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0

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As mentioned in the title, could you mention any integral with such properties?

I have tried to figure it out in several ways, but failed. Could anyone help?

real-analysis integration complex-analysis limits analysis

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asked Mar 13 at 9:05

kalulakalula

32

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closed as off-topic by José Carlos Santos, mrtaurho, Hanul Jeon, Delta-u, Shailesh Mar 13 at 10:53

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, mrtaurho, Hanul Jeon, Delta-u, Shailesh

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See: math.stackexchange.com/questions/658664/…
  $endgroup$
  – Robert Z
  Mar 13 at 9:16
  
  

  
  
  
  

add a comment | 

0

$begingroup$

As mentioned in the title, could you mention any integral with such properties?

I have tried to figure it out in several ways, but failed. Could anyone help?

real-analysis integration complex-analysis limits analysis

share|cite|improve this question

asked Mar 13 at 9:05

kalulakalula

32

New contributor

kalula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

closed as off-topic by José Carlos Santos, mrtaurho, Hanul Jeon, Delta-u, Shailesh Mar 13 at 10:53

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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, mrtaurho, Hanul Jeon, Delta-u, Shailesh

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See: math.stackexchange.com/questions/658664/…
  $endgroup$
  – Robert Z
  Mar 13 at 9:16
  
  

  
  
  
  

add a comment | 

$begingroup$

As mentioned in the title, could you mention any integral with such properties?

I have tried to figure it out in several ways, but failed. Could anyone help?

real-analysis integration complex-analysis limits analysis

share|cite|improve this question

asked Mar 13 at 9:05

kalulakalula

32

New contributor

kalula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked Mar 13 at 9:05

kalulakalula

32

New contributor

kalula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked Mar 13 at 9:05

kalulakalula

32

New contributor

kalula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 13 at 9:05

kalulakalula

32

New contributor

kalula is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 13 at 9:05

kalulakalula

32

3 Answers
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$int_- infty^ infty cos(t^2) dt=int_- infty^ infty sin(t^2) dt= sqrtpi/2.$

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answered Mar 13 at 9:09

FredFred

48.5k1849

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add a comment | 

0

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$int_0^1frac 1 sqrt x, dx$ is convergent even though the integrand tends to $infty$ as $x to 0$. If you don't want to allow infinite limits also consider $int_0^1frac sin (frac 1 x) sqrt x, dx$

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edited Mar 13 at 9:10

answered Mar 13 at 9:07

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Gibbs I have edited the answer. You may note that the integrand is even continuous on $(0,1)$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 13 at 9:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. Sorry, I just wanted to let you know that because I am used to think that "limit equal to $infty$" is not the same as "limit does not exist".
  $endgroup$
  – Gibbs
  Mar 13 at 9:19
  

  
  
  
  

add a comment | 

0

$begingroup$

On the positive real numbers take a sequence of isosceles triangles of area $1/2^n$ and constant height $2$. For example, construct a first triangle like that on $[0,1]$, which will have then area $1 = 1/2^0$ (step $n=0$). Do the same on $[2,2+1/2^1]$, on $[4,4+1/2^2]$, on $[6,6+1/2^3]$, and so forth. Notice that you need to shorten the base length to get area $1/2^n$ at step $n$. In the remaining intervals define the function to be $0$. The limit of this function for $x to +infty$ does not exist because the height of each triangle is constant and equals $1$, but the integral over the positive real numbers is just the sum of the areas of the triangles, which is finite: $sum_n=0^infty 1/2^n=1$.

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answered Mar 13 at 9:17

GibbsGibbs

5,4103827

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add a comment | 

0

$begingroup$

$int_- infty^ infty cos(t^2) dt=int_- infty^ infty sin(t^2) dt= sqrtpi/2.$

share|cite|improve this answer

answered Mar 13 at 9:09

FredFred

48.5k1849

$endgroup$

add a comment | 

0

$begingroup$

$int_- infty^ infty cos(t^2) dt=int_- infty^ infty sin(t^2) dt= sqrtpi/2.$

share|cite|improve this answer

answered Mar 13 at 9:09

FredFred

48.5k1849

$endgroup$

add a comment | 

$begingroup$

$int_- infty^ infty cos(t^2) dt=int_- infty^ infty sin(t^2) dt= sqrtpi/2.$

share|cite|improve this answer

answered Mar 13 at 9:09

FredFred

48.5k1849

$endgroup$

share|cite|improve this answer

answered Mar 13 at 9:09

FredFred

48.5k1849

answered Mar 13 at 9:09

FredFred

48.5k1849

answered Mar 13 at 9:09

FredFred

48.5k1849

0

$begingroup$

$int_0^1frac 1 sqrt x, dx$ is convergent even though the integrand tends to $infty$ as $x to 0$. If you don't want to allow infinite limits also consider $int_0^1frac sin (frac 1 x) sqrt x, dx$

share|cite|improve this answer

edited Mar 13 at 9:10

answered Mar 13 at 9:07

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Gibbs I have edited the answer. You may note that the integrand is even continuous on $(0,1)$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 13 at 9:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. Sorry, I just wanted to let you know that because I am used to think that "limit equal to $infty$" is not the same as "limit does not exist".
  $endgroup$
  – Gibbs
  Mar 13 at 9:19
  

