What do improper integrals represent? [closed]

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So I was wondering what is the "meaning" behind the improper integral, say $int_0^inftyf(x) dx$. Usually, an integral of a function over a certain region is simply the area under the function, however, in this case, it doesn't seem fully the case. After all, if it was the sum of the areas then $int_0^infty sin(x)$ would converge, as the areas form a telescoping series. Hence my confusion, what does the definition of this kind of improper integral really mean?

improper-integrals

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asked Jul 31 at 19:24

Sorfosh

910616

closed as off-topic by Mark Viola, amWhy, Xander Henderson, Leucippus, max_zorn Aug 1 at 5:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The integrals still represent areas. By definition $int_0^infty sin x dx = lim_n to infty int_0^n sin x dx$, which does not converge just as $lim_n to infty cos n$ or $lim_n to infty (-1)^n$ do not converge.
  – JavaMan
  Jul 31 at 19:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, did you try looking anything up online before asking here? What did you find?
  – JavaMan
  Jul 31 at 19:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JavaMan I know what the definition means. I know why it does not converge. But if you consider the telescoping series of the areas under and above $f(x)=sin(x)$ you get a convergent series.
  – Sorfosh
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you understand the definition of the improper integral, then your misunderstanding is coming from the definition of convergence. Improper integrals still represent area. Divergence can mean that the quantity tends to infinity, or it can mean that the limit never settles around a single number. The second type of divergence is what happens here.
  – JavaMan
  Jul 31 at 19:35
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @Sorfosh The "areas under and above $sin(x)$" do NOT represent a telescoping series. And even if they did, telescoping does not imply convergence.
  – Mark Viola
  Jul 31 at 19:36
  
  

  
  
  
  

 | 
show 2 more comments

up vote
-1
down vote

favorite

So I was wondering what is the "meaning" behind the improper integral, say $int_0^inftyf(x) dx$. Usually, an integral of a function over a certain region is simply the area under the function, however, in this case, it doesn't seem fully the case. After all, if it was the sum of the areas then $int_0^infty sin(x)$ would converge, as the areas form a telescoping series. Hence my confusion, what does the definition of this kind of improper integral really mean?

improper-integrals

share|cite|improve this question

asked Jul 31 at 19:24

Sorfosh

910616

closed as off-topic by Mark Viola, amWhy, Xander Henderson, Leucippus, max_zorn Aug 1 at 5:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The integrals still represent areas. By definition $int_0^infty sin x dx = lim_n to infty int_0^n sin x dx$, which does not converge just as $lim_n to infty cos n$ or $lim_n to infty (-1)^n$ do not converge.
  – JavaMan
  Jul 31 at 19:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, did you try looking anything up online before asking here? What did you find?
  – JavaMan
  Jul 31 at 19:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JavaMan I know what the definition means. I know why it does not converge. But if you consider the telescoping series of the areas under and above $f(x)=sin(x)$ you get a convergent series.
  – Sorfosh
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you understand the definition of the improper integral, then your misunderstanding is coming from the definition of convergence. Improper integrals still represent area. Divergence can mean that the quantity tends to infinity, or it can mean that the limit never settles around a single number. The second type of divergence is what happens here.
  – JavaMan
  Jul 31 at 19:35
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @Sorfosh The "areas under and above $sin(x)$" do NOT represent a telescoping series. And even if they did, telescoping does not imply convergence.
  – Mark Viola
  Jul 31 at 19:36
  
  

  
  
  
  

 | 
show 2 more comments

up vote
-1
down vote

favorite

up vote
-1
down vote

favorite

So I was wondering what is the "meaning" behind the improper integral, say $int_0^inftyf(x) dx$. Usually, an integral of a function over a certain region is simply the area under the function, however, in this case, it doesn't seem fully the case. After all, if it was the sum of the areas then $int_0^infty sin(x)$ would converge, as the areas form a telescoping series. Hence my confusion, what does the definition of this kind of improper integral really mean?

