$int_{0}^{infty}frac{sin(x)}{x}dx$ exists, but $int_{mathbb{R^+}}frac{sin(x)}{x}dlambda$ doesn't exist?
$begingroup$
$int_{0}^{infty}frac{sin(x)}{x}dx$ is riemann integrable.
But how to prove that it's not Lebesgue integrable ?
(I tried contradiction).
If $f(x)=frac{sin(x)}{x}$, I supppose that $fin L^1(mathbb{R^+})$.
then, I' ll define the function sequence : $f_n(x)=frac{sin( x)}{x}mathbb{1}_{[0,n]}(x).$
We have $f_n$ converges $lambda$-a.e to $f$.
and $|f_n|leq|f|$, for all $1leq n$.
and $f$ integrable (hypothesis).
then with the dominated convergence theorem
$$lim_nint_{0}^{n}f_ndlambda=int_{mathbb{R^+}}fdlambda$$
which is equivalent to $$int_{0}^{infty}frac{sin(x)}{x}dx=int_{mathbb{R^+}}f
dlambda$$
Thats basically what I did.
integration lebesgue-integral
edited Dec 23 '18 at 7:37
DavidG
2,1321724
asked Dec 23 '18 at 7:15
Anas BOUALIIAnas BOUALII
1438
$endgroup$
|
show 1 more comment
$begingroup$
$int_{0}^{infty}frac{sin(x)}{x}dx$ is riemann integrable.
But how to prove that it's not Lebesgue integrable ?
(I tried contradiction).
If $f(x)=frac{sin(x)}{x}$, I supppose that $fin L^1(mathbb{R^+})$.
then, I' ll define the function sequence : $f_n(x)=frac{sin( x)}{x}mathbb{1}_{[0,n]}(x).$
We have $f_n$ converges $lambda$-a.e to $f$.
and $|f_n|leq|f|$, for all $1leq n$.
and $f$ integrable (hypothesis).
then with the dominated convergence theorem
$$lim_nint_{0}^{n}f_ndlambda=int_{mathbb{R^+}}fdlambda$$
which is equivalent to $$int_{0}^{infty}frac{sin(x)}{x}dx=int_{mathbb{R^+}}f
dlambda$$
Thats basically what I did.
integration lebesgue-integral
edited Dec 23 '18 at 7:37
DavidG
2,1321724
asked Dec 23 '18 at 7:15
Anas BOUALIIAnas BOUALII
1438
$endgroup$
1
$begingroup$
Hi Anas, as part of the community guidelines you need to provide some more context and background to your question. Furthermore if you have attempted the problem you need to provide your working. Just a heads up :-)
$endgroup$
– DavidG
Dec 23 '18 at 7:19
$begingroup$
@DavidG, check .
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:31
$begingroup$
Looks good to me! Just wanted to let you know. This page has very specific rules about posting. It annoyed me to start with, but once you get use to it you realise it's a great format to work with. Sorry I can't help you wing your question though.
$endgroup$
– DavidG
Dec 23 '18 at 7:35
$begingroup$
Well, i can't find the contradiction with the last equality. but thanks anyway.
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:38
$begingroup$
Actually it is not Riemann integrable either. It is only improperly Riemann integrable.
$endgroup$
– metamorphy
Dec 23 '18 at 7:43
|
show 1 more comment
$begingroup$
$int_{0}^{infty}frac{sin(x)}{x}dx$ is riemann integrable.
But how to prove that it's not Lebesgue integrable ?
(I tried contradiction).
If $f(x)=frac{sin(x)}{x}$, I supppose that $fin L^1(mathbb{R^+})$.
then, I' ll define the function sequence : $f_n(x)=frac{sin( x)}{x}mathbb{1}_{[0,n]}(x).$
We have $f_n$ converges $lambda$-a.e to $f$.
and $|f_n|leq|f|$, for all $1leq n$.
and $f$ integrable (hypothesis).
then with the dominated convergence theorem
$$lim_nint_{0}^{n}f_ndlambda=int_{mathbb{R^+}}fdlambda$$
which is equivalent to $$int_{0}^{infty}frac{sin(x)}{x}dx=int_{mathbb{R^+}}f
dlambda$$
Thats basically what I did.
integration lebesgue-integral
edited Dec 23 '18 at 7:37
DavidG
2,1321724
asked Dec 23 '18 at 7:15
Anas BOUALIIAnas BOUALII
1438
$endgroup$
$int_{0}^{infty}frac{sin(x)}{x}dx$ is riemann integrable.
But how to prove that it's not Lebesgue integrable ?
(I tried contradiction).
