Difference between revisions of "Improper Integrals"

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The definition of a definite integral:
 +
:<math>\int\limits_a^b f(x)dx</math>
 +
 +
requires the interval <math>[a,b]</math> be finite. The Fundamental Theorem of Calculus requires that <math>f</math> be continuous on <math>[a,b]</math> . In this section, you will be studying a method of evaluating integrals that fail these requirements&mdash;either because their limits of integration are infinite, or because a finite number of discontinuities exist on the interval <math>[a,b]</math> . Integrals that fail either of these requirements are '''improper integrals'''. (If you are not familiar with [[Calculus/L'Hôpital's rule|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
 +
:<math>\int\limits_1^\infty\frac{dx}{x^2}</math>
 +
Assigning a finite upper bound <math>b</math> in place of infinity gives
 +
:<math>\lim_{b\to\infty}\int\limits_1^b\frac{dx}{x^2}=\lim_{b\to\infty}\left(\frac11-\frac{1}{b}\right)=\lim_{b\to\infty}\left(1-\frac{1}{b}\right)=1</math>
 +
This improper integral can be interpreted as the area of the unbounded region between <math>f(x)=\frac{1}{x^2}</math> , <math>y=0</math> (the <math>x</math>-axis), and <math>x=1</math> .
 +
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
===Definition===
 +
1. Suppose <math>\int\limits_a^b f(x)dx</math> exists for all <math>b\ge a</math> . Then we define
 +
:<math>\int\limits_a^\infty f(x)dx=\lim_{b\to\infty}\int\limits_a^b f(x)dx</math> , 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 <math>\int\limits_a^b f(x)dx</math> exists for all <math>a\le b</math> we define
 +
:<math>\int\limits_{-\infty}^b f(x)dx=\lim_{a\to-\infty}\int\limits_a^b f(x)dx</math>
 +
3. Finally suppose <math>c</math> is a fixed real number and that <math>\int\limits_{-\infty}^c f(x)dx</math> and <math>\int\limits_c^\infty f(x)dx</math> are both convergent. Then we define
 +
:<math>\int\limits_{-\infty}^\infty f(x)dx=\int\limits_{-\infty}^c f(x)dx+\int\limits_c^\infty f(x)dx</math>
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</blockquote>
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 +
{{ExampleRobox|title=Example: Convergent Improper Integral}}
 +
 +
We claim that
 +
:<math>\int\limits_0^\infty e^{-x}dx=1</math>
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 +
To do this we calculate
 +
:{|
 +
|<math>\int\limits_0^\infty e^{-x}dx</math>
 +
|<math>=\lim_{b\to\infty}\int\limits_0^b e^{-x}dx</math>
 +
|-
 +
|
 +
|<math>=\lim_{b\to\infty}(-e^{-x})\Bigr|_0^b</math>
 +
|-
 +
|
 +
|<math>=\lim_{b\to\infty}\left(-e^{-b}+1 \right)</math>
 +
|-
 +
|
 +
|<math>=1</math>
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|}
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{{Robox/Close}}
 +
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{{ExampleRobox|title=Example: Divergent Improper Integral}}
 +
 +
We claim that the integral
 +
:<math>\int\limits_1^\infty \frac{dx}{x}</math> diverges.
 +
 +
This follows as
 +
:{|
 +
|<math>\int\limits_1^\infty\frac{dx}{x}</math>
 +
|<math>=\lim_{b\to\infty}\int\limits_1^b \frac{dx}{x}</math>
 +
|-
 +
|
 +
|<math>=\lim_{b\to\infty}\ln(x)\Bigr|_1^b</math>
 +
|-
 +
|
 +
|<math>=\lim_{b\to\infty}\left(\ln(b)-0\right)</math>
 +
|-
 +
|
 +
|<math>=\infty</math>
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|}
 +
 +
Therefore
 +
:<math>\int\limits_1^\infty\frac{dx}{x}</math> diverges.
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{{Robox/Close}}
