Show: $lim_{nrightarrow infty} left|int_{1}^{e}left[ln(x)right]^n:dx right|= 0 $












3












$begingroup$


This is probably an unremarkable result, but I was curious about exploring the following family of definite integrals to see what asymptotic behaviour it as as $n rightarrow infty$



$$I_n = int_{1}^{e}left[ln(x)right]^n:dx mbox{ for } n in mathbb{N}$$



The method I took:



Employ integration by parts:



begin{align}
v'(x) &= 1 & u(x) &= left[ln(x)right]^n \
v(x) &= x & u'(x) &= nleft[ln(x)right]^{n - 1}frac{1}{x}
end{align}



Thus,



begin{align}
I_n &= int_{1}^{e}left[ln(x)right]^n:dx = x cdot left[ln(x)right]^n Big|_{1}^{e} - int_{1}^{e} x cdot nleft[ln(x)right]^{n - 1}frac{1}{x} \
&= e - nint_{1}^{e}left[ln(x)right]^{n - 1}:dx
end{align}



Thus,



begin{equation}
I_n = e - nI_{n - 1}
end{equation}



Where



begin{equation}
I_1 = int_{1}^{e} ln(x):dx = xleft(ln(x) - 1right)Big|_{1}^{e} = 1
end{equation}



This yields the solution:



begin{align}
I_n &= n!left[(-1)^{n - 1} + esum_{i = 0}^{n - 2} frac{(-1)^i}{(n - i)!} right] = n!left(-1right)^nleft[-1 + esum_{i = 2}^{n } frac{(-1)^i}{i!} right] \
&= n!left(-1right)^nleft[-1 + esum_{i = 0}^{n } frac{(-1)^i}{i!} right]
end{align}



We see that:



begin{equation}
lim_{nrightarrow infty} I_n = n! (-1)^n left( -1 + ecdot e^{-1} right) = 0
end{equation}



Or



begin{equation}
lim_{nrightarrow infty}int_{1}^{e}left[ln(x)right]^n:dx = 0
end{equation}



I was wondering if I could have a more technical eye have a look over this to see if the result I've obtained is correct.



Edit - Final result was incorrect and now corrected as per Song's pickup.






integration limits proof-verification definite-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Your mistake is $-1+ecdot e^{-1} = 1$. It should be $0$. Actually, by dominated convergence theorem, $int_1^e (log x)^n dx to 0$. In the same spirit, we can show $int_1^A (log x)^n dx = mathcal{o}((log A)^n)$.
    $endgroup$
    – Song
    Dec 10 '18 at 23:22












  • $begingroup$
    Hi @Song - Yes, indeed- Thank you very much. I will edit my post.
    $endgroup$
    – DavidG
    Dec 10 '18 at 23:23






  • 2




    $begingroup$
    For a more direct approach leading to the same optimal result, note that, by concavity of the logarithm, for every $1<x<e$, $ln x<e^{-1}x$ hence $$I_n<e^{-n}int_1^ex^ndx<e^{-n}int_0^ex^ndx=frac e{n+1}$$
    $endgroup$
    – Did
    Dec 10 '18 at 23:39
















3












$begingroup$


This is probably an unremarkable result, but I was curious about exploring the following family of definite integrals to see what asymptotic behaviour it as as $n rightarrow infty$



$$I_n = int_{1}^{e}left[ln(x)right]^n:dx mbox{ for } n in mathbb{N}$$



The method I took:



Employ integration by parts:



begin{align}
v'(x) &= 1 & u(x) &= left[ln(x)right]^n \
v(x) &= x & u'(x) &= nleft[ln(x)right]^{n - 1}frac{1}{x}
end{align}



Thus,



begin{align}
I_n &= int_{1}^{e}left[ln(x)right]^n:dx = x cdot left[ln(x)right]^n Big|_{1}^{e} - int_{1}^{e} x cdot nleft[ln(x)right]^{n - 1}frac{1}{x} \
&= e - nint_{1}^{e}left[ln(x)right]^{n - 1}:dx
end{align}



Thus,



begin{equation}
I_n = e - nI_{n - 1}
end{equation}



Where



begin{equation}
I_1 = int_{1}^{e} ln(x):dx = xleft(ln(x) - 1right)Big|_{1}^{e} = 1
end{equation}



This yields the solution:



begin{align}
I_n &= n!left[(-1)^{n - 1} + esum_{i = 0}^{n - 2} frac{(-1)^i}{(n - i)!} right] = n!left(-1right)^nleft[-1 + esum_{i = 2}^{n } frac{(-1)^i}{i!} right] \
&= n!left(-1right)^nleft[-1 + esum_{i = 0}^{n } frac{(-1)^i}{i!} right]
end{align}



We see that:



begin{equation}
lim_{nrightarrow infty} I_n = n! (-1)^n left( -1 + ecdot e^{-1} right) = 0
end{equation}



Or



begin{equation}
lim_{nrightarrow infty}int_{1}^{e}left[ln(x)right]^n:dx = 0
end{equation}



I was wondering if I could have a more technical eye have a look over this to see if the result I've obtained is correct.



