Improper Integrals

The definition of a definite integral:

$\int \limits _{a}^{b}f(x)dx$

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. (If you are not familiar with L'Hôpital's rule, it is a good idea to review it before reading this section.)

Improper Integrals with Infinite Limits of Integration

Consider the integral

$\int \limits _{1}^{\infty }{\frac {dx}{x^{2}}}$

Assigning a finite upper bound $b$ in place of infinity gives

$\lim _{b\to \infty }\int \limits _{1}^{b}{\frac {dx}{x^{2}}}=\lim _{b\to \infty }\left({\frac {1}{1}}-{\frac {1}{b}}\right)=\lim _{b\to \infty }\left(1-{\frac {1}{b}}\right)=1$

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

Definition

1. Suppose $\int \limits _{a}^{b}f(x)dx$ exists for all $b\geq a$ . Then we define

$\int \limits _{a}^{\infty }f(x)dx=\lim _{b\to \infty }\int \limits _{a}^{b}f(x)dx$ , 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 $\int \limits _{a}^{b}f(x)dx$ exists for all $a\leq b$ we define

$\int \limits _{-\infty }^{b}f(x)dx=\lim _{a\to -\infty }\int \limits _{a}^{b}f(x)dx$

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

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

We claim that

$\int \limits _{0}^{\infty }e^{-x}dx=1$

To do this we calculate

$\int \limits _{0}^{\infty }e^{-x}dx$	$=\lim _{b\to \infty }\int \limits _{0}^{b}e^{-x}dx$
	$=\lim _{b\to \infty }(-e^{-x}){\Bigr |}_{0}^{b}$
	$=\lim _{b\to \infty }\left(-e^{-b}+1\right)$
	$=1$

We claim that the integral

$\int \limits _{1}^{\infty }{\frac {dx}{x}}$ diverges.

This follows as

$\int \limits _{1}^{\infty }{\frac {dx}{x}}$	$=\lim _{b\to \infty }\int \limits _{1}^{b}{\frac {dx}{x}}$
	$=\lim _{b\to \infty }\ln(x){\Bigr |}_{1}^{b}$
	$=\lim _{b\to \infty }\left(\ln(b)-0\right)$
	$=\infty$

Therefore

$\int \limits _{1}^{\infty }{\frac {dx}{x}}$ diverges.

Find $\int \limits _{0}^{\infty }x^{2}e^{-x}dx$ .

To calculate the integral use integration by parts twice to get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_0^b x^2e^{-x}dx}	$=(-x^{2}e^{-x}){\Bigr |}_{0}^{b}+2\int \limits _{0}^{b}xe^{-x}dx$
	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =-b^2e^{-b}+2\left((-x e^{-x})\Bigr|_0^b+\int\limits_0^b e^{-x}dx\right)}
	$=-b^{2}e^{-b}+2\left(-be^{-b}-(e^{-x}){\Bigr |}_{0}^{b}\right)$
	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =-b^2 e^{-b}+2(-be^{-b}-e^{-b}+1)}

Now Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{b\to\infty}e^{-b}=0} and because exponentials overpower polynomials, we see that $\lim _{b\to \infty }b^{2}e^{-b}=0$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{b\to\infty}be^{-b}=0} as well. Hence,

$\int \limits _{0}^{\infty }x^{2}e^{-x}dx=\lim _{b\to \infty }\int \limits _{0}^{b}x^{2}e^{-x}dx=0+2(0-0+1)=2$

Show Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_1^\infty\frac{dx}{x^p}=\begin{cases}\frac{1}{p-1},&\text{if }p>1\\ \text{diverges},&\text{if }p\le1\end{cases}}

If Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p\ne 1} then

$\int \limits _{1}^{\infty }{\frac {dx}{x^{p}}}$	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\lim_{b\to\infty}\int\limits_1^b x^{-p}dx}
	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\lim_{b\to\infty}\left(\frac{x^{-p+1}}{-p+1}\right)\Bigg|_1^b}
	$=-{\frac {1}{1-p}}\lim _{b\to \infty }\left(b^{-p+1}-1\right)$
	Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\begin{cases}\frac{1}{p-1},&\text{if }p>1\\ \text{diverges},&\text{if }p<1\end{cases}}