 +
 +
{{ExampleRobox|title=Example: Improper Integral}}
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 +
Find <math>\int\limits_0^\infty x^2e^{-x}dx</math> .
 +
 +
To calculate the integral use integration by parts twice to get
 +
:{|
 +
|<math>\int\limits_0^b x^2e^{-x}dx</math>
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|<math>=(-x^2e^{-x})\Bigr|_0^b+2\int\limits_0^b xe^{-x}dx</math>
 +
|-
 +
|
 +
|<math>=-b^2e^{-b}+2\left((-x e^{-x})\Bigr|_0^b+\int\limits_0^b e^{-x}dx\right)</math>
 +
|-
 +
|
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|<math>=-b^2e^{-b}+2\left(-be^{-b}-(e^{-x})\Bigr|_0^b\right)</math>
 +
|-
 +
|
 +
|<math>=-b^2 e^{-b}+2(-be^{-b}-e^{-b}+1)</math>
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|}
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 +
Now <math>\lim_{b\to\infty}e^{-b}=0</math> and because exponentials overpower polynomials, we see that <math>\lim_{b\to\infty}b^2e^{-b}=0</math> and <math>\lim_{b\to\infty}be^{-b}=0</math> as well. Hence,
 +
:<math>\int\limits_0^\infty x^2e^{-x}dx=\lim_{b\to\infty}\int\limits_0^b x^2e^{-x}dx=0+2(0-0+1)=2</math>
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{{Robox/Close}}
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{{ExampleRobox|title=Example: Powers}}
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Show <math>\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}</math>
 +
 +
If <math>p\ne 1</math> then
 +
:{|
 +
|<math>\int\limits_1^\infty \frac{dx}{x^p}</math>
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|<math>=\lim_{b\to\infty}\int\limits_1^b x^{-p}dx</math>
 +
|-
 +
|
 +
|<math>=\lim_{b\to\infty}\left(\frac{x^{-p+1}}{-p+1}\right)\Bigg|_1^b</math>
 +
|-
 +
|
 +
|<math>=-\frac{1}{1-p}\lim_{b\to\infty}\left(b^{-p+1}-1\right)</math>
 +
|-
 +
|
 +
|<math>=\begin{cases}\frac{1}{p-1},&\text{if }p>1\\ \text{diverges},&\text{if }p<1\end{cases}</math>
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|}
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 +
Notice that we had to assume that <math>p\ne 1</math> to avoid dividing by 0. However the <math>p=1</math> case was done in a previous example.
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{{Robox/Close}}
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==Improper Integrals with a Finite Number Discontinuities==
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First we give a definition for the integral of functions which have a discontinuity at one point.
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 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
 +
===Definition of improper integrals with a single discontinuity===
 +
If <math>f</math> is continuous on the interval <math>[a,b)</math> and is discontinuous at <math>b</math> , we define
 +
:<math>\int\limits_a^b f(x)dx=\lim_{c\to b^-}\int\limits_a^c f(x)dx</math>
 +
If the limit in question exists we say the integral '''converges''' and otherwise we say it '''diverges'''.
 +
 +
Similarly if <math>f</math> is continuous on the interval <math>(a,b]</math> and is discontinuous at <math>a</math> , we define
 +
:<math>\int\limits_a^b f(x)dx=\lim_{c\to a^+}\int\limits_c^b f(x)dx</math>
 +
Finally suppose <math>f</math> has an discontinuity at a point <math>c\in(a,b)</math> and is continuous at all other points in <math>[a,b]</math> . If <math>\int\limits_a^c f(x)dx</math> and <math>\int\limits_c^b f(x)dx</math> converge we define
 +