improper-integrals

share|cite|improve this question

asked Jul 31 at 19:24

Sorfosh

910616

So I was wondering what is the "meaning" behind the improper integral, say $int_0^inftyf(x) dx$. Usually, an integral of a function over a certain region is simply the area under the function, however, in this case, it doesn't seem fully the case. After all, if it was the sum of the areas then $int_0^infty sin(x)$ would converge, as the areas form a telescoping series. Hence my confusion, what does the definition of this kind of improper integral really mean?

improper-integrals

share|cite|improve this question

asked Jul 31 at 19:24

Sorfosh

910616

share|cite|improve this question

share|cite|improve this question

asked Jul 31 at 19:24

Sorfosh

910616

asked Jul 31 at 19:24

Sorfosh

910616

asked Jul 31 at 19:24

Sorfosh

910616

910616

closed as off-topic by Mark Viola, amWhy, Xander Henderson, Leucippus, max_zorn Aug 1 at 5:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by Mark Viola, amWhy, Xander Henderson, Leucippus, max_zorn Aug 1 at 5:52

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Xander Henderson, Leucippus

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The integrals still represent areas. By definition $int_0^infty sin x dx = lim_n to infty int_0^n sin x dx$, which does not converge just as $lim_n to infty cos n$ or $lim_n to infty (-1)^n$ do not converge.
  – JavaMan
  Jul 31 at 19:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, did you try looking anything up online before asking here? What did you find?
  – JavaMan
  Jul 31 at 19:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JavaMan I know what the definition means. I know why it does not converge. But if you consider the telescoping series of the areas under and above $f(x)=sin(x)$ you get a convergent series.
  – Sorfosh
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you understand the definition of the improper integral, then your misunderstanding is coming from the definition of convergence. Improper integrals still represent area. Divergence can mean that the quantity tends to infinity, or it can mean that the limit never settles around a single number. The second type of divergence is what happens here.
  – JavaMan
  Jul 31 at 19:35
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @Sorfosh The "areas under and above $sin(x)$" do NOT represent a telescoping series. And even if they did, telescoping does not imply convergence.
  – Mark Viola
  Jul 31 at 19:36
  
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The integrals still represent areas. By definition $int_0^infty sin x dx = lim_n to infty int_0^n sin x dx$, which does not converge just as $lim_n to infty cos n$ or $lim_n to infty (-1)^n$ do not converge.
  – JavaMan
  Jul 31 at 19:27
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, did you try looking anything up online before asking here? What did you find?
  – JavaMan
  Jul 31 at 19:28
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JavaMan I know what the definition means. I know why it does not converge. But if you consider the telescoping series of the areas under and above $f(x)=sin(x)$ you get a convergent series.
  – Sorfosh
  Jul 31 at 19:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  If you understand the definition of the improper integral, then your misunderstanding is coming from the definition of convergence. Improper integrals still represent area. Divergence can mean that the quantity tends to infinity, or it can mean that the limit never settles around a single number. The second type of divergence is what happens here.
  – JavaMan
  Jul 31 at 19:35
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  @Sorfosh The "areas under and above $sin(x)$" do NOT represent a telescoping series. And even if they did, telescoping does not imply convergence.
  – Mark Viola
  Jul 31 at 19:36
  
  

  
  
  
  

The integrals still represent areas. By definition $int_0^infty sin x dx = lim_n to infty int_0^n sin x dx$, which does not converge just as $lim_n to infty cos n$ or $lim_n to infty (-1)^n$ do not converge.
– JavaMan
Jul 31 at 19:27

The integrals still represent areas. By definition $int_0^infty sin x dx = lim_n to infty int_0^n sin x dx$, which does not converge just as $lim_n to infty cos n$ or $lim_n to infty (-1)^n$ do not converge.
– JavaMan
Jul 31 at 19:27

Also, did you try looking anything up online before asking here? What did you find?
– JavaMan
Jul 31 at 19:28

Also, did you try looking anything up online before asking here? What did you find?
– JavaMan
Jul 31 at 19:28

@JavaMan I know what the definition means. I know why it does not converge. But if you consider the telescoping series of the areas under and above $f(x)=sin(x)$ you get a convergent series.
– Sorfosh
Jul 31 at 19:32

