Confusion about the Evaluation of an Integral.

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I have a confusion about the following question:

Question. $rho:mathbbR to mathbbR$ be a continuous function such that $rho(x) ge 0 ~forall ~x in mathbbR$, and $rho(x)=0$ if $|x|ge 1$ and $$int_-infty^inftyrho(t)dt=1$$
Let $f:mathbbRtomathbbR$ be a continuous function. Evaluate $$lim_epsilon to 0frac1epsilonint_-infty^inftyrho(fracxepsilon)~f(x) dx$$

My Solution. The last integral in the question reduces to :
$$
lim_epsilon to 0frac1epsilonint_-1^1rho(fracxepsilon)~f(x)dx
$$

Now we substitute $x/epsilon=t$ in the last integral so that we have:

$lim_epsilon to 0frac1epsilonint_-1^1rho(fracxepsilon)~f(x)dx = lim_epsilon to 0int_-1/ epsilon^1/ epsilonrho(t)~f(epsilon t)dt
$

Is this right?...First, I cannot understand where the continuity of $rho$ has been used. Is it even necessary?

Second, Is the interchanging of limit and integral permissible here? or does it require something more for this interchanging?

I'm not sure how to answer these question here...Please help me to understand what is going on here....!! Thank you so much.

integration proof-verification continuity

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edited Jul 26 at 9:49

asked Jul 26 at 9:25

Indrajit Ghosh

561415

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ä°ntegral domain is $int_-epsilon^epsilon$.
  – Jo'
  Jul 26 at 9:35
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

I have a confusion about the following question:

Question. $rho:mathbbR to mathbbR$ be a continuous function such that $rho(x) ge 0 ~forall ~x in mathbbR$, and $rho(x)=0$ if $|x|ge 1$ and $$int_-infty^inftyrho(t)dt=1$$
Let $f:mathbbRtomathbbR$ be a continuous function. Evaluate $$lim_epsilon to 0frac1epsilonint_-infty^inftyrho(fracxepsilon)~f(x) dx$$

My Solution. The last integral in the question reduces to :
$$
lim_epsilon to 0frac1epsilonint_-1^1rho(fracxepsilon)~f(x)dx
$$

Now we substitute $x/epsilon=t$ in the last integral so that we have:

$lim_epsilon to 0frac1epsilonint_-1^1rho(fracxepsilon)~f(x)dx = lim_epsilon to 0int_-1/ epsilon^1/ epsilonrho(t)~f(epsilon t)dt
$

Is this right?...First, I cannot understand where the continuity of $rho$ has been used. Is it even necessary?

Second, Is the interchanging of limit and integral permissible here? or does it require something more for this interchanging?

I'm not sure how to answer these question here...Please help me to understand what is going on here....!! Thank you so much.

integration proof-verification continuity

share|cite|improve this question

edited Jul 26 at 9:49

asked Jul 26 at 9:25

Indrajit Ghosh

561415

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ä°ntegral domain is $int_-epsilon^epsilon$.
  – Jo'
  Jul 26 at 9:35
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I have a confusion about the following question:

Question. $rho:mathbbR to mathbbR$ be a continuous function such that $rho(x) ge 0 ~forall ~x in mathbbR$, and $rho(x)=0$ if $|x|ge 1$ and $$int_-infty^inftyrho(t)dt=1$$
Let $f:mathbbRtomathbbR$ be a continuous function. Evaluate $$lim_epsilon to 0frac1epsilonint_-infty^inftyrho(fracxepsilon)~f(x) dx$$

My Solution. The last integral in the question reduces to :
$$
lim_epsilon to 0frac1epsilonint_-1^1rho(fracxepsilon)~f(x)dx
$$

Now we substitute $x/epsilon=t$ in the last integral so that we have:

$lim_epsilon to 0frac1epsilonint_-1^1rho(fracxepsilon)~f(x)dx = lim_epsilon to 0int_-1/ epsilon^1/ epsilonrho(t)~f(epsilon t)dt
$

Is this right?...First, I cannot understand where the continuity of $rho$ has been used. Is it even necessary?

Second, Is the interchanging of limit and integral permissible here? or does it require something more for this interchanging?

