Mixing
Finite integral and limit
Let $L:mathbb{R} to [0,infty)$ a decreasing function such that
$int_{0}^{infty}x^{n-1}L(x)dx < infty$
Prove that $lim_{x to infty}x^{n}L(x) = 0$
Can anybody help me?
limits improper-integrals
asked Nov 21 '18 at 22:59
ZAFZAF
4307
add a comment |
Let $L:mathbb{R} to [0,infty)$ a decreasing function such that
$int_{0}^{infty}x^{n-1}L(x)dx < infty$
Prove that $lim_{x to infty}x^{n}L(x) = 0$
Can anybody help me?
limits improper-integrals
asked Nov 21 '18 at 22:59
ZAFZAF
4307
I guess you have to use induction. For $n=1$, convergence of the integral plus decresing of $L$ tells you that $lim_{xtoinfty}L(x)=0$. Maybe you can prove that $exists lim_{xtoinfty} x L(x)ge0$ and by the way of contradiction, this limit cannot be positive nor $+infty$.
– Tito Eliatron
Nov 21 '18 at 23:14
add a comment |
Let $L:mathbb{R} to [0,infty)$ a decreasing function such that
$int_{0}^{infty}x^{n-1}L(x)dx < infty$
Prove that $lim_{x to infty}x^{n}L(x) = 0$
Can anybody help me?
limits improper-integrals
asked Nov 21 '18 at 22:59
ZAFZAF
4307
Let $L:mathbb{R} to [0,infty)$ a decreasing function such that
$int_{0}^{infty}x^{n-1}L(x)dx < infty$
Prove that $lim_{x to infty}x^{n}L(x) = 0$
Can anybody help me?
limits improper-integrals
limits improper-integrals
asked Nov 21 '18 at 22:59
ZAFZAF
4307
asked Nov 21 '18 at 22:59
ZAFZAF
4307
edited Nov 21 '18 at 23:13
ZAF
asked Nov 21 '18 at 22:59
ZAFZAF
4307
asked Nov 21 '18 at 22:59
ZAFZAF
4307
asked Nov 21 '18 at 22:59
ZAFZAF
4307
4307
I guess you have to use induction. For $n=1$, convergence of the integral plus decresing of $L$ tells you that $lim_{xtoinfty}L(x)=0$. Maybe you can prove that $exists lim_{xtoinfty} x L(x)ge0$ and by the way of contradiction, this limit cannot be positive nor $+infty$.
– Tito Eliatron
Nov 21 '18 at 23:14
add a comment |
I guess you have to use induction. For $n=1$, convergence of the integral plus decresing of $L$ tells you that $lim_{xtoinfty}L(x)=0$. Maybe you can prove that $exists lim_{xtoinfty} x L(x)ge0$ and by the way of contradiction, this limit cannot be positive nor $+infty$.
– Tito Eliatron
Nov 21 '18 at 23:14
I guess you have to use induction. For $n=1$, convergence of the integral plus decresing of $L$ tells you that $lim_{xtoinfty}L(x)=0$. Maybe you can prove that $exists lim_{xtoinfty} x L(x)ge0$ and by the way of contradiction, this limit cannot be positive nor $+infty$.
– Tito Eliatron
Nov 21 '18 at 23:14
I guess you have to use induction. For $n=1$, convergence of the integral plus decresing of $L$ tells you that $lim_{xtoinfty}L(x)=0$. Maybe you can prove that $exists lim_{xtoinfty} x L(x)ge0$ and by the way of contradiction, this limit cannot be positive nor $+infty$.
– Tito Eliatron
Nov 21 '18 at 23:14
add a comment |
1 Answer
1
active
oldest
votes
Suppose $x^{n}L(x)$ does not tend to $0$. Then there is a sequence $x_j$ increasing to $infty$ and a $delta >0$ such that $x_j^{n}L(x_j)geq delta$ for all $j$. We can find a subsequence $(y_k)$ of $(x_j)$ such that $y_{k+1}^{n} >2y_k^{n}$. [Use induction for this]. Now $int_0^{infty} x^{n-1} L(x), dxgeq sum_k int_{y_k}^{y_{k+1}} x^{n-1} L(x), dx geq sum_k L(y_{k+1})frac {y_{k+1}^{n}-y_k^{n}} n geq delta sum_k frac {y_{k+1}^{n}-y_k^{n}} {ny_{k+1}^{n}} =sum_k frac {1-frac {y_k^{n}} {y_{k+1}^{n}}} n=infty$ because $y_{k+1}^{n} >2y_k^{n}$.
