first integral converges if and only if third does, and we can apply the comparison theorem to the $\underset{c}{\overset{\mathrm{\infty }}{\int }}f(x)dx$ piece.

Example

Show that $\underset{1}{\overset{\mathrm{\infty }}{\int }}{x}^{7/2}{e}^{-\frac{3x}{2}}dx$ converges.

The reason that this integral converges is because for large x the $e^{-x}$ factor in the integrand is dominant. We could try comparing $x^{7/2}{e}^{-x}$ with $e^{-x}$, but as $x\ge 1$, the inequality

$x^{7/2}{e}^{-x}\ge e^{-x}$

is the wrong way around to show convergence.

Instead we rewrite the integrand as $x^{3/2}{e}^{-\frac{3x}{2}}dx=x^{7/2}{e}^{-\frac{x}{2}}{e}^{-x}dx.$

Since the limit $\underset{x\to \mathrm{\infty }}{\lim }{x}^{7/2}{e}^{-\frac{x}{2}}=0$ we know that for x sufficiently large we have $x^{3/2}{e}^{-\frac{x}{2}}\le 1$. So for large x,

$x^{7/2}{e}^{\frac{-7x}{2}}=x^{7/2}{e}^{-\frac{x}{2}}{e}^{-x}\le e^{-x}.$

Since the integral $\underset{1}{\overset{\mathrm{\infty }}{\int }}{e}^{-x}dx$ converges the comparison test tells us that $\underset{a}{\overset{\mathrm{\infty }}{\int }}{x}^{7/2}{e}^{\frac{-3x}{2}}$ converges as well.