@JavaMan I know what the definition means. I know why it does not converge. But if you consider the telescoping series of the areas under and above $f(x)=sin(x)$ you get a convergent series.
– Sorfosh
Jul 31 at 19:32

If you understand the definition of the improper integral, then your misunderstanding is coming from the definition of convergence. Improper integrals still represent area. Divergence can mean that the quantity tends to infinity, or it can mean that the limit never settles around a single number. The second type of divergence is what happens here.
– JavaMan
Jul 31 at 19:35

If you understand the definition of the improper integral, then your misunderstanding is coming from the definition of convergence. Improper integrals still represent area. Divergence can mean that the quantity tends to infinity, or it can mean that the limit never settles around a single number. The second type of divergence is what happens here.
– JavaMan
Jul 31 at 19:35

3

3

@Sorfosh The "areas under and above $sin(x)$" do NOT represent a telescoping series. And even if they did, telescoping does not imply convergence.
– Mark Viola
Jul 31 at 19:36

@Sorfosh The "areas under and above $sin(x)$" do NOT represent a telescoping series. And even if they did, telescoping does not imply convergence.
– Mark Viola
Jul 31 at 19:36

 | 
show 2 more comments

3 Answers
3

active

oldest

votes

up vote
2
down vote



accepted

I still assert that $int_0^infty f(x),dx$ represents the signed area between $f$ and the $x$ axis. You complain that this cannot be right, because the signed area between $sin x$ and the positive $x$ axis is zero, being a telescoping sum, while $int_0^infty sin x,dx$ does not converge. However, this is not a telescoping sum any more than
$$
1-1+1-1+1-dots
$$
is. The above sum does not converge, as the partial sums are $1,0,1,0,dots$. In the same way that this series does not have a sum, the region under $sin x$ and the positive $x$ axis does not have an area, which agrees with the statement that $int_0^infty sin x,dx$ diverges.

share|cite|improve this answer

answered Jul 31 at 19:56

Mike Earnest

14.7k11644

add a comment | 

up vote
2
down vote

If $f : [0,infty) to Bbb R$ then: $$int_0^infty f(x) mathrm dx := lim_b to infty int_0^b f(x) mathrm dx$$

share|cite|improve this answer

answered Jul 31 at 19:27

Kenny Lau

17.7k2156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I know what the definition is. I am confused what the definition represents since it does not seem to represent the area.
  – Sorfosh
  Jul 31 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sorfosh : What's the difficulty? If the area up to $b$ is $A(b)$, then of course $A(infty)$ would mean the limiting area as $b$ increases without bound (if that exists). It's completely analogous to understanding $sum_n=0^inftya_n$ as a limit of partial sums, since there's no such thing as an infinite sum. Do you have problems understanding infinite series?
  – MPW
  Jul 31 at 19:44
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

If the integral converges then for all $n> N implies int_0^n f(x) dx$ is
within $epsilon$ of some constant.

But $int_0^2npi sin x dx = 0$ while $int_0^(2n+1)pi sin x dx = 2$

share|cite|improve this answer

answered Jul 31 at 19:53

Doug M

39k31749

add a comment | 

3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

I still assert that $int_0^infty f(x),dx$ represents the signed area between $f$ and the $x$ axis. You complain that this cannot be right, because the signed area between $sin x$ and the positive $x$ axis is zero, being a telescoping sum, while $int_0^infty sin x,dx$ does not converge. However, this is not a telescoping sum any more than
$$
1-1+1-1+1-dots
$$
is. The above sum does not converge, as the partial sums are $1,0,1,0,dots$. In the same way that this series does not have a sum, the region under $sin x$ and the positive $x$ axis does not have an area, which agrees with the statement that $int_0^infty sin x,dx$ diverges.

