Convergence of $int_1^2 frac{1}{sqrt ln x} mathrm { d}x$ by Asymptotic Comparison Test.
$begingroup$
According to the material I have,
If $f:[a,b)to mathbb{R} ,bin mathbb{R},fgeq0 $
Then:
i) If $f$ is an infinite function of real order $alphalt1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}xltinfty$
ii) If $f$ is an infinite function of real order $alphageq1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}x=infty$
Based on this information I am trying to evaluate the convergence of $int_1^2 frac{1}{sqrt ln x} mathrm { d}x$
In order to determine the order of my function I consider $$lim_{xto1+}{frac{1}{sqrtln x overfrac{1}{(x-2)^alpha}}}=lim_{xto1+}frac {(x-2)^alpha}{sqrtln x}$$
Despite trying various methods I haven't been able to obtain a real number as a limit value for some $alphainmathbb{R}$
asymptotics improper-integrals
asked Jan 8 at 14:23
Turan NəsibliTuran Nəsibli
294
$endgroup$
add a comment |
$begingroup$
According to the material I have,
If $f:[a,b)to mathbb{R} ,bin mathbb{R},fgeq0 $
Then:
i) If $f$ is an infinite function of real order $alphalt1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}xltinfty$
ii) If $f$ is an infinite function of real order $alphageq1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}x=infty$
Based on this information I am trying to evaluate the convergence of $int_1^2 frac{1}{sqrt ln x} mathrm { d}x$
In order to determine the order of my function I consider $$lim_{xto1+}{frac{1}{sqrtln x overfrac{1}{(x-2)^alpha}}}=lim_{xto1+}frac {(x-2)^alpha}{sqrtln x}$$
Despite trying various methods I haven't been able to obtain a real number as a limit value for some $alphainmathbb{R}$
asymptotics improper-integrals
asked Jan 8 at 14:23
Turan NəsibliTuran Nəsibli
294
$endgroup$
add a comment |
$begingroup$
According to the material I have,
If $f:[a,b)to mathbb{R} ,bin mathbb{R},fgeq0 $
Then:
i) If $f$ is an infinite function of real order $alphalt1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}xltinfty$
ii) If $f$ is an infinite function of real order $alphageq1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}x=infty$
Based on this information I am trying to evaluate the convergence of $int_1^2 frac{1}{sqrt ln x} mathrm { d}x$
In order to determine the order of my function I consider $$lim_{xto1+}{frac{1}{sqrtln x overfrac{1}{(x-2)^alpha}}}=lim_{xto1+}frac {(x-2)^alpha}{sqrtln x}$$
Despite trying various methods I haven't been able to obtain a real number as a limit value for some $alphainmathbb{R}$
asymptotics improper-integrals
asked Jan 8 at 14:23
Turan NəsibliTuran Nəsibli
294
$endgroup$
According to the material I have,
If $f:[a,b)to mathbb{R} ,bin mathbb{R},fgeq0 $
Then:
i) If $f$ is an infinite function of real order $alphalt1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}xltinfty$
ii) If $f$ is an infinite function of real order $alphageq1$ with respect to $frac {1}{x-b}$ $implies$ $int_a^bf(x)mathrm { d}x=infty$
Based on this information I am trying to evaluate the convergence of $int_1^2 frac{1}{sqrt ln x} mathrm { d}x$
In order to determine the order of my function I consider $$lim_{xto1+}{frac{1}{sqrtln x overfrac{1}{(x-2)^alpha}}}=lim_{xto1+}frac {(x-2)^alpha}{sqrtln x}$$
Despite trying various methods I haven't been able to obtain a real number as a limit value for some $alphainmathbb{R}$
asymptotics improper-integrals
asymptotics improper-integrals
asked Jan 8 at 14:23
Turan NəsibliTuran Nəsibli
294
asked Jan 8 at 14:23
Turan NəsibliTuran Nəsibli
294
asked Jan 8 at 14:23
Turan NəsibliTuran Nəsibli
294
asked Jan 8 at 14:23
Turan NəsibliTuran Nəsibli
294
asked Jan 8 at 14:23
Turan NəsibliTuran Nəsibli
294
294
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An issue is that you look at $b=2$. You should consider the fact that the "problem" is at $1$, and adapt your theorem (it's easy) for $fcolon (a,b]tomathbb{R}$ and look at the $a=1$ end.