If $f(x)=frac{sin(x)}{x}$, I supppose that $fin L^1(mathbb{R^+})$.
then, I' ll define the function sequence : $f_n(x)=frac{sin( x)}{x}mathbb{1}_{[0,n]}(x).$
We have $f_n$ converges $lambda$-a.e to $f$.
and $|f_n|leq|f|$, for all $1leq n$.
and $f$ integrable (hypothesis).
then with the dominated convergence theorem
$$lim_nint_{0}^{n}f_ndlambda=int_{mathbb{R^+}}fdlambda$$
which is equivalent to $$int_{0}^{infty}frac{sin(x)}{x}dx=int_{mathbb{R^+}}f
dlambda$$
Thats basically what I did.
integration lebesgue-integral
integration lebesgue-integral
edited Dec 23 '18 at 7:37
DavidG
2,1321724
asked Dec 23 '18 at 7:15
Anas BOUALIIAnas BOUALII
1438
edited Dec 23 '18 at 7:37
DavidG
2,1321724
asked Dec 23 '18 at 7:15
Anas BOUALIIAnas BOUALII
1438
edited Dec 23 '18 at 7:37
DavidG
2,1321724
edited Dec 23 '18 at 7:37
DavidG
2,1321724
edited Dec 23 '18 at 7:37
DavidG
2,1321724
2,1321724
asked Dec 23 '18 at 7:15
Anas BOUALIIAnas BOUALII
1438
asked Dec 23 '18 at 7:15
Anas BOUALIIAnas BOUALII
1438
asked Dec 23 '18 at 7:15
Anas BOUALIIAnas BOUALII
1438
1438
1
$begingroup$
Hi Anas, as part of the community guidelines you need to provide some more context and background to your question. Furthermore if you have attempted the problem you need to provide your working. Just a heads up :-)
$endgroup$
– DavidG
Dec 23 '18 at 7:19
$begingroup$
@DavidG, check .
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:31
$begingroup$
Looks good to me! Just wanted to let you know. This page has very specific rules about posting. It annoyed me to start with, but once you get use to it you realise it's a great format to work with. Sorry I can't help you wing your question though.
$endgroup$
– DavidG
Dec 23 '18 at 7:35
$begingroup$
Well, i can't find the contradiction with the last equality. but thanks anyway.
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:38
$begingroup$
Actually it is not Riemann integrable either. It is only improperly Riemann integrable.
$endgroup$
– metamorphy
Dec 23 '18 at 7:43
|
show 1 more comment
1
$begingroup$
Hi Anas, as part of the community guidelines you need to provide some more context and background to your question. Furthermore if you have attempted the problem you need to provide your working. Just a heads up :-)
$endgroup$
– DavidG
Dec 23 '18 at 7:19
$begingroup$
@DavidG, check .
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:31
$begingroup$
Looks good to me! Just wanted to let you know. This page has very specific rules about posting. It annoyed me to start with, but once you get use to it you realise it's a great format to work with. Sorry I can't help you wing your question though.
$endgroup$
– DavidG
Dec 23 '18 at 7:35
$begingroup$
Well, i can't find the contradiction with the last equality. but thanks anyway.
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:38
$begingroup$
Actually it is not Riemann integrable either. It is only improperly Riemann integrable.
$endgroup$
– metamorphy
Dec 23 '18 at 7:43
1
1
$begingroup$
Hi Anas, as part of the community guidelines you need to provide some more context and background to your question. Furthermore if you have attempted the problem you need to provide your working. Just a heads up :-)
$endgroup$
– DavidG
Dec 23 '18 at 7:19
$begingroup$
Hi Anas, as part of the community guidelines you need to provide some more context and background to your question. Furthermore if you have attempted the problem you need to provide your working. Just a heads up :-)
$endgroup$
– DavidG
Dec 23 '18 at 7:19
$begingroup$
@DavidG, check .
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:31
$begingroup$
@DavidG, check .
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:31
$begingroup$
Looks good to me! Just wanted to let you know. This page has very specific rules about posting. It annoyed me to start with, but once you get use to it you realise it's a great format to work with. Sorry I can't help you wing your question though.
$endgroup$
– DavidG
Dec 23 '18 at 7:35
$begingroup$
Looks good to me! Just wanted to let you know. This page has very specific rules about posting. It annoyed me to start with, but once you get use to it you realise it's a great format to work with. Sorry I can't help you wing your question though.
$endgroup$
– DavidG
Dec 23 '18 at 7:35
$begingroup$
Well, i can't find the contradiction with the last equality. but thanks anyway.
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:38
$begingroup$
Well, i can't find the contradiction with the last equality. but thanks anyway.
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:38
$begingroup$
Actually it is not Riemann integrable either. It is only improperly Riemann integrable.
$endgroup$
– metamorphy
Dec 23 '18 at 7:43
$begingroup$
Actually it is not Riemann integrable either. It is only improperly Riemann integrable.