I'm not sure how to answer these question here...Please help me to understand what is going on here....!! Thank you so much.

integration proof-verification continuity

share|cite|improve this question

edited Jul 26 at 9:49

asked Jul 26 at 9:25

Indrajit Ghosh

561415

I have a confusion about the following question:

Question. $rho:mathbbR to mathbbR$ be a continuous function such that $rho(x) ge 0 ~forall ~x in mathbbR$, and $rho(x)=0$ if $|x|ge 1$ and $$int_-infty^inftyrho(t)dt=1$$
Let $f:mathbbRtomathbbR$ be a continuous function. Evaluate $$lim_epsilon to 0frac1epsilonint_-infty^inftyrho(fracxepsilon)~f(x) dx$$

My Solution. The last integral in the question reduces to :
$$
lim_epsilon to 0frac1epsilonint_-1^1rho(fracxepsilon)~f(x)dx
$$

Now we substitute $x/epsilon=t$ in the last integral so that we have:

$lim_epsilon to 0frac1epsilonint_-1^1rho(fracxepsilon)~f(x)dx = lim_epsilon to 0int_-1/ epsilon^1/ epsilonrho(t)~f(epsilon t)dt
$

Is this right?...First, I cannot understand where the continuity of $rho$ has been used. Is it even necessary?

Second, Is the interchanging of limit and integral permissible here? or does it require something more for this interchanging?

I'm not sure how to answer these question here...Please help me to understand what is going on here....!! Thank you so much.

integration proof-verification continuity

share|cite|improve this question

edited Jul 26 at 9:49

asked Jul 26 at 9:25

Indrajit Ghosh

561415

share|cite|improve this question

share|cite|improve this question

edited Jul 26 at 9:49

edited Jul 26 at 9:49

edited Jul 26 at 9:49

asked Jul 26 at 9:25

Indrajit Ghosh

561415

asked Jul 26 at 9:25

Indrajit Ghosh

561415

asked Jul 26 at 9:25

Indrajit Ghosh

561415

561415

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ä°ntegral domain is $int_-epsilon^epsilon$.
  – Jo'
  Jul 26 at 9:35
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Ä°ntegral domain is $int_-epsilon^epsilon$.
  – Jo'
  Jul 26 at 9:35
  
  

  
  
  
  

Ä°ntegral domain is $int_-epsilon^epsilon$.
– Jo'
Jul 26 at 9:35

Ä°ntegral domain is $int_-epsilon^epsilon$.
– Jo'
Jul 26 at 9:35

add a comment | 

2 Answers
2

active

oldest

votes

up vote
1
down vote



accepted

First point to be noted: integral with respect to $t$ is from $-frac 1 epsilon$ to $-frac 1 epsilon$ but since $rho $ is supported by $[-1,1]$ this does not cause any problem Secondly you have to justify interchnage of limit and integral. Consider $int_1^1 rho (t)f(epsilon t), dt -f(0)$ which you can write as $int_1^1 rho (t)[f(epsilon t)-f(0)], dt$. Given $eta >0$ we can find $r$ such that $|f(epsilon t)-f(0)|<eta $ if $epsilon <r$ (because $|epsilon t | leq epsilon)$ Now can you complete the proof?

share|cite|improve this answer

answered Jul 26 at 9:43

Kavi Rama Murthy

20k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I understand .....but where the continuity of $rho$ has been used...
  – Indrajit Ghosh
  Jul 26 at 10:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @IndrajitGhosh Continuity of $rho $ is not required. It can be just a non-negative mesurable function such that $int rho (x), dx=1$ and $rho (x)=0$ for $|x| geq 1$.
  – Kavi Rama Murthy
  Jul 26 at 10:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KaviRamaMurthy...I understand the rest ,,,,thank you sir..
  – Indrajit Ghosh
  Jul 26 at 10:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi You can make the change of variable even when $rho $ is not continuous. See my comment for sufficient conditions on $rho $.
  – Kavi Rama Murthy
  Jul 26 at 10:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi HAve you studied measure theory? If you know about induced measures and change of variables in measure theory you would know that continuity is not required.
  – Kavi Rama Murthy
  Jul 26 at 10:10
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote

In order to legitimize the interchange of limit and integral, consider the difference
$$left|int_-1/epsilon^1/epsilonrho(t)~f(epsilon t)dt-f(0)right|=left|int_-1^1rho(t)~(f(epsilon t)-f(0)) dtright|leq int_-1^1rho(t)~left|f(epsilon t)-f(0)right| dt$$
where $epsilonleq 1$.
Now, by the continuity of $f$ at $0$, for all $n>0$ there is $r>0$ such that $|f(x)-f(0)|<1/n$ for all $|x|<r$. Hence, for $|epsilon|<r$, then $|epsilon t|<r$ and
$$left|int_-1^1rho(t)~f(epsilon t)dt-f(0)right|leq int_-1^1rho(t)(1/n) dt=frac1n$$
which goes to $0$ as $nto+infty$.

