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Evaluating $I = lim_varepsilon to 0 int_-varepsilon^varepsilon frac1varepsilon u(x) dx$ through a change of variables?
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I want to evaluate the following integral using change of variables but I'm not sure if what I'm doing is correct. It 'feels' like I should get the answer $I = 2 u(0)$.
$$
I = lim_varepsilon to 0 int_-varepsilon^varepsilon frac1varepsilon u(x) dx.
$$
Ok so here is what I'm doing. Let $hat x = x/varepsilon$ and $hat u(hat x) = u(x)$. We have $dx = varepsilon dhat x$ and when $x=pm varepsilon$ we have $hat x = pm 1$. So this means that
$$
beginalign
I & = lim_varepsilon to 0 int_-varepsilon^varepsilon frac1varepsilon u(x) dx \
& = lim_varepsilon to 0 int_-1^1 frac1varepsilon hat u(hat x) varepsilon dhat x \
& = lim_varepsilon to 0 int_-1^1 hat u(hat x) dhat x.
endalign
$$
The limits no longer depends on $varepsilon$ so we can write
beginalign
I & = int_-1^1 lim_varepsilon to 0 hat u(hat x) dhat x \
& = int_-1^1 lim_varepsilon to 0 u(varepsilon hat x) dhat x \
& = int_-1^1u(0) dhat x \
& = 2 u(0).
endalign
Is this all correct and are all the steps fully rigorous?
real-analysis limits distribution-theory dirac-delta change-of-variable
asked Jul 18 at 16:39
csss
1,22811221
 |Â
show 2 more comments
up vote
0
down vote
favorite
I want to evaluate the following integral using change of variables but I'm not sure if what I'm doing is correct. It 'feels' like I should get the answer $I = 2 u(0)$.
$$
I = lim_varepsilon to 0 int_-varepsilon^varepsilon frac1varepsilon u(x) dx.
$$
Ok so here is what I'm doing. Let $hat x = x/varepsilon$ and $hat u(hat x) = u(x)$. We have $dx = varepsilon dhat x$ and when $x=pm varepsilon$ we have $hat x = pm 1$. So this means that
$$
beginalign
I & = lim_varepsilon to 0 int_-varepsilon^varepsilon frac1varepsilon u(x) dx \
& = lim_varepsilon to 0 int_-1^1 frac1varepsilon hat u(hat x) varepsilon dhat x \
& = lim_varepsilon to 0 int_-1^1 hat u(hat x) dhat x.
endalign
$$
The limits no longer depends on $varepsilon$ so we can write
beginalign
I & = int_-1^1 lim_varepsilon to 0 hat u(hat x) dhat x \
& = int_-1^1 lim_varepsilon to 0 u(varepsilon hat x) dhat x \
& = int_-1^1u(0) dhat x \
& = 2 u(0).
endalign
Is this all correct and are all the steps fully rigorous?
real-analysis limits distribution-theory dirac-delta change-of-variable
asked Jul 18 at 16:39
csss
1,22811221
1
What do you know about $u$?
– Fimpellizieri
Jul 18 at 16:41
1
The step $$ lim_epsilon to 0 int_-1^1 hat u(hat x), d hat x = int_-1^1 lim_epsilon to 0 hat u(hat x), d hat x $$ needs to be justified
– Omnomnomnom
Jul 18 at 16:46
1
Is there a reason you tagged this question with distribution-theory and dirac-delta?
– Omnomnomnom
Jul 18 at 16:48
1
The desired limit is $2u(0)$ if $u$ is continuous at $0$. This follows by an immediate application of Fundamental Theorem of Calculus. Your steps are OK but moving the limit operator inside integral is tricky and in general requires more analysis than a typical usage of Fundamental Theorem of Calculus.
– Paramanand Singh
Jul 19 at 0:26
1
Also any limit of the form $lim_epsilonto 0f(epsilon)$ is independent of $epsilon$. It is the expression $f(epsilon) $ after the limit operator which depends on $epsilon$.
