Mixing
Proof that $int_{0}^{1}frac{1}{x-ln(x)}mathbb{d}x$ is convergent.
$begingroup$
We're asked to prove whether the above integral is convergent or divergent. And I would love to have some feedback on whether my proof is right or wrong, or how can I improve it.
Proof:
Notice that for $xin{}(0,1]$ we have that $x-ln(x)>0$ since $ln(x)leq0$ for $0<xleq{}1$. And since $x<sqrt{x}$ on this interval, it holds that $x-ln(x)>sqrt{x}$ such that $frac{1}{sqrt{x}}>frac{1}{x-ln(x)}>0$.
Now consider $int_0^{1}frac{1}{sqrt{x}}mathbb{d}x$. This integral converges since $lim_{cto0^+}int_{c}^{1}frac{1}{sqrt{x}}mathbb{d}x=0<infty$ hence by the comparison lemma, $int_{0}^{1}frac{1}{x-ln(x)}mathbb{d}x$ converges as well. QED
Thanks in advance
calculus integration proof-verification proof-writing improper-integrals
asked Jan 17 at 13:57
kareem bokaikareem bokai
536
$endgroup$
add a comment |
$begingroup$
We're asked to prove whether the above integral is convergent or divergent. And I would love to have some feedback on whether my proof is right or wrong, or how can I improve it.
Proof:
Notice that for $xin{}(0,1]$ we have that $x-ln(x)>0$ since $ln(x)leq0$ for $0<xleq{}1$. And since $x<sqrt{x}$ on this interval, it holds that $x-ln(x)>sqrt{x}$ such that $frac{1}{sqrt{x}}>frac{1}{x-ln(x)}>0$.
Now consider $int_0^{1}frac{1}{sqrt{x}}mathbb{d}x$. This integral converges since $lim_{cto0^+}int_{c}^{1}frac{1}{sqrt{x}}mathbb{d}x=0<infty$ hence by the comparison lemma, $int_{0}^{1}frac{1}{x-ln(x)}mathbb{d}x$ converges as well. QED
Thanks in advance
calculus integration proof-verification proof-writing improper-integrals
asked Jan 17 at 13:57
kareem bokaikareem bokai
536
$endgroup$
1
$begingroup$
Slight correction; $$lim_{cto0^+}int_{c}^{1}frac{1}{sqrt{x}}mathbb{d}x=color{red}{2}<infty$$ But other than this, I do believe the proof is correct.
$endgroup$
– John Doe
Jan 17 at 14:00
$begingroup$
@JohnDoe Ah yes of course, my bad. Awesome. Thanks :)
$endgroup$
– kareem bokai
Jan 17 at 14:03
add a comment |
$begingroup$
We're asked to prove whether the above integral is convergent or divergent. And I would love to have some feedback on whether my proof is right or wrong, or how can I improve it.
Proof:
Notice that for $xin{}(0,1]$ we have that $x-ln(x)>0$ since $ln(x)leq0$ for $0<xleq{}1$. And since $x<sqrt{x}$ on this interval, it holds that $x-ln(x)>sqrt{x}$ such that $frac{1}{sqrt{x}}>frac{1}{x-ln(x)}>0$.
Now consider $int_0^{1}frac{1}{sqrt{x}}mathbb{d}x$. This integral converges since $lim_{cto0^+}int_{c}^{1}frac{1}{sqrt{x}}mathbb{d}x=0<infty$ hence by the comparison lemma, $int_{0}^{1}frac{1}{x-ln(x)}mathbb{d}x$ converges as well. QED
Thanks in advance
calculus integration proof-verification proof-writing improper-integrals
asked Jan 17 at 13:57
kareem bokaikareem bokai
536
$endgroup$
We're asked to prove whether the above integral is convergent or divergent. And I would love to have some feedback on whether my proof is right or wrong, or how can I improve it.
Proof:
Notice that for $xin{}(0,1]$ we have that $x-ln(x)>0$ since $ln(x)leq0$ for $0<xleq{}1$. And since $x<sqrt{x}$ on this interval, it holds that $x-ln(x)>sqrt{x}$ such that $frac{1}{sqrt{x}}>frac{1}{x-ln(x)}>0$.
Now consider $int_0^{1}frac{1}{sqrt{x}}mathbb{d}x$. This integral converges since $lim_{cto0^+}int_{c}^{1}frac{1}{sqrt{x}}mathbb{d}x=0<infty$ hence by the comparison lemma, $int_{0}^{1}frac{1}{x-ln(x)}mathbb{d}x$ converges as well. QED
Thanks in advance
calculus integration proof-verification proof-writing improper-integrals
calculus integration proof-verification proof-writing improper-integrals
asked Jan 17 at 13:57
kareem bokaikareem bokai
536
asked Jan 17 at 13:57
kareem bokaikareem bokai
536
asked Jan 17 at 13:57
kareem bokaikareem bokai
536
asked Jan 17 at 13:57
kareem bokaikareem bokai
536
asked Jan 17 at 13:57
kareem bokaikareem bokai
536
536
1
$begingroup$
Slight correction; $$lim_{cto0^+}int_{c}^{1}frac{1}{sqrt{x}}mathbb{d}x=color{red}{2}<infty$$ But other than this, I do believe the proof is correct.
$endgroup$
– John Doe
Jan 17 at 14:00
$begingroup$
@JohnDoe Ah yes of course, my bad. Awesome. Thanks :)
$endgroup$
– kareem bokai
Jan 17 at 14:03
add a comment |
1
$begingroup$
Slight correction; $$lim_{cto0^+}int_{c}^{1}frac{1}{sqrt{x}}mathbb{d}x=color{red}{2}<infty$$ But other than this, I do believe the proof is correct.
$endgroup$
– John Doe
Jan 17 at 14:00
$begingroup$
@JohnDoe Ah yes of course, my bad. Awesome. Thanks :)
$endgroup$
– kareem bokai
Jan 17 at 14:03
1
1
$begingroup$
Slight correction; $$lim_{cto0^+}int_{c}^{1}frac{1}{sqrt{x}}mathbb{d}x=color{red}{2}<infty$$ But other than this, I do believe the proof is correct.
$endgroup$
– John Doe
Jan 17 at 14:00
$begingroup$
Slight correction; $$lim_{cto0^+}int_{c}^{1}frac{1}{sqrt{x}}mathbb{d}x=color{red}{2}<infty$$ But other than this, I do believe the proof is correct.
$endgroup$
– John Doe
Jan 17 at 14:00
$begingroup$
@JohnDoe Ah yes of course, my bad. Awesome. Thanks :)
$endgroup$
– kareem bokai
Jan 17 at 14:03
$begingroup$
@JohnDoe Ah yes of course, my bad. Awesome. Thanks :)
$endgroup$
– kareem bokai
Jan 17 at 14:03
add a comment |
0
active
oldest
votes
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%2f3077003%2fproof-that-int-01-frac1x-lnx-mathbbdx-is-convergent%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3077003%2fproof-that-int-01-frac1x-lnx-mathbbdx-is-convergent%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
1
$begingroup$
Slight correction; $$lim_{cto0^+}int_{c}^{1}frac{1}{sqrt{x}}mathbb{d}x=color{red}{2}<infty$$ But other than this, I do believe the proof is correct.
$endgroup$
– John Doe
Jan 17 at 14:00
$begingroup$
@JohnDoe Ah yes of course, my bad. Awesome. Thanks :)
$endgroup$
– kareem bokai
Jan 17 at 14:03