share|cite|improve this answer

edited Jul 26 at 9:59

answered Jul 26 at 9:43

Robert Z

83.8k954122

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

First point to be noted: integral with respect to $t$ is from $-frac 1 epsilon$ to $-frac 1 epsilon$ but since $rho $ is supported by $[-1,1]$ this does not cause any problem Secondly you have to justify interchnage of limit and integral. Consider $int_1^1 rho (t)f(epsilon t), dt -f(0)$ which you can write as $int_1^1 rho (t)[f(epsilon t)-f(0)], dt$. Given $eta >0$ we can find $r$ such that $|f(epsilon t)-f(0)|<eta $ if $epsilon <r$ (because $|epsilon t | leq epsilon)$ Now can you complete the proof?

share|cite|improve this answer

answered Jul 26 at 9:43

Kavi Rama Murthy

20k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I understand .....but where the continuity of $rho$ has been used...
  – Indrajit Ghosh
  Jul 26 at 10:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @IndrajitGhosh Continuity of $rho $ is not required. It can be just a non-negative mesurable function such that $int rho (x), dx=1$ and $rho (x)=0$ for $|x| geq 1$.
  – Kavi Rama Murthy
  Jul 26 at 10:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KaviRamaMurthy...I understand the rest ,,,,thank you sir..
  – Indrajit Ghosh
  Jul 26 at 10:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi You can make the change of variable even when $rho $ is not continuous. See my comment for sufficient conditions on $rho $.
  – Kavi Rama Murthy
  Jul 26 at 10:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi HAve you studied measure theory? If you know about induced measures and change of variables in measure theory you would know that continuity is not required.
  – Kavi Rama Murthy
  Jul 26 at 10:10
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote



accepted

First point to be noted: integral with respect to $t$ is from $-frac 1 epsilon$ to $-frac 1 epsilon$ but since $rho $ is supported by $[-1,1]$ this does not cause any problem Secondly you have to justify interchnage of limit and integral. Consider $int_1^1 rho (t)f(epsilon t), dt -f(0)$ which you can write as $int_1^1 rho (t)[f(epsilon t)-f(0)], dt$. Given $eta >0$ we can find $r$ such that $|f(epsilon t)-f(0)|<eta $ if $epsilon <r$ (because $|epsilon t | leq epsilon)$ Now can you complete the proof?

share|cite|improve this answer

answered Jul 26 at 9:43

Kavi Rama Murthy

20k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I understand .....but where the continuity of $rho$ has been used...
  – Indrajit Ghosh
  Jul 26 at 10:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @IndrajitGhosh Continuity of $rho $ is not required. It can be just a non-negative mesurable function such that $int rho (x), dx=1$ and $rho (x)=0$ for $|x| geq 1$.
  – Kavi Rama Murthy
  Jul 26 at 10:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KaviRamaMurthy...I understand the rest ,,,,thank you sir..
  – Indrajit Ghosh
  Jul 26 at 10:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi You can make the change of variable even when $rho $ is not continuous. See my comment for sufficient conditions on $rho $.
  – Kavi Rama Murthy
  Jul 26 at 10:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi HAve you studied measure theory? If you know about induced measures and change of variables in measure theory you would know that continuity is not required.
  – Kavi Rama Murthy
  Jul 26 at 10:10
  

  
  
  
  

 | 
show 1 more comment

up vote
1
down vote



accepted

up vote
1
down vote



accepted

First point to be noted: integral with respect to $t$ is from $-frac 1 epsilon$ to $-frac 1 epsilon$ but since $rho $ is supported by $[-1,1]$ this does not cause any problem Secondly you have to justify interchnage of limit and integral. Consider $int_1^1 rho (t)f(epsilon t), dt -f(0)$ which you can write as $int_1^1 rho (t)[f(epsilon t)-f(0)], dt$. Given $eta >0$ we can find $r$ such that $|f(epsilon t)-f(0)|<eta $ if $epsilon <r$ (because $|epsilon t | leq epsilon)$ Now can you complete the proof?