– Paramanand Singh
Jul 19 at 0:28
 |Â
show 2 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to evaluate the following integral using change of variables but I'm not sure if what I'm doing is correct. It 'feels' like I should get the answer $I = 2 u(0)$.
$$
I = lim_varepsilon to 0 int_-varepsilon^varepsilon frac1varepsilon u(x) dx.
$$
Ok so here is what I'm doing. Let $hat x = x/varepsilon$ and $hat u(hat x) = u(x)$. We have $dx = varepsilon dhat x$ and when $x=pm varepsilon$ we have $hat x = pm 1$. So this means that
$$
beginalign
I & = lim_varepsilon to 0 int_-varepsilon^varepsilon frac1varepsilon u(x) dx \
& = lim_varepsilon to 0 int_-1^1 frac1varepsilon hat u(hat x) varepsilon dhat x \
& = lim_varepsilon to 0 int_-1^1 hat u(hat x) dhat x.
endalign
$$
The limits no longer depends on $varepsilon$ so we can write
beginalign
I & = int_-1^1 lim_varepsilon to 0 hat u(hat x) dhat x \
& = int_-1^1 lim_varepsilon to 0 u(varepsilon hat x) dhat x \
& = int_-1^1u(0) dhat x \
& = 2 u(0).
endalign
Is this all correct and are all the steps fully rigorous?
real-analysis limits distribution-theory dirac-delta change-of-variable
asked Jul 18 at 16:39
csss
1,22811221
I want to evaluate the following integral using change of variables but I'm not sure if what I'm doing is correct. It 'feels' like I should get the answer $I = 2 u(0)$.
$$
I = lim_varepsilon to 0 int_-varepsilon^varepsilon frac1varepsilon u(x) dx.
$$
Ok so here is what I'm doing. Let $hat x = x/varepsilon$ and $hat u(hat x) = u(x)$. We have $dx = varepsilon dhat x$ and when $x=pm varepsilon$ we have $hat x = pm 1$. So this means that
$$
beginalign
I & = lim_varepsilon to 0 int_-varepsilon^varepsilon frac1varepsilon u(x) dx \
& = lim_varepsilon to 0 int_-1^1 frac1varepsilon hat u(hat x) varepsilon dhat x \
& = lim_varepsilon to 0 int_-1^1 hat u(hat x) dhat x.
endalign
$$
The limits no longer depends on $varepsilon$ so we can write
beginalign
I & = int_-1^1 lim_varepsilon to 0 hat u(hat x) dhat x \
& = int_-1^1 lim_varepsilon to 0 u(varepsilon hat x) dhat x \
& = int_-1^1u(0) dhat x \
& = 2 u(0).
endalign
Is this all correct and are all the steps fully rigorous?
real-analysis limits distribution-theory dirac-delta change-of-variable
asked Jul 18 at 16:39
csss
1,22811221
asked Jul 18 at 16:39
csss
1,22811221
asked Jul 18 at 16:39
csss
1,22811221
asked Jul 18 at 16:39
csss
1,22811221
1,22811221
1
What do you know about $u$?
– Fimpellizieri
Jul 18 at 16:41
1
The step $$ lim_epsilon to 0 int_-1^1 hat u(hat x), d hat x = int_-1^1 lim_epsilon to 0 hat u(hat x), d hat x $$ needs to be justified
– Omnomnomnom
Jul 18 at 16:46
1
Is there a reason you tagged this question with distribution-theory and dirac-delta?
– Omnomnomnom
Jul 18 at 16:48
1
The desired limit is $2u(0)$ if $u$ is continuous at $0$. This follows by an immediate application of Fundamental Theorem of Calculus. Your steps are OK but moving the limit operator inside integral is tricky and in general requires more analysis than a typical usage of Fundamental Theorem of Calculus.
– Paramanand Singh
Jul 19 at 0:26
1
Also any limit of the form $lim_epsilonto 0f(epsilon)$ is independent of $epsilon$. It is the expression $f(epsilon) $ after the limit operator which depends on $epsilon$.
– Paramanand Singh
Jul 19 at 0:28
 |Â
show 2 more comments
1
What do you know about $u$?