:<math>\int\limits_a^b f(x)dx</math>=<math>\int\limits_a^c f(x)dx+\int\limits_c^b f(x)dx</math>
 +
</blockquote>
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{{ExampleRobox|title=Example 1}}
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Show <math>\int\limits_0^1\frac{dx}{x^p}=\begin{cases}\frac{1}{1-p},&\text{if }p<1\\ \text{diverges},&\text{if }p\ge1\end{cases}</math>
 +
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If <math>p\ne1</math> then
 +
:{|
 +
|<math>\int\limits_0^1\frac{dx}{x^p}</math>
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|<math>=\lim_{a\to 0^+}\int\limits_a^1 x^{-p}dx</math>
 +
|-
 +
|
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|<math>=\lim_{a\to 0^+}\left(\frac{x^{-p+1}}{-p+1}\right)\Bigg|_a^1</math>
 +
|-
 +
|
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|<math>=-\frac{1}{1-p}\lim_{a\to 0^+}\left(1-a^{-p+1}\right)</math>
 +
|-
 +
|
 +
|<math>=\begin{cases}\frac{1}{1-p},&\mbox{if }p<1\\ \text{diverges},&\text{if }p>1\end{cases}</math>
 +
|}
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Notice that we had to assume that <math>p\ne1</math> do avoid dividing by 0. So instead we do the <math>p=1</math> case separately,
 +
:<math>\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]</math>
 +
which diverges.
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{{Robox/Close}}
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 +
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{{ExampleRobox|title=Example 2}}
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The integral <math>\int\limits_{-1}^3\frac{dx}{x-2}</math> is improper because the integrand is not continuous at <math>x=2</math> . However had we not noticed that we might have been tempted to apply the fundamental theorem of calculus and conclude that it equals
 +
:<math>\ln\big(|x-2|\big)\Big|_{-1}^3=\ln(5)-\ln(3)=\ln\left(\tfrac{5}{3}\right)</math>
 +
which is not correct. In fact the integral diverges since
 +
:<math>\begin{align}\int\limits_{-1}^3\frac{dx}{x-2}&=\lim_{b\to2^-}\int\limits_{-1}^b \frac{dx}{x-2}+\lim_{a\to2^+}\int\limits_a^3\frac{dx}{x-2}\\
 +
&=\lim_{b\to2^-}\ln\big(|x-2|\big)\Big|_{-1}^b+\lim_{a\to2^+}\ln\big(|x-2|\big)\Big|_a^3\\
 +
&=\lim_{b\to2^-}\Big[\ln(2-b)-\ln(3)\Big]+\lim_{a\to2^+}\Big[\ln(1)-\ln(a-2)\Big]\\
 +
&=\lim_{b\to2^-}\Big[\ln(2-b)\Big]-\ln(3)+\lim_{a\to2^+}\Big[-\ln(a-2)\Big]\\
 +
\end{align}</math>
 +
and <math>\lim_{b\to2^-}\Big[\ln(2-b)\Big]</math> and <math>\lim_{a\to2^+}\Big[-\ln(a-2)\Big]</math> both diverge.
 +
{{Robox/Close}}
 +
 +
We can also give a definition of the integral of a function with a finite number of discontinuities.
 +
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
 +
===Definition: Improper integrals with finite number of discontinuities===
 +
Suppose <math>f</math> is continuous on <math>[a,b]</math> except at points <math>c_1<c_2<\ldots<c_n</math> in <math>[a,b]</math> . We define
 +
<math>\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}</math>
 +
as long as each integral on the right converges.
 +
</blockquote>
 +
 +
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.
 +
 +
<blockquote style="background: white; border: 1px solid black; padding: 1em;">
 +
'''Theorem (Comparison Test)'''
 +
Let <math>f,g</math> be continuous functions defined for all <math>x\ge a</math> .
 +
#Suppose <math>g(x)\ge f(x)\ge0</math> for all <math>x\ge a</math> . Then if <math>\int\limits_a^\infty g(x)dx</math> converges so does <math>\int\limits_a^\infty f(x)dx</math> .
 +