Rewrite, for $x>1$,
$$
sqrt{ln x} = sqrt{ln(1+(x-1))},.
$$
This seems silly, but now, recall that
$$
lim_{uto 0} frac{ln(1+u)}{u} = 1
$$
so that
$$
lim_{xto 1} frac{sqrt{ln x}}{(x-1)^{1/2}} = lim_{xto 1} sqrt{frac{ln x}{x-1}}= lim_{xto 1} sqrt{frac{ln(1+(x-1))}{x-1}} = sqrt{lim_{xto 1} frac{ln(1+(x-1))}{x-1}} = sqrt{1}=1
$$
Can you conclude?
answered Jan 8 at 14:49
Clement C.Clement C.
49.8k33886
$endgroup$
$begingroup$
Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
$endgroup$
– Turan Nəsibli
Jan 8 at 15:42
$begingroup$
You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
$endgroup$
– Clement C.
Jan 8 at 15:44
$begingroup$
I'm trying to fix the problem,and thank you for the help.
$endgroup$
– Turan Nəsibli
Jan 8 at 15:48
$begingroup$
@TuranNəsibli You're welcome! (for the Latex issue: usa^{-1}, with braces.)
$endgroup$
– Clement C.
Jan 8 at 15:48
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%2f3066228%2fconvergence-of-int-12-frac1-sqrt-ln-x-mathrm-dx-by-asymptotic-comp%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
$begingroup$
An issue is that you look at $b=2$. You should consider the fact that the "problem" is at $1$, and adapt your theorem (it's easy) for $fcolon (a,b]tomathbb{R}$ and look at the $a=1$ end.
Rewrite, for $x>1$,
$$
sqrt{ln x} = sqrt{ln(1+(x-1))},.
$$
This seems silly, but now, recall that
$$
lim_{uto 0} frac{ln(1+u)}{u} = 1
$$
so that
$$
lim_{xto 1} frac{sqrt{ln x}}{(x-1)^{1/2}} = lim_{xto 1} sqrt{frac{ln x}{x-1}}= lim_{xto 1} sqrt{frac{ln(1+(x-1))}{x-1}} = sqrt{lim_{xto 1} frac{ln(1+(x-1))}{x-1}} = sqrt{1}=1
$$
Can you conclude?
answered Jan 8 at 14:49
Clement C.Clement C.
49.8k33886
$endgroup$
$begingroup$
Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
$endgroup$
– Turan Nəsibli
Jan 8 at 15:42
$begingroup$
You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
$endgroup$
– Clement C.
Jan 8 at 15:44
$begingroup$
I'm trying to fix the problem,and thank you for the help.
$endgroup$
– Turan Nəsibli
Jan 8 at 15:48
$begingroup$
@TuranNəsibli You're welcome! (for the Latex issue: usa^{-1}, with braces.)
$endgroup$
– Clement C.
Jan 8 at 15:48
add a comment |
$begingroup$
An issue is that you look at $b=2$. You should consider the fact that the "problem" is at $1$, and adapt your theorem (it's easy) for $fcolon (a,b]tomathbb{R}$ and look at the $a=1$ end.
Rewrite, for $x>1$,
$$
sqrt{ln x} = sqrt{ln(1+(x-1))},.
$$
This seems silly, but now, recall that
$$
lim_{uto 0} frac{ln(1+u)}{u} = 1
$$
so that
$$
lim_{xto 1} frac{sqrt{ln x}}{(x-1)^{1/2}} = lim_{xto 1} sqrt{frac{ln x}{x-1}}= lim_{xto 1} sqrt{frac{ln(1+(x-1))}{x-1}} = sqrt{lim_{xto 1} frac{ln(1+(x-1))}{x-1}} = sqrt{1}=1
$$
Can you conclude?
answered Jan 8 at 14:49
Clement C.Clement C.
49.8k33886
$endgroup$
$begingroup$
Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
$endgroup$
– Turan Nəsibli
Jan 8 at 15:42
$begingroup$
You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
$endgroup$
– Clement C.
Jan 8 at 15:44
$begingroup$
I'm trying to fix the problem,and thank you for the help.
$endgroup$
– Turan Nəsibli
Jan 8 at 15:48
$begingroup$
@TuranNəsibli You're welcome! (for the Latex issue: usa^{-1}, with braces.)
$endgroup$
– Clement C.
Jan 8 at 15:48
add a comment |
$begingroup$
An issue is that you look at $b=2$. You should consider the fact that the "problem" is at $1$, and adapt your theorem (it's easy) for $fcolon (a,b]tomathbb{R}$ and look at the $a=1$ end.