share|cite|improve this answer

answered Jul 26 at 9:43

Kavi Rama Murthy

20k2829

First point to be noted: integral with respect to $t$ is from $-frac 1 epsilon$ to $-frac 1 epsilon$ but since $rho $ is supported by $[-1,1]$ this does not cause any problem Secondly you have to justify interchnage of limit and integral. Consider $int_1^1 rho (t)f(epsilon t), dt -f(0)$ which you can write as $int_1^1 rho (t)[f(epsilon t)-f(0)], dt$. Given $eta >0$ we can find $r$ such that $|f(epsilon t)-f(0)|<eta $ if $epsilon <r$ (because $|epsilon t | leq epsilon)$ Now can you complete the proof?

share|cite|improve this answer

answered Jul 26 at 9:43

Kavi Rama Murthy

20k2829

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 9:43

Kavi Rama Murthy

20k2829

answered Jul 26 at 9:43

Kavi Rama Murthy

20k2829

answered Jul 26 at 9:43

Kavi Rama Murthy

20k2829

20k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I understand .....but where the continuity of $rho$ has been used...
  – Indrajit Ghosh
  Jul 26 at 10:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @IndrajitGhosh Continuity of $rho $ is not required. It can be just a non-negative mesurable function such that $int rho (x), dx=1$ and $rho (x)=0$ for $|x| geq 1$.
  – Kavi Rama Murthy
  Jul 26 at 10:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KaviRamaMurthy...I understand the rest ,,,,thank you sir..
  – Indrajit Ghosh
  Jul 26 at 10:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi You can make the change of variable even when $rho $ is not continuous. See my comment for sufficient conditions on $rho $.
  – Kavi Rama Murthy
  Jul 26 at 10:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi HAve you studied measure theory? If you know about induced measures and change of variables in measure theory you would know that continuity is not required.
  – Kavi Rama Murthy
  Jul 26 at 10:10
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I understand .....but where the continuity of $rho$ has been used...
  – Indrajit Ghosh
  Jul 26 at 10:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @IndrajitGhosh Continuity of $rho $ is not required. It can be just a non-negative mesurable function such that $int rho (x), dx=1$ and $rho (x)=0$ for $|x| geq 1$.
  – Kavi Rama Murthy
  Jul 26 at 10:03
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @KaviRamaMurthy...I understand the rest ,,,,thank you sir..
  – Indrajit Ghosh
  Jul 26 at 10:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi You can make the change of variable even when $rho $ is not continuous. See my comment for sufficient conditions on $rho $.
  – Kavi Rama Murthy
  Jul 26 at 10:05
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @nicomezi HAve you studied measure theory? If you know about induced measures and change of variables in measure theory you would know that continuity is not required.
  – Kavi Rama Murthy
  Jul 26 at 10:10
  

  
  
  
  

I understand .....but where the continuity of $rho$ has been used...
– Indrajit Ghosh
Jul 26 at 10:01

I understand .....but where the continuity of $rho$ has been used...
– Indrajit Ghosh
Jul 26 at 10:01

@IndrajitGhosh Continuity of $rho $ is not required. It can be just a non-negative mesurable function such that $int rho (x), dx=1$ and $rho (x)=0$ for $|x| geq 1$.
– Kavi Rama Murthy
Jul 26 at 10:03

@IndrajitGhosh Continuity of $rho $ is not required. It can be just a non-negative mesurable function such that $int rho (x), dx=1$ and $rho (x)=0$ for $|x| geq 1$.
– Kavi Rama Murthy
Jul 26 at 10:03

@KaviRamaMurthy...I understand the rest ,,,,thank you sir..
– Indrajit Ghosh
Jul 26 at 10:04

@KaviRamaMurthy...I understand the rest ,,,,thank you sir..
– Indrajit Ghosh
Jul 26 at 10:04

@nicomezi You can make the change of variable even when $rho $ is not continuous. See my comment for sufficient conditions on $rho $.
– Kavi Rama Murthy
Jul 26 at 10:05

@nicomezi You can make the change of variable even when $rho $ is not continuous. See my comment for sufficient conditions on $rho $.
– Kavi Rama Murthy
Jul 26 at 10:05

@nicomezi HAve you studied measure theory? If you know about induced measures and change of variables in measure theory you would know that continuity is not required.
– Kavi Rama Murthy
Jul 26 at 10:10