#Suppose <math>f(x)\ge h(x)\ge0</math> for all <math>x\ge a</math> . Then if <math>\int\limits_a^\infty h(x)dx</math> diverges so does <math>\int\limits_a^\infty f(x)dx</math> .
 +
</blockquote>
 +
 +
A similar theorem holds for improper integrals of the form <math>\int\limits_{-\infty}^b f(x)dx</math> and for improper integrals with discontinuities.
 +
 +
{{ExampleRobox|title=Example: Use of comparsion test to show convergence}}
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 +
Show that <math>\int\limits_1^\infty\frac{\sin(x)+2}{x^2}dx</math> converges.
 +
 +
For all <math>x</math> we know that <math>-1\le\sin(x)\le 1</math> so <math>1\le\sin(x)+2\le3</math> . This implies that
 +
:<math>0\le\frac{\sin(x)+2}{x^2}\le\frac{3}{x^2}</math> .
 +
We have seen that <math>\int\limits_1^\infty\frac{3}{x^2}dx=3\int\limits_1^\infty\frac{dx}{x^2}</math> converges. So putting <math>f(x)=\frac{\sin(x)+2}{x^2}</math> and <math>g(x)=\frac{3}{x^2}</math> into the comparison test we get that the integral <math>\int\limits_1^\infty\frac{\sin(x)+2}{x^2}dx</math> converges as well.
 +
{{Robox/Close}}
 +
 +
{{ExampleRobox|title=Example: Use of Comparsion Test to show divergence}}
 +
 +
Show that <math>\int\limits_1^\infty\frac{\sin(x)+2}{x}dx</math> diverges.
 +
 +
Just as in the previous example we know that <math>\sin(x)+2\ge1</math> for all <math>x</math> . Thus
 +
:<math>\frac{\sin(x)+2}{x}\ge\frac{1}{x}\ge0</math>
 +
We have seen that <math>\int\limits_1^\infty\frac{dx}{x}</math> diverges. So putting <math>f(x)=\frac{\sin(x)+2}{x}</math> and <math>g(x)=\frac{1}{x}</math> into the comparison test we get that <math>\int\limits_1^\infty\frac{\sin(x)+2}{x}dx</math> diverges as well.
 +
{{Robox/Close}}
 +
 +
===An extension of the comparison theorem===
 +
To apply the comparison theorem you do not really need <math>g(x)\ge f(x)\ge0</math> for ''all'' <math>x\ge a</math> . What we actually need is this inequality holds for sufficiently large <math>x</math> (i.e. there is a number <math>c</math> such that <math>g(x)\ge f(x)</math> for all <math>x\ge c</math>). For then
 +
:<math>\int\limits_a^\infty f(x)dx=\int\limits_a^c f(x)dx+\int\limits_c^\infty f(x)dx</math>
 +
so the first integral converges if and only if third does, and we can apply the comparison theorem to the <math>\int\limits_c^\infty f(x)dx</math> piece.
 +
 +
 +
{{ExampleRobox|title=Example}}
 +
 +
Show that <math>\int\limits_1^\infty\sqrt{\frac{x^7}{e^{3x}}}dx</math> converges.
 +
 +
The reason that this integral converges is because for large <math>x</math> the <math>e^{-x}</math> factor in the integrand is dominant. We could try comparing <math>x^\frac{7}{2}e^{-x}</math> with <math>e^{-x}</math> , but as <math>x\ge1</math> , the inequality
 +
:<math>x^\frac{7}{2}e^{-x}\ge e^{-x}</math>
 +
is the wrong way around to show convergence.
 +
 +
Instead we rewrite the integrand as <math>x^\frac{7}{2}e^{-\frac{3x}{2}}dx=x^\frac{7}{2}e^{-\frac{x}{2}}e^{-x}dx</math> .
 +
 +
Since the limit <math>\lim_{x\to\infty}\Big[x^\frac{7}{2}e^{-\frac{x}{2}}\Big]=0</math> we know that for <math>x</math> sufficiently large we have <math>x^\frac{7}{2}e^{-\frac{x}{2}}\le1</math> . So for large <math>x</math> ,
 +
:<math>x^\frac{7}{2}e^{-\frac{7x}{2}}=x^\frac{7}{2}e^{-\frac{x}{2}}e^{-x}\le e^{-x}</math>
 +
Since the integral <math>\int\limits_1^\infty e^{-x} dx</math> converges the comparison test tells us that <math>\int\limits_1^\infty\sqrt{\frac{x^7}{e^{3x}}}dx</math> converges as well.
 +
 +
 +
==Resources==
 
