Conditionally convergent integral
$begingroup$
It is known that in a conditionally convergent series, the terms can be rearranged so as to output any desired value. Thus, a conditionally convergent series is said to be undefined.
My question is about the following integral:
$intlimits_{ - infty }^infty {{{sin left( x right)} over x}dx} $
It's known that this integral is conditionally convergent: it converges, but no absolutely.
Can one say that if ${a_n}buildrel {n to infty } over
longrightarrow infty $ and ${b_n}buildrel {n to infty } over
longrightarrow - infty $ that the value of
$$mathop {lim }limits_{n to infty } intlimits_{{b_n}}^{{a_n}} {{{sin left( x right)} over x}dx} $$
depends on the choice of the sequences ${a_n},{b_n}$ ?
I have experimented a bit in wolfram alpha but no matter what I tried, the result was $pi $.
I do remember hearing and in fact I've read also here:
Is there a rearrangement theorem for conditionally convergent improper integrals?
That the value of the conditionally convergent integral depends on the rearrangement.
improper-integrals
asked Dec 7 '18 at 8:52
zokomokozokomoko
174214
$endgroup$
add a comment |
$begingroup$
It is known that in a conditionally convergent series, the terms can be rearranged so as to output any desired value. Thus, a conditionally convergent series is said to be undefined.
My question is about the following integral:
$intlimits_{ - infty }^infty {{{sin left( x right)} over x}dx} $
It's known that this integral is conditionally convergent: it converges, but no absolutely.
Can one say that if ${a_n}buildrel {n to infty } over
longrightarrow infty $ and ${b_n}buildrel {n to infty } over
longrightarrow - infty $ that the value of
$$mathop {lim }limits_{n to infty } intlimits_{{b_n}}^{{a_n}} {{{sin left( x right)} over x}dx} $$
depends on the choice of the sequences ${a_n},{b_n}$ ?
I have experimented a bit in wolfram alpha but no matter what I tried, the result was $pi $.
I do remember hearing and in fact I've read also here:
Is there a rearrangement theorem for conditionally convergent improper integrals?
That the value of the conditionally convergent integral depends on the rearrangement.
improper-integrals
asked Dec 7 '18 at 8:52
zokomokozokomoko
174214
$endgroup$
add a comment |
$begingroup$
It is known that in a conditionally convergent series, the terms can be rearranged so as to output any desired value. Thus, a conditionally convergent series is said to be undefined.
My question is about the following integral:
$intlimits_{ - infty }^infty {{{sin left( x right)} over x}dx} $
It's known that this integral is conditionally convergent: it converges, but no absolutely.
Can one say that if ${a_n}buildrel {n to infty } over
longrightarrow infty $ and ${b_n}buildrel {n to infty } over
longrightarrow - infty $ that the value of
$$mathop {lim }limits_{n to infty } intlimits_{{b_n}}^{{a_n}} {{{sin left( x right)} over x}dx} $$
depends on the choice of the sequences ${a_n},{b_n}$ ?
I have experimented a bit in wolfram alpha but no matter what I tried, the result was $pi $.
I do remember hearing and in fact I've read also here:
Is there a rearrangement theorem for conditionally convergent improper integrals?
That the value of the conditionally convergent integral depends on the rearrangement.
improper-integrals
asked Dec 7 '18 at 8:52
zokomokozokomoko
174214
$endgroup$
It is known that in a conditionally convergent series, the terms can be rearranged so as to output any desired value. Thus, a conditionally convergent series is said to be undefined.
My question is about the following integral:
$intlimits_{ - infty }^infty {{{sin left( x right)} over x}dx} $
It's known that this integral is conditionally convergent: it converges, but no absolutely.
Can one say that if ${a_n}buildrel {n to infty } over
longrightarrow infty $ and ${b_n}buildrel {n to infty } over
longrightarrow - infty $ that the value of
$$mathop {lim }limits_{n to infty } intlimits_{{b_n}}^{{a_n}} {{{sin left( x right)} over x}dx} $$
depends on the choice of the sequences ${a_n},{b_n}$ ?
I have experimented a bit in wolfram alpha but no matter what I tried, the result was $pi $.
I do remember hearing and in fact I've read also here:
Is there a rearrangement theorem for conditionally convergent improper integrals?
That the value of the conditionally convergent integral depends on the rearrangement.
improper-integrals
improper-integrals
asked Dec 7 '18 at 8:52
zokomokozokomoko
174214
asked Dec 7 '18 at 8:52
zokomokozokomoko
174214
asked Dec 7 '18 at 8:52
zokomokozokomoko
174214
asked Dec 7 '18 at 8:52
zokomokozokomoko
174214
asked Dec 7 '18 at 8:52
zokomokozokomoko
174214
174214
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The limit is $pi$ irrespective of what those sequences are. It is a standard fact that $lim _{tto infty}int_0^{t} frac {sin , x} x , dx=pi/2$ from which my claim follows.
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029668%2fconditionally-convergent-integral%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
$begingroup$
The limit is $pi$ irrespective of what those sequences are. It is a standard fact that $lim _{tto infty}int_0^{t} frac {sin , x} x , dx=pi/2$ from which my claim follows.
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
$endgroup$
add a comment |
$begingroup$
The limit is $pi$ irrespective of what those sequences are. It is a standard fact that $lim _{tto infty}int_0^{t} frac {sin , x} x , dx=pi/2$ from which my claim follows.
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
$endgroup$
add a comment |
$begingroup$
The limit is $pi$ irrespective of what those sequences are. It is a standard fact that $lim _{tto infty}int_0^{t} frac {sin , x} x , dx=pi/2$ from which my claim follows.
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
$endgroup$
The limit is $pi$ irrespective of what those sequences are. It is a standard fact that $lim _{tto infty}int_0^{t} frac {sin , x} x , dx=pi/2$ from which my claim follows.
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
answered Dec 7 '18 at 8:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42161
58.2k42161
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029668%2fconditionally-convergent-integral%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