Rewrite, for $x>1$,
$$
sqrt{ln x} = sqrt{ln(1+(x-1))},.
$$
This seems silly, but now, recall that
$$
lim_{uto 0} frac{ln(1+u)}{u} = 1
$$
so that
$$
lim_{xto 1} frac{sqrt{ln x}}{(x-1)^{1/2}} = lim_{xto 1} sqrt{frac{ln x}{x-1}}= lim_{xto 1} sqrt{frac{ln(1+(x-1))}{x-1}} = sqrt{lim_{xto 1} frac{ln(1+(x-1))}{x-1}} = sqrt{1}=1
$$
Can you conclude?
answered Jan 8 at 14:49
Clement C.Clement C.
49.8k33886
$endgroup$
An issue is that you look at $b=2$. You should consider the fact that the "problem" is at $1$, and adapt your theorem (it's easy) for $fcolon (a,b]tomathbb{R}$ and look at the $a=1$ end.
Rewrite, for $x>1$,
$$
sqrt{ln x} = sqrt{ln(1+(x-1))},.
$$
This seems silly, but now, recall that
$$
lim_{uto 0} frac{ln(1+u)}{u} = 1
$$
so that
$$
lim_{xto 1} frac{sqrt{ln x}}{(x-1)^{1/2}} = lim_{xto 1} sqrt{frac{ln x}{x-1}}= lim_{xto 1} sqrt{frac{ln(1+(x-1))}{x-1}} = sqrt{lim_{xto 1} frac{ln(1+(x-1))}{x-1}} = sqrt{1}=1
$$
Can you conclude?
answered Jan 8 at 14:49
Clement C.Clement C.
49.8k33886
answered Jan 8 at 14:49
Clement C.Clement C.
49.8k33886
answered Jan 8 at 14:49
Clement C.Clement C.
49.8k33886
answered Jan 8 at 14:49
Clement C.Clement C.
49.8k33886
49.8k33886
$begingroup$
Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
$endgroup$
– Turan Nəsibli
Jan 8 at 15:42
$begingroup$
You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
$endgroup$
– Clement C.
Jan 8 at 15:44
$begingroup$
I'm trying to fix the problem,and thank you for the help.
$endgroup$
– Turan Nəsibli
Jan 8 at 15:48
$begingroup$
@TuranNəsibli You're welcome! (for the Latex issue: usa^{-1}, with braces.)
$endgroup$
– Clement C.
Jan 8 at 15:48
add a comment |
$begingroup$
Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
$endgroup$
– Turan Nəsibli
Jan 8 at 15:42
$begingroup$
You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
$endgroup$
– Clement C.
Jan 8 at 15:44
$begingroup$
I'm trying to fix the problem,and thank you for the help.
$endgroup$
– Turan Nəsibli
Jan 8 at 15:48
$begingroup$
@TuranNəsibli You're welcome! (for the Latex issue: usa^{-1}, with braces.)
$endgroup$
– Clement C.
Jan 8 at 15:48
$begingroup$
Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
$endgroup$
– Turan Nəsibli
Jan 8 at 15:42
$begingroup$
Well,yes, problem is indeed at $a=1$,so it seemed weird to me trying to determine the order of function with respect to ${1over x-2}$.So, to conclude, if c is a point where my function is problematic, I should try to determine the order with respect to ${1over x-c}$.Moreover, if I am determining the order of $f(x)={1over sqrt ln x}$ ,shouldn't I write $$lim_{xto1+}{(x-1)^{1over2}over sqrt{ln (1+(x-1))}}=sqrt{lim_{xto1+}frac{x-1}{ln(1+(x-1))}}=sqrt{lim_{xto1+}(frac {ln(1+(x-1))}{x-1})^-1}=sqrt1=1$$ ?
$endgroup$
– Turan Nəsibli
Jan 8 at 15:42
$begingroup$
You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
$endgroup$
– Clement C.
Jan 8 at 15:44
$begingroup$
You seem to have a problem with the superscript, but yes. But that's the same... if $f/gto 1$ and $g>0$, then $g/fto 1$ and $f>0$. @TuranNəsibli
$endgroup$
– Clement C.
Jan 8 at 15:44
$begingroup$
I'm trying to fix the problem,and thank you for the help.
$endgroup$
– Turan Nəsibli
Jan 8 at 15:48
$begingroup$
I'm trying to fix the problem,and thank you for the help.
$endgroup$
– Turan Nəsibli
Jan 8 at 15:48
$begingroup$
@TuranNəsibli You're welcome! (for the Latex issue: us
a^{-1}, with braces.)$endgroup$
– Clement C.
Jan 8 at 15:48
$begingroup$
@TuranNəsibli You're welcome! (for the Latex issue: us
a^{-1}, with braces.)$endgroup$
– Clement C.
Jan 8 at 15:48
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%2f3066228%2fconvergence-of-int-12-frac1-sqrt-ln-x-mathrm-dx-by-asymptotic-comp%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