  
  
  
  

add a comment | 

0

$begingroup$

$int_0^1frac 1 sqrt x, dx$ is convergent even though the integrand tends to $infty$ as $x to 0$. If you don't want to allow infinite limits also consider $int_0^1frac sin (frac 1 x) sqrt x, dx$

share|cite|improve this answer

edited Mar 13 at 9:10

answered Mar 13 at 9:07

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Gibbs I have edited the answer. You may note that the integrand is even continuous on $(0,1)$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 13 at 9:11
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. Sorry, I just wanted to let you know that because I am used to think that "limit equal to $infty$" is not the same as "limit does not exist".
  $endgroup$
  – Gibbs
  Mar 13 at 9:19
  

  
  
  
  

add a comment | 

$begingroup$

$int_0^1frac 1 sqrt x, dx$ is convergent even though the integrand tends to $infty$ as $x to 0$. If you don't want to allow infinite limits also consider $int_0^1frac sin (frac 1 x) sqrt x, dx$

share|cite|improve this answer

edited Mar 13 at 9:10

answered Mar 13 at 9:07

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

$endgroup$

share|cite|improve this answer

edited Mar 13 at 9:10

answered Mar 13 at 9:07

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

answered Mar 13 at 9:07

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

answered Mar 13 at 9:07

Kavi Rama MurthyKavi Rama Murthy

68.1k53068

0

$begingroup$

On the positive real numbers take a sequence of isosceles triangles of area $1/2^n$ and constant height $2$. For example, construct a first triangle like that on $[0,1]$, which will have then area $1 = 1/2^0$ (step $n=0$). Do the same on $[2,2+1/2^1]$, on $[4,4+1/2^2]$, on $[6,6+1/2^3]$, and so forth. Notice that you need to shorten the base length to get area $1/2^n$ at step $n$. In the remaining intervals define the function to be $0$. The limit of this function for $x to +infty$ does not exist because the height of each triangle is constant and equals $1$, but the integral over the positive real numbers is just the sum of the areas of the triangles, which is finite: $sum_n=0^infty 1/2^n=1$.

share|cite|improve this answer

answered Mar 13 at 9:17

GibbsGibbs

5,4103827

$endgroup$

add a comment | 

0

$begingroup$

On the positive real numbers take a sequence of isosceles triangles of area $1/2^n$ and constant height $2$. For example, construct a first triangle like that on $[0,1]$, which will have then area $1 = 1/2^0$ (step $n=0$). Do the same on $[2,2+1/2^1]$, on $[4,4+1/2^2]$, on $[6,6+1/2^3]$, and so forth. Notice that you need to shorten the base length to get area $1/2^n$ at step $n$. In the remaining intervals define the function to be $0$. The limit of this function for $x to +infty$ does not exist because the height of each triangle is constant and equals $1$, but the integral over the positive real numbers is just the sum of the areas of the triangles, which is finite: $sum_n=0^infty 1/2^n=1$.

share|cite|improve this answer

answered Mar 13 at 9:17

GibbsGibbs

5,4103827

$endgroup$

add a comment | 

$begingroup$

On the positive real numbers take a sequence of isosceles triangles of area $1/2^n$ and constant height $2$. For example, construct a first triangle like that on $[0,1]$, which will have then area $1 = 1/2^0$ (step $n=0$). Do the same on $[2,2+1/2^1]$, on $[4,4+1/2^2]$, on $[6,6+1/2^3]$, and so forth. Notice that you need to shorten the base length to get area $1/2^n$ at step $n$. In the remaining intervals define the function to be $0$. The limit of this function for $x to +infty$ does not exist because the height of each triangle is constant and equals $1$, but the integral over the positive real numbers is just the sum of the areas of the triangles, which is finite: $sum_n=0^infty 1/2^n=1$.

share|cite|improve this answer

answered Mar 13 at 9:17

GibbsGibbs

5,4103827

$endgroup$

share|cite|improve this answer

answered Mar 13 at 9:17

GibbsGibbs

5,4103827

answered Mar 13 at 9:17

GibbsGibbs

5,4103827

answered Mar 13 at 9:17

GibbsGibbs

5,4103827