share|cite|improve this answer

answered Jul 31 at 19:56

Mike Earnest

14.7k11644

add a comment | 

up vote
2
down vote



accepted

I still assert that $int_0^infty f(x),dx$ represents the signed area between $f$ and the $x$ axis. You complain that this cannot be right, because the signed area between $sin x$ and the positive $x$ axis is zero, being a telescoping sum, while $int_0^infty sin x,dx$ does not converge. However, this is not a telescoping sum any more than
$$
1-1+1-1+1-dots
$$
is. The above sum does not converge, as the partial sums are $1,0,1,0,dots$. In the same way that this series does not have a sum, the region under $sin x$ and the positive $x$ axis does not have an area, which agrees with the statement that $int_0^infty sin x,dx$ diverges.

share|cite|improve this answer

answered Jul 31 at 19:56

Mike Earnest

14.7k11644

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

I still assert that $int_0^infty f(x),dx$ represents the signed area between $f$ and the $x$ axis. You complain that this cannot be right, because the signed area between $sin x$ and the positive $x$ axis is zero, being a telescoping sum, while $int_0^infty sin x,dx$ does not converge. However, this is not a telescoping sum any more than
$$
1-1+1-1+1-dots
$$
is. The above sum does not converge, as the partial sums are $1,0,1,0,dots$. In the same way that this series does not have a sum, the region under $sin x$ and the positive $x$ axis does not have an area, which agrees with the statement that $int_0^infty sin x,dx$ diverges.

share|cite|improve this answer

answered Jul 31 at 19:56

Mike Earnest

14.7k11644

I still assert that $int_0^infty f(x),dx$ represents the signed area between $f$ and the $x$ axis. You complain that this cannot be right, because the signed area between $sin x$ and the positive $x$ axis is zero, being a telescoping sum, while $int_0^infty sin x,dx$ does not converge. However, this is not a telescoping sum any more than
$$
1-1+1-1+1-dots
$$
is. The above sum does not converge, as the partial sums are $1,0,1,0,dots$. In the same way that this series does not have a sum, the region under $sin x$ and the positive $x$ axis does not have an area, which agrees with the statement that $int_0^infty sin x,dx$ diverges.

share|cite|improve this answer

answered Jul 31 at 19:56

Mike Earnest

14.7k11644

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 19:56

Mike Earnest

14.7k11644

answered Jul 31 at 19:56

Mike Earnest

14.7k11644

answered Jul 31 at 19:56

Mike Earnest

14.7k11644

14.7k11644

add a comment | 

add a comment | 

up vote
2
down vote

If $f : [0,infty) to Bbb R$ then: $$int_0^infty f(x) mathrm dx := lim_b to infty int_0^b f(x) mathrm dx$$

share|cite|improve this answer

answered Jul 31 at 19:27

Kenny Lau

17.7k2156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I know what the definition is. I am confused what the definition represents since it does not seem to represent the area.
  – Sorfosh
  Jul 31 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sorfosh : What's the difficulty? If the area up to $b$ is $A(b)$, then of course $A(infty)$ would mean the limiting area as $b$ increases without bound (if that exists). It's completely analogous to understanding $sum_n=0^inftya_n$ as a limit of partial sums, since there's no such thing as an infinite sum. Do you have problems understanding infinite series?
  – MPW
  Jul 31 at 19:44
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

If $f : [0,infty) to Bbb R$ then: $$int_0^infty f(x) mathrm dx := lim_b to infty int_0^b f(x) mathrm dx$$

share|cite|improve this answer

answered Jul 31 at 19:27

Kenny Lau

17.7k2156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I know what the definition is. I am confused what the definition represents since it does not seem to represent the area.
  – Sorfosh
  Jul 31 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sorfosh : What's the difficulty? If the area up to $b$ is $A(b)$, then of course $A(infty)$ would mean the limiting area as $b$ increases without bound (if that exists). It's completely analogous to understanding $sum_n=0^inftya_n$ as a limit of partial sums, since there's no such thing as an infinite sum. Do you have problems understanding infinite series?
  – MPW
  Jul 31 at 19:44
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