$endgroup$
– metamorphy
Dec 23 '18 at 7:43
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Hint. If such function is Lebesgue integrable then also its absolute value is Lebesgue integrable. Let $n$ a positive integer, then
$$
int_0^{2pi n} frac{|sin x|}{x} ,dx = sum_{k=0}^{n-1} int_{2pi k}^{2pi(k+1)} frac{|sin x|}{x},dx
ge sum_{k=0}^{n-1} frac{1}{2pi (k+1)} int_{0}^{2pi} |sin x|,dx
$$
Can you take it from here?
answered Dec 23 '18 at 7:40
Robert ZRobert Z
98.5k1068139
$endgroup$
$begingroup$
Yes, I get it. thank you. I knew the proposition , but i didn't know how to use it correctly.
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:47
add a comment |
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1 Answer
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$begingroup$
Hint. If such function is Lebesgue integrable then also its absolute value is Lebesgue integrable. Let $n$ a positive integer, then
$$
int_0^{2pi n} frac{|sin x|}{x} ,dx = sum_{k=0}^{n-1} int_{2pi k}^{2pi(k+1)} frac{|sin x|}{x},dx
ge sum_{k=0}^{n-1} frac{1}{2pi (k+1)} int_{0}^{2pi} |sin x|,dx
$$
Can you take it from here?
answered Dec 23 '18 at 7:40
Robert ZRobert Z
98.5k1068139
$endgroup$
$begingroup$
Yes, I get it. thank you. I knew the proposition , but i didn't know how to use it correctly.
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:47
add a comment |
$begingroup$
Hint. If such function is Lebesgue integrable then also its absolute value is Lebesgue integrable. Let $n$ a positive integer, then
$$
int_0^{2pi n} frac{|sin x|}{x} ,dx = sum_{k=0}^{n-1} int_{2pi k}^{2pi(k+1)} frac{|sin x|}{x},dx
ge sum_{k=0}^{n-1} frac{1}{2pi (k+1)} int_{0}^{2pi} |sin x|,dx
$$
Can you take it from here?
answered Dec 23 '18 at 7:40
Robert ZRobert Z
98.5k1068139
$endgroup$
$begingroup$
Yes, I get it. thank you. I knew the proposition , but i didn't know how to use it correctly.
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:47
add a comment |
$begingroup$
Hint. If such function is Lebesgue integrable then also its absolute value is Lebesgue integrable. Let $n$ a positive integer, then
$$
int_0^{2pi n} frac{|sin x|}{x} ,dx = sum_{k=0}^{n-1} int_{2pi k}^{2pi(k+1)} frac{|sin x|}{x},dx
ge sum_{k=0}^{n-1} frac{1}{2pi (k+1)} int_{0}^{2pi} |sin x|,dx
$$
Can you take it from here?
answered Dec 23 '18 at 7:40
Robert ZRobert Z
98.5k1068139
$endgroup$
Hint. If such function is Lebesgue integrable then also its absolute value is Lebesgue integrable. Let $n$ a positive integer, then
$$
int_0^{2pi n} frac{|sin x|}{x} ,dx = sum_{k=0}^{n-1} int_{2pi k}^{2pi(k+1)} frac{|sin x|}{x},dx
ge sum_{k=0}^{n-1} frac{1}{2pi (k+1)} int_{0}^{2pi} |sin x|,dx
$$
Can you take it from here?
answered Dec 23 '18 at 7:40
Robert ZRobert Z
98.5k1068139
answered Dec 23 '18 at 7:40
Robert ZRobert Z
98.5k1068139
answered Dec 23 '18 at 7:40
Robert ZRobert Z
98.5k1068139
answered Dec 23 '18 at 7:40
Robert ZRobert Z
98.5k1068139
98.5k1068139
$begingroup$
Yes, I get it. thank you. I knew the proposition , but i didn't know how to use it correctly.
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:47
add a comment |
$begingroup$
Yes, I get it. thank you. I knew the proposition , but i didn't know how to use it correctly.
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:47
$begingroup$
Yes, I get it. thank you. I knew the proposition , but i didn't know how to use it correctly.
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:47
$begingroup$
Yes, I get it. thank you. I knew the proposition , but i didn't know how to use it correctly.
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:47
add a comment |
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1
$begingroup$
Hi Anas, as part of the community guidelines you need to provide some more context and background to your question. Furthermore if you have attempted the problem you need to provide your working. Just a heads up :-)
$endgroup$
– DavidG
Dec 23 '18 at 7:19
$begingroup$
@DavidG, check .
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:31
$begingroup$
Looks good to me! Just wanted to let you know. This page has very specific rules about posting. It annoyed me to start with, but once you get use to it you realise it's a great format to work with. Sorry I can't help you wing your question though.
$endgroup$
– DavidG
Dec 23 '18 at 7:35
$begingroup$
Well, i can't find the contradiction with the last equality. but thanks anyway.
$endgroup$
– Anas BOUALII
Dec 23 '18 at 7:38
$begingroup$
Actually it is not Riemann integrable either. It is only improperly Riemann integrable.
$endgroup$
– metamorphy
Dec 23 '18 at 7:43