Notice that we had to assume that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle p\ne 1} to avoid dividing by 0. However the $p=1$ case was done in a previous example. Template:Robox/Close

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

$\int \limits _{a}^{b}f(x)dx=\lim _{c\to b^{-}}\int \limits _{a}^{c}f(x)dx$

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

$\int \limits _{a}^{b}f(x)dx=\lim _{c\to a^{+}}\int \limits _{c}^{b}f(x)dx$

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

$\int \limits _{a}^{b}f(x)dx$=$\int \limits _{a}^{c}f(x)dx+\int \limits _{c}^{b}f(x)dx$

Show $\int \limits _{0}^{1}{\frac {dx}{x^{p}}}={\begin{cases}{\frac {1}{1-p}},&{\text{if }}p<1\\{\text{diverges}},&{\text{if }}p\geq 1\end{cases}}$

If $p\neq 1$ then

$\int \limits _{0}^{1}{\frac {dx}{x^{p}}}$	$=\lim _{a\to 0^{+}}\int \limits _{a}^{1}x^{-p}dx$
	$=\lim _{a\to 0^{+}}\left({\frac {x^{-p+1}}{-p+1}}\right){\Bigg |}_{a}^{1}$
	$=-{\frac {1}{1-p}}\lim _{a\to 0^{+}}\left(1-a^{-p+1}\right)$
	$={\begin{cases}{\frac {1}{1-p}},&{\mbox{if }}p<1\\{\text{diverges}},&{\text{if }}p>1\end{cases}}$

Notice that we had to assume that $p\neq 1$ do avoid dividing by 0. So instead we do the $p=1$ case separately,

$\int \limits _{0}^{1}{\frac {dx}{x}}=\lim _{a\to 0^{+}}\left[\ln {\big (}|x|{\big )}{\Big |}_{a}^{1}\right]=\lim _{a\to 0^{+}}{\Big [}-\ln(a){\Big ]}$

which diverges. Template:Robox/Close

Template:ExampleRobox

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

$\ln {\big (}|x-2|{\big )}{\Big |}_{-1}^{3}=\ln(5)-\ln(3)=\ln \left({\tfrac {5}{3}}\right)$

which is not correct. In fact the integral diverges since

${\begin{aligned}\int \limits _{-1}^{3}{\frac {dx}{x-2}}&=\lim _{b\to 2^{-}}\int \limits _{-1}^{b}{\frac {dx}{x-2}}+\lim _{a\to 2^{+}}\int \limits _{a}^{3}{\frac {dx}{x-2}}\\&=\lim _{b\to 2^{-}}\ln {\big (}|x-2|{\big )}{\Big |}_{-1}^{b}+\lim _{a\to 2^{+}}\ln {\big (}|x-2|{\big )}{\Big |}_{a}^{3}\\&=\lim _{b\to 2^{-}}{\Big [}\ln(2-b)-\ln(3){\Big ]}+\lim _{a\to 2^{+}}{\Big [}\ln(1)-\ln(a-2){\Big ]}\\&=\lim _{b\to 2^{-}}{\Big [}\ln(2-b){\Big ]}-\ln(3)+\lim _{a\to 2^{+}}{\Big [}-\ln(a-2){\Big ]}\\\end{aligned}}$

and $\lim _{b\to 2^{-}}{\Big [}\ln(2-b){\Big ]}$ and $\lim _{a\to 2^{+}}{\Big [}-\ln(a-2){\Big ]}$ both diverge. Template:Robox/Close

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 $c_{1}<c_{2}<\ldots <c_{n}$ in $[a,b]$ . We define Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align}\int\limits_a^b f(x)dx=\int\limits_a^{c_1}f(x)dx+\int\limits_{c_1}^{c_2}f(x)dx+\int\limits_{c_2}^{c_3}f(x)dx+\cdots+\int\limits_{c_{n-1}}^{c_n} f(x)dx+\int\limits_{c_n}^b f(x)dx\end{align}} 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.

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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f,g} be continuous functions defined for all $x\geq a$ .

1. Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)\ge f(x)\ge0} for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x\ge a} . Then if $\int \limits _{a}^{\infty }g(x)dx$ converges so does Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^\infty f(x)dx} .
2. Suppose Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)\ge h(x)\ge0} for all $x\geq a$ . Then if Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^\infty h(x)dx} diverges so does Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^\infty f(x)dx} .

A similar theorem holds for improper integrals of the form Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_{-\infty}^b f(x)dx} and for improper integrals with discontinuities.

Show that $\int \limits _{1}^{\infty }{\frac {\sin(x)+2}{x^{2}}}dx$ converges.

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

$0\leq {\frac {\sin(x)+2}{x^{2}}}\leq {\frac {3}{x^{2}}}$ .

We have seen that $\int \limits _{1}^{\infty }{\frac {3}{x^{2}}}dx=3\int \limits _{1}^{\infty }{\frac {dx}{x^{2}}}$ 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 $\int \limits _{1}^{\infty }{\frac {\sin(x)+2}{x^{2}}}dx$ converges as well. Template:Robox/Close

Show that $\int \limits _{1}^{\infty }{\frac {\sin(x)+2}{x}}dx$ diverges.

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

${\frac {\sin(x)+2}{x}}\geq {\frac {1}{x}}\geq 0$

We have seen that $\int \limits _{1}^{\infty }{\frac {dx}{x}}$ diverges. So putting $f(x)={\frac {\sin(x)+2}{x}}$ and $g(x)={\frac {1}{x}}$ into the comparison test we get that $\int \limits _{1}^{\infty }{\frac {\sin(x)+2}{x}}dx$ diverges as well. Template:Robox/Close

An extension of the comparison theorem

To apply the comparison theorem you do not really need $g(x)\geq f(x)\geq 0$ for all $x\geq a$ . What we actually need is this inequality holds for sufficiently large $x$ (i.e. there is a number Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} such that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)\ge f(x)} for all $x\geq c$). For then

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_a^\infty f(x)dx=\int\limits_a^c f(x)dx+\int\limits_c^\infty f(x)dx}

so the first integral converges if and only if third does, and we can apply the comparison theorem to the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_c^\infty f(x)dx} piece.

Template:ExampleRobox

Show that $\int \limits _{1}^{\infty }{\sqrt {\frac {x^{7}}{e^{3x}}}}dx$ converges.

The reason that this integral converges is because for large Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-x}} factor in the integrand is dominant. We could try comparing Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^\frac{7}{2}e^{-x}} with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{-x}} , but as $x\geq 1$ , the inequality

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^\frac{7}{2}e^{-x}\ge e^{-x}}

is the wrong way around to show convergence.

Instead we rewrite the integrand as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^\frac{7}{2}e^{-\frac{3x}{2}}dx=x^\frac{7}{2}e^{-\frac{x}{2}}e^{-x}dx} .

Since the limit $\lim _{x\to \infty }{\Big [}x^{\frac {7}{2}}e^{-{\frac {x}{2}}}{\Big ]}=0$ we know that for Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} sufficiently large we have Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^\frac{7}{2}e^{-\frac{x}{2}}\le1} . So for large $x$ ,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x^\frac{7}{2}e^{-\frac{7x}{2}}=x^\frac{7}{2}e^{-\frac{x}{2}}e^{-x}\le e^{-x}}

Since the integral Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int\limits_1^\infty e^{-x} dx} converges the comparison test tells us that $\int \limits _{1}^{\infty }{\sqrt {\frac {x^{7}}{e^{3x}}}}dx$ converges as well.

Resources

Improper Integrals by James Sousa

Ex 1: Improper Integral - Infinite Interval (-inf,+inf) by James Sousa

Ex 2: Improper Integral - Infinite Interval (-inf, constant) by James Sousa

Ex 3: Improper Integral - Infinite Interval (-inf,+inf) by James Sousa

Improper Integral - Basic Idea and Example by patrickJMT

Improper Integral - More Complicated Example by patrickJMT

Improper Integral - Infinity in Upper and Lower Limits by patrickJMT

What makes an integral improper? by Krista King

What makes an integral improper? by Krista King

Improper Integrals by Krista King

Improper Integral Example 2 PART 1/2 by Krista King

Improper Integral Example 2 PART 2/2 by Krista King