If $f : [0,infty) to Bbb R$ then: $$int_0^infty f(x) mathrm dx := lim_b to infty int_0^b f(x) mathrm dx$$

share|cite|improve this answer

answered Jul 31 at 19:27

Kenny Lau

17.7k2156

If $f : [0,infty) to Bbb R$ then: $$int_0^infty f(x) mathrm dx := lim_b to infty int_0^b f(x) mathrm dx$$

share|cite|improve this answer

answered Jul 31 at 19:27

Kenny Lau

17.7k2156

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 19:27

Kenny Lau

17.7k2156

answered Jul 31 at 19:27

Kenny Lau

17.7k2156

answered Jul 31 at 19:27

Kenny Lau

17.7k2156

17.7k2156

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I know what the definition is. I am confused what the definition represents since it does not seem to represent the area.
  – Sorfosh
  Jul 31 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sorfosh : What's the difficulty? If the area up to $b$ is $A(b)$, then of course $A(infty)$ would mean the limiting area as $b$ increases without bound (if that exists). It's completely analogous to understanding $sum_n=0^inftya_n$ as a limit of partial sums, since there's no such thing as an infinite sum. Do you have problems understanding infinite series?
  – MPW
  Jul 31 at 19:44
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I know what the definition is. I am confused what the definition represents since it does not seem to represent the area.
  – Sorfosh
  Jul 31 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Sorfosh : What's the difficulty? If the area up to $b$ is $A(b)$, then of course $A(infty)$ would mean the limiting area as $b$ increases without bound (if that exists). It's completely analogous to understanding $sum_n=0^inftya_n$ as a limit of partial sums, since there's no such thing as an infinite sum. Do you have problems understanding infinite series?
  – MPW
  Jul 31 at 19:44
  
  

  
  
  
  

I know what the definition is. I am confused what the definition represents since it does not seem to represent the area.
– Sorfosh
Jul 31 at 19:31

I know what the definition is. I am confused what the definition represents since it does not seem to represent the area.
– Sorfosh
Jul 31 at 19:31

@Sorfosh : What's the difficulty? If the area up to $b$ is $A(b)$, then of course $A(infty)$ would mean the limiting area as $b$ increases without bound (if that exists). It's completely analogous to understanding $sum_n=0^inftya_n$ as a limit of partial sums, since there's no such thing as an infinite sum. Do you have problems understanding infinite series?
– MPW
Jul 31 at 19:44

@Sorfosh : What's the difficulty? If the area up to $b$ is $A(b)$, then of course $A(infty)$ would mean the limiting area as $b$ increases without bound (if that exists). It's completely analogous to understanding $sum_n=0^inftya_n$ as a limit of partial sums, since there's no such thing as an infinite sum. Do you have problems understanding infinite series?
– MPW
Jul 31 at 19:44

add a comment | 

up vote
1
down vote

If the integral converges then for all $n> N implies int_0^n f(x) dx$ is
within $epsilon$ of some constant.

But $int_0^2npi sin x dx = 0$ while $int_0^(2n+1)pi sin x dx = 2$

share|cite|improve this answer

answered Jul 31 at 19:53

Doug M

39k31749

add a comment | 

up vote
1
down vote

If the integral converges then for all $n> N implies int_0^n f(x) dx$ is
within $epsilon$ of some constant.

But $int_0^2npi sin x dx = 0$ while $int_0^(2n+1)pi sin x dx = 2$

share|cite|improve this answer

answered Jul 31 at 19:53

Doug M

39k31749

add a comment | 

up vote
1
down vote

up vote
1
down vote

If the integral converges then for all $n> N implies int_0^n f(x) dx$ is
within $epsilon$ of some constant.

But $int_0^2npi sin x dx = 0$ while $int_0^(2n+1)pi sin x dx = 2$

share|cite|improve this answer

answered Jul 31 at 19:53

Doug M

39k31749

If the integral converges then for all $n> N implies int_0^n f(x) dx$ is
within $epsilon$ of some constant.

But $int_0^2npi sin x dx = 0$ while $int_0^(2n+1)pi sin x dx = 2$

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answered Jul 31 at 19:53

Doug M

39k31749

share|cite|improve this answer

share|cite|improve this answer

answered Jul 31 at 19:53

Doug M

39k31749

answered Jul 31 at 19:53

Doug M

39k31749

answered Jul 31 at 19:53

Doug M

39k31749

39k31749

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