@nicomezi HAve you studied measure theory? If you know about induced measures and change of variables in measure theory you would know that continuity is not required.
– Kavi Rama Murthy
Jul 26 at 10:10

 | 
show 1 more comment

up vote
1
down vote

In order to legitimize the interchange of limit and integral, consider the difference
$$left|int_-1/epsilon^1/epsilonrho(t)~f(epsilon t)dt-f(0)right|=left|int_-1^1rho(t)~(f(epsilon t)-f(0)) dtright|leq int_-1^1rho(t)~left|f(epsilon t)-f(0)right| dt$$
where $epsilonleq 1$.
Now, by the continuity of $f$ at $0$, for all $n>0$ there is $r>0$ such that $|f(x)-f(0)|<1/n$ for all $|x|<r$. Hence, for $|epsilon|<r$, then $|epsilon t|<r$ and
$$left|int_-1^1rho(t)~f(epsilon t)dt-f(0)right|leq int_-1^1rho(t)(1/n) dt=frac1n$$
which goes to $0$ as $nto+infty$.

share|cite|improve this answer

edited Jul 26 at 9:59

answered Jul 26 at 9:43

Robert Z

83.8k954122

add a comment | 

up vote
1
down vote

In order to legitimize the interchange of limit and integral, consider the difference
$$left|int_-1/epsilon^1/epsilonrho(t)~f(epsilon t)dt-f(0)right|=left|int_-1^1rho(t)~(f(epsilon t)-f(0)) dtright|leq int_-1^1rho(t)~left|f(epsilon t)-f(0)right| dt$$
where $epsilonleq 1$.
Now, by the continuity of $f$ at $0$, for all $n>0$ there is $r>0$ such that $|f(x)-f(0)|<1/n$ for all $|x|<r$. Hence, for $|epsilon|<r$, then $|epsilon t|<r$ and
$$left|int_-1^1rho(t)~f(epsilon t)dt-f(0)right|leq int_-1^1rho(t)(1/n) dt=frac1n$$
which goes to $0$ as $nto+infty$.

share|cite|improve this answer

edited Jul 26 at 9:59

answered Jul 26 at 9:43

Robert Z

83.8k954122

add a comment | 

up vote
1
down vote

up vote
1
down vote

In order to legitimize the interchange of limit and integral, consider the difference
$$left|int_-1/epsilon^1/epsilonrho(t)~f(epsilon t)dt-f(0)right|=left|int_-1^1rho(t)~(f(epsilon t)-f(0)) dtright|leq int_-1^1rho(t)~left|f(epsilon t)-f(0)right| dt$$
where $epsilonleq 1$.
Now, by the continuity of $f$ at $0$, for all $n>0$ there is $r>0$ such that $|f(x)-f(0)|<1/n$ for all $|x|<r$. Hence, for $|epsilon|<r$, then $|epsilon t|<r$ and
$$left|int_-1^1rho(t)~f(epsilon t)dt-f(0)right|leq int_-1^1rho(t)(1/n) dt=frac1n$$
which goes to $0$ as $nto+infty$.

share|cite|improve this answer

edited Jul 26 at 9:59

answered Jul 26 at 9:43

Robert Z

83.8k954122

In order to legitimize the interchange of limit and integral, consider the difference
$$left|int_-1/epsilon^1/epsilonrho(t)~f(epsilon t)dt-f(0)right|=left|int_-1^1rho(t)~(f(epsilon t)-f(0)) dtright|leq int_-1^1rho(t)~left|f(epsilon t)-f(0)right| dt$$
where $epsilonleq 1$.
Now, by the continuity of $f$ at $0$, for all $n>0$ there is $r>0$ such that $|f(x)-f(0)|<1/n$ for all $|x|<r$. Hence, for $|epsilon|<r$, then $|epsilon t|<r$ and
$$left|int_-1^1rho(t)~f(epsilon t)dt-f(0)right|leq int_-1^1rho(t)(1/n) dt=frac1n$$
which goes to $0$ as $nto+infty$.

share|cite|improve this answer

edited Jul 26 at 9:59

answered Jul 26 at 9:43

Robert Z

83.8k954122

share|cite|improve this answer

share|cite|improve this answer

edited Jul 26 at 9:59

edited Jul 26 at 9:59

edited Jul 26 at 9:59

answered Jul 26 at 9:43

Robert Z

83.8k954122

answered Jul 26 at 9:43

Robert Z

83.8k954122

answered Jul 26 at 9:43

Robert Z

83.8k954122

83.8k954122

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