– Fimpellizieri
Jul 18 at 16:41
1
The step $$ lim_epsilon to 0 int_-1^1 hat u(hat x), d hat x = int_-1^1 lim_epsilon to 0 hat u(hat x), d hat x $$ needs to be justified
– Omnomnomnom
Jul 18 at 16:46
1
Is there a reason you tagged this question with distribution-theory and dirac-delta?
– Omnomnomnom
Jul 18 at 16:48
1
The desired limit is $2u(0)$ if $u$ is continuous at $0$. This follows by an immediate application of Fundamental Theorem of Calculus. Your steps are OK but moving the limit operator inside integral is tricky and in general requires more analysis than a typical usage of Fundamental Theorem of Calculus.
– Paramanand Singh
Jul 19 at 0:26
1
Also any limit of the form $lim_epsilonto 0f(epsilon)$ is independent of $epsilon$. It is the expression $f(epsilon) $ after the limit operator which depends on $epsilon$.
– Paramanand Singh
Jul 19 at 0:28
1
1
What do you know about $u$?
– Fimpellizieri
Jul 18 at 16:41
What do you know about $u$?
– Fimpellizieri
Jul 18 at 16:41
1
1
The step $$ lim_epsilon to 0 int_-1^1 hat u(hat x), d hat x = int_-1^1 lim_epsilon to 0 hat u(hat x), d hat x $$ needs to be justified
– Omnomnomnom
Jul 18 at 16:46
The step $$ lim_epsilon to 0 int_-1^1 hat u(hat x), d hat x = int_-1^1 lim_epsilon to 0 hat u(hat x), d hat x $$ needs to be justified
– Omnomnomnom
Jul 18 at 16:46
1
1
Is there a reason you tagged this question with distribution-theory and dirac-delta?
– Omnomnomnom
Jul 18 at 16:48
Is there a reason you tagged this question with distribution-theory and dirac-delta?
– Omnomnomnom
Jul 18 at 16:48
1
1
The desired limit is $2u(0)$ if $u$ is continuous at $0$. This follows by an immediate application of Fundamental Theorem of Calculus. Your steps are OK but moving the limit operator inside integral is tricky and in general requires more analysis than a typical usage of Fundamental Theorem of Calculus.
– Paramanand Singh
Jul 19 at 0:26
The desired limit is $2u(0)$ if $u$ is continuous at $0$. This follows by an immediate application of Fundamental Theorem of Calculus. Your steps are OK but moving the limit operator inside integral is tricky and in general requires more analysis than a typical usage of Fundamental Theorem of Calculus.
– Paramanand Singh
Jul 19 at 0:26
1
1
Also any limit of the form $lim_epsilonto 0f(epsilon)$ is independent of $epsilon$. It is the expression $f(epsilon) $ after the limit operator which depends on $epsilon$.
– Paramanand Singh
Jul 19 at 0:28
Also any limit of the form $lim_epsilonto 0f(epsilon)$ is independent of $epsilon$. It is the expression $f(epsilon) $ after the limit operator which depends on $epsilon$.
– Paramanand Singh
Jul 19 at 0:28
 |Â
show 2 more comments
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1
What do you know about $u$?
– Fimpellizieri
Jul 18 at 16:41
1
The step $$ lim_epsilon to 0 int_-1^1 hat u(hat x), d hat x = int_-1^1 lim_epsilon to 0 hat u(hat x), d hat x $$ needs to be justified
– Omnomnomnom
Jul 18 at 16:46
1
Is there a reason you tagged this question with distribution-theory and dirac-delta?
– Omnomnomnom
Jul 18 at 16:48
1
The desired limit is $2u(0)$ if $u$ is continuous at $0$. This follows by an immediate application of Fundamental Theorem of Calculus. Your steps are OK but moving the limit operator inside integral is tricky and in general requires more analysis than a typical usage of Fundamental Theorem of Calculus.
– Paramanand Singh
Jul 19 at 0:26
1
Also any limit of the form $lim_epsilonto 0f(epsilon)$ is independent of $epsilon$. It is the expression $f(epsilon) $ after the limit operator which depends on $epsilon$.
– Paramanand Singh
Jul 19 at 0:28