$int_{0}^{infty}frac{sin(x)}{x}dx$ exists, but $int_{mathbb{R^+}}frac{sin(x)}{x}dlambda$ doesn't exist?












1












$begingroup$


$int_{0}^{infty}frac{sin(x)}{x}dx$ 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







share|cite|improve this question











$endgroup$








  • 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 :-)
    $endgroup$
    – user150203
    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.
    $endgroup$
    – user150203
    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












$begingroup$


$int_{0}^{infty}frac{sin(x)}{x}dx$ 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







share|cite|improve this question











$endgroup$








  • 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 :-)
    $endgroup$
    – user150203
    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.
    $endgroup$
    – user150203
    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








1





$begingroup$


$int_{0}^{infty}frac{sin(x)}{x}dx$ 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







share|cite|improve this question











$endgroup$




$int_{0}^{infty}frac{sin(x)}{x}dx$ 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 7:37







user150203

















asked Dec 23 '18 at 7:15









Anas BOUALIIAnas BOUALII

1408




1408








  • 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 :-)
    $endgroup$
    – user150203
    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.
    $endgroup$
    – user150203
    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




    $begingroup$
    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 :-)
    $endgroup$
    – user150203
    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.
    $endgroup$
    – user150203
    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 :-)
$endgroup$
– user150203
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 :-)
$endgroup$
– user150203
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.
$endgroup$
– user150203
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.
$endgroup$
– user150203
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










1 Answer
1






active

oldest

votes


















1












$begingroup$

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?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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












Your Answer








StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050129%2fint-0-infty-frac-sinxxdx-exists-but-int-mathbbr-frac-si%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















1












$begingroup$

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?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














1












1








1





$begingroup$

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?






share|cite|improve this answer









$endgroup$



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?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 23 '18 at 7:40









Robert ZRobert Z

102k1072145




102k1072145












  • $begingroup$
    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




$begingroup$
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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3050129%2fint-0-infty-frac-sinxxdx-exists-but-int-mathbbr-frac-si%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more