[https://youtu.be/F19y9IiH-cM Improper Integrals] by James Sousa
 
[https://youtu.be/F19y9IiH-cM Improper Integrals] by James Sousa
  

Revision as of 17:21, 6 October 2021

The definition of a definite integral:

requires the interval be finite. The Fundamental Theorem of Calculus requires that be continuous on . 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 . 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

Assigning a finite upper bound in place of infinity gives

This improper integral can be interpreted as the area of the unbounded region between , (the -axis), and .

Definition

1. Suppose exists for all . Then we define

, 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 exists for all we define

3. Finally suppose is a fixed real number and that and are both convergent. Then we define

Template:ExampleRobox

We claim 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 \int\limits_0^\infty e^{-x}dx=1}

To do this we calculate

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^\infty e^{-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 =\lim_{b\to\infty}\int\limits_0^b e^{-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 =\lim_{b\to\infty}(-e^{-x})\Bigr|_0^b}
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(-e^{-b}+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 =1}

Template:Robox/Close

Template:ExampleRobox

We claim that 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 \frac{dx}{x}} diverges.

This follows 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 \int\limits_1^\infty\frac{dx}{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 =\lim_{b\to\infty}\int\limits_1^b \frac{dx}{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 =\lim_{b\to\infty}\ln(x)\Bigr|_1^b}
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(\ln(b)-0\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 =\infty}

Therefore

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}} diverges.

Template:Robox/Close

Template:ExampleRobox

Find 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^\infty x^2e^{-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} 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^2e^{-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)}
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(-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 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}b^2e^{-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,

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^\infty x^2e^{-x}dx=\lim_{b\to\infty}\int\limits_0^b x^2e^{-x}dx=0+2(0-0+1)=2}

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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

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}} 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}
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 =-\frac{1}{1-p}\lim_{b\to\infty}\left(b^{-p+1}-1\right)}

Notice that we had to assume that to avoid dividing by 0. However 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 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 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} is continuous on the interval 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 [a,b)} and is discontinuous at 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} , 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 \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 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} is continuous on the interval 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 (a,b]} and is discontinuous at 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 a} , 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 \int\limits_a^b f(x)dx=\lim_{c\to a^+}\int\limits_c^b f(x)dx}

Finally 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} has an discontinuity at a point 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\in(a,b)} and is continuous at all other points in 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 [a,b]} . 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^c f(x)dx} 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 \int\limits_c^b f(x)dx} converge 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 \int\limits_a^b f(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 \int\limits_a^c f(x)dx+\int\limits_c^b f(x)dx}

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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_0^1\frac{dx}{x^p}=\begin{cases}\frac{1}{1-p},&\text{if }p<1\\ \text{diverges},&\text{if }p\ge1\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\ne1} 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_0^1\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_{a\to 0^+}\int\limits_a^1 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_{a\to 0^+}\left(\frac{x^{-p+1}}{-p+1}\right)\Bigg|_a^1}
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 =-\frac{1}{1-p}\lim_{a\to 0^+}\left(1-a^{-p+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}{1-p},&\mbox{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\ne1} do avoid dividing by 0. So instead we do 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 p=1} case separately,

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_0^1 \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


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

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 \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

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_{-1}^3\frac{dx}{x-2}&=\lim_{b\to2^-}\int\limits_{-1}^b \frac{dx}{x-2}+\lim_{a\to2^+}\int\limits_a^3\frac{dx}{x-2}\\ &=\lim_{b\to2^-}\ln\big(|x-2|\big)\Big|_{-1}^b+\lim_{a\to2^+}\ln\big(|x-2|\big)\Big|_a^3\\ &=\lim_{b\to2^-}\Big[\ln(2-b)-\ln(3)\Big]+\lim_{a\to2^+}\Big[\ln(1)-\ln(a-2)\Big]\\ &=\lim_{b\to2^-}\Big[\ln(2-b)\Big]-\ln(3)+\lim_{a\to2^+}\Big[-\ln(a-2)\Big]\\ \end{align}}

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\to2^-}\Big[\ln(2-b)\Big]} 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_{a\to2^+}\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 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} is continuous on 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 [a,b]} except at points 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_1<c_2<\ldots<c_n} in 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 [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 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} .

  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 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 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 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 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 and for improper integrals with discontinuities.

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Show that converges.

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} we know 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 -1\le\sin(x)\le 1} so 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 1\le\sin(x)+2\le3} . This implies 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 0\le\frac{\sin(x)+2}{x^2}\le\frac{3}{x^2}} .

We have seen 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 \int\limits_1^\infty\frac{3}{x^2}dx=3\int\limits_1^\infty\frac{dx}{x^2}} converges. So putting 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)=\frac{\sin(x)+2}{x^2}} 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 g(x)=\frac{3}{x^2}} into the comparison test we get that 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\frac{\sin(x)+2}{x^2}dx} converges as well. Template:Robox/Close

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Show 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 \int\limits_1^\infty\frac{\sin(x)+2}{x}dx} diverges.

Just as in the previous example we know 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 \sin(x)+2\ge1} 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} . Thus

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 \frac{\sin(x)+2}{x}\ge\frac{1}{x}\ge0}

We have seen 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 \int\limits_1^\infty\frac{dx}{x}} diverges. So putting 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)=\frac{\sin(x)+2}{x}} 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 g(x)=\frac{1}{x}} into the comparison test we get 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 \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 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} . What we actually need is this inequality holds for sufficiently 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} (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 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 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.


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Show 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 \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 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\ge1} , 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 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_{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 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} ,

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 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\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

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