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












2












$begingroup$


$int_{0}^{infty}frac{sin(x)}{x}dx$ 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 :-)
    $endgroup$
    – 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
















2












$begingroup$


$int_{0}^{infty}frac{sin(x)}{x}dx$ 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 :-)
    $endgroup$
    – 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














2












2








2





$begingroup$


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









DavidG

2,1321724




2,1321724










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














  • 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










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.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

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

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


















  • $begingroup$
    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

Ellipse (mathématiques)

Image
Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
Read more

Quarter-circle Tiles

Image
up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
Read more

Mont Emei

Image
Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
Read more