answered Nov 21 '18 at 23:36
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
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%2f3008499%2ffinite-integral-and-limit%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
Suppose $x^{n}L(x)$ does not tend to $0$. Then there is a sequence $x_j$ increasing to $infty$ and a $delta >0$ such that $x_j^{n}L(x_j)geq delta$ for all $j$. We can find a subsequence $(y_k)$ of $(x_j)$ such that $y_{k+1}^{n} >2y_k^{n}$. [Use induction for this]. Now $int_0^{infty} x^{n-1} L(x), dxgeq sum_k int_{y_k}^{y_{k+1}} x^{n-1} L(x), dx geq sum_k L(y_{k+1})frac {y_{k+1}^{n}-y_k^{n}} n geq delta sum_k frac {y_{k+1}^{n}-y_k^{n}} {ny_{k+1}^{n}} =sum_k frac {1-frac {y_k^{n}} {y_{k+1}^{n}}} n=infty$ because $y_{k+1}^{n} >2y_k^{n}$.
answered Nov 21 '18 at 23:36
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
add a comment |
Suppose $x^{n}L(x)$ does not tend to $0$. Then there is a sequence $x_j$ increasing to $infty$ and a $delta >0$ such that $x_j^{n}L(x_j)geq delta$ for all $j$. We can find a subsequence $(y_k)$ of $(x_j)$ such that $y_{k+1}^{n} >2y_k^{n}$. [Use induction for this]. Now $int_0^{infty} x^{n-1} L(x), dxgeq sum_k int_{y_k}^{y_{k+1}} x^{n-1} L(x), dx geq sum_k L(y_{k+1})frac {y_{k+1}^{n}-y_k^{n}} n geq delta sum_k frac {y_{k+1}^{n}-y_k^{n}} {ny_{k+1}^{n}} =sum_k frac {1-frac {y_k^{n}} {y_{k+1}^{n}}} n=infty$ because $y_{k+1}^{n} >2y_k^{n}$.
answered Nov 21 '18 at 23:36
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
add a comment |
Suppose $x^{n}L(x)$ does not tend to $0$. Then there is a sequence $x_j$ increasing to $infty$ and a $delta >0$ such that $x_j^{n}L(x_j)geq delta$ for all $j$. We can find a subsequence $(y_k)$ of $(x_j)$ such that $y_{k+1}^{n} >2y_k^{n}$. [Use induction for this]. Now $int_0^{infty} x^{n-1} L(x), dxgeq sum_k int_{y_k}^{y_{k+1}} x^{n-1} L(x), dx geq sum_k L(y_{k+1})frac {y_{k+1}^{n}-y_k^{n}} n geq delta sum_k frac {y_{k+1}^{n}-y_k^{n}} {ny_{k+1}^{n}} =sum_k frac {1-frac {y_k^{n}} {y_{k+1}^{n}}} n=infty$ because $y_{k+1}^{n} >2y_k^{n}$.
answered Nov 21 '18 at 23:36
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
Suppose $x^{n}L(x)$ does not tend to $0$. Then there is a sequence $x_j$ increasing to $infty$ and a $delta >0$ such that $x_j^{n}L(x_j)geq delta$ for all $j$. We can find a subsequence $(y_k)$ of $(x_j)$ such that $y_{k+1}^{n} >2y_k^{n}$. [Use induction for this]. Now $int_0^{infty} x^{n-1} L(x), dxgeq sum_k int_{y_k}^{y_{k+1}} x^{n-1} L(x), dx geq sum_k L(y_{k+1})frac {y_{k+1}^{n}-y_k^{n}} n geq delta sum_k frac {y_{k+1}^{n}-y_k^{n}} {ny_{k+1}^{n}} =sum_k frac {1-frac {y_k^{n}} {y_{k+1}^{n}}} n=infty$ because $y_{k+1}^{n} >2y_k^{n}$.
answered Nov 21 '18 at 23:36
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 21 '18 at 23:36
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 21 '18 at 23:36
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
answered Nov 21 '18 at 23:36
Kavi Rama MurthyKavi Rama Murthy
51.8k32055
51.8k32055
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3008499%2ffinite-integral-and-limit%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
I guess you have to use induction. For $n=1$, convergence of the integral plus decresing of $L$ tells you that $lim_{xtoinfty}L(x)=0$. Maybe you can prove that $exists lim_{xtoinfty} x L(x)ge0$ and by the way of contradiction, this limit cannot be positive nor $+infty$.
– Tito Eliatron
Nov 21 '18 at 23:14