Edit - Final result was incorrect and now corrected as per Song's pickup.






integration limits proof-verification definite-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Your mistake is $-1+ecdot e^{-1} = 1$. It should be $0$. Actually, by dominated convergence theorem, $int_1^e (log x)^n dx to 0$. In the same spirit, we can show $int_1^A (log x)^n dx = mathcal{o}((log A)^n)$.
    $endgroup$
    – Song
    Dec 10 '18 at 23:22












  • $begingroup$
    Hi @Song - Yes, indeed- Thank you very much. I will edit my post.
    $endgroup$
    – DavidG
    Dec 10 '18 at 23:23






  • 2




    $begingroup$
    For a more direct approach leading to the same optimal result, note that, by concavity of the logarithm, for every $1<x<e$, $ln x<e^{-1}x$ hence $$I_n<e^{-n}int_1^ex^ndx<e^{-n}int_0^ex^ndx=frac e{n+1}$$
    $endgroup$
    – Did
    Dec 10 '18 at 23:39














3












3








3





$begingroup$


This is probably an unremarkable result, but I was curious about exploring the following family of definite integrals to see what asymptotic behaviour it as as $n rightarrow infty$



$$I_n = int_{1}^{e}left[ln(x)right]^n:dx mbox{ for } n in mathbb{N}$$



The method I took:



Employ integration by parts:



begin{align}
v'(x) &= 1 & u(x) &= left[ln(x)right]^n \
v(x) &= x & u'(x) &= nleft[ln(x)right]^{n - 1}frac{1}{x}
end{align}



Thus,



begin{align}
I_n &= int_{1}^{e}left[ln(x)right]^n:dx = x cdot left[ln(x)right]^n Big|_{1}^{e} - int_{1}^{e} x cdot nleft[ln(x)right]^{n - 1}frac{1}{x} \
&= e - nint_{1}^{e}left[ln(x)right]^{n - 1}:dx
end{align}



Thus,



begin{equation}
I_n = e - nI_{n - 1}
end{equation}



Where



begin{equation}
I_1 = int_{1}^{e} ln(x):dx = xleft(ln(x) - 1right)Big|_{1}^{e} = 1
end{equation}



This yields the solution:



begin{align}
I_n &= n!left[(-1)^{n - 1} + esum_{i = 0}^{n - 2} frac{(-1)^i}{(n - i)!} right] = n!left(-1right)^nleft[-1 + esum_{i = 2}^{n } frac{(-1)^i}{i!} right] \
&= n!left(-1right)^nleft[-1 + esum_{i = 0}^{n } frac{(-1)^i}{i!} right]
end{align}



We see that:



begin{equation}
lim_{nrightarrow infty} I_n = n! (-1)^n left( -1 + ecdot e^{-1} right) = 0
end{equation}



Or



begin{equation}
lim_{nrightarrow infty}int_{1}^{e}left[ln(x)right]^n:dx = 0
end{equation}



I was wondering if I could have a more technical eye have a look over this to see if the result I've obtained is correct.



Edit - Final result was incorrect and now corrected as per Song's pickup.






integration limits proof-verification definite-integrals







share|cite|improve this question











$endgroup$




This is probably an unremarkable result, but I was curious about exploring the following family of definite integrals to see what asymptotic behaviour it as as $n rightarrow infty$



$$I_n = int_{1}^{e}left[ln(x)right]^n:dx mbox{ for } n in mathbb{N}$$



The method I took:



Employ integration by parts:



begin{align}
v'(x) &= 1 & u(x) &= left[ln(x)right]^n \
v(x) &= x & u'(x) &= nleft[ln(x)right]^{n - 1}frac{1}{x}
end{align}



Thus,



begin{align}
I_n &= int_{1}^{e}left[ln(x)right]^n:dx = x cdot left[ln(x)right]^n Big|_{1}^{e} - int_{1}^{e} x cdot nleft[ln(x)right]^{n - 1}frac{1}{x} \
&= e - nint_{1}^{e}left[ln(x)right]^{n - 1}:dx
end{align}



Thus,



begin{equation}
I_n = e - nI_{n - 1}
end{equation}



Where



begin{equation}
I_1 = int_{1}^{e} ln(x):dx = xleft(ln(x) - 1right)Big|_{1}^{e} = 1
end{equation}



This yields the solution:



begin{align}
I_n &= n!left[(-1)^{n - 1} + esum_{i = 0}^{n - 2} frac{(-1)^i}{(n - i)!} right] = n!left(-1right)^nleft[-1 + esum_{i = 2}^{n } frac{(-1)^i}{i!} right] \
&= n!left(-1right)^nleft[-1 + esum_{i = 0}^{n } frac{(-1)^i}{i!} right]
end{align}



We see that:



begin{equation}
lim_{nrightarrow infty} I_n = n! (-1)^n left( -1 + ecdot e^{-1} right) = 0
end{equation}



Or



begin{equation}
lim_{nrightarrow infty}int_{1}^{e}left[ln(x)right]^n:dx = 0
end{equation}



I was wondering if I could have a more technical eye have a look over this to see if the result I've obtained is correct.



Edit - Final result was incorrect and now corrected as per Song's pickup.






integration limits proof-verification definite-integrals




integration limits proof-verification definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 23:24







DavidG

















asked Dec 10 '18 at 23:17









DavidGDavidG

2,2641721




2,2641721








  • 1




    $begingroup$
    Your mistake is $-1+ecdot e^{-1} = 1$. It should be $0$. Actually, by dominated convergence theorem, $int_1^e (log x)^n dx to 0$. In the same spirit, we can show $int_1^A (log x)^n dx = mathcal{o}((log A)^n)$.
    $endgroup$
    – Song
    Dec 10 '18 at 23:22












  • $begingroup$
    Hi @Song - Yes, indeed- Thank you very much. I will edit my post.
    $endgroup$
    – DavidG
    Dec 10 '18 at 23:23






  • 2




    $begingroup$
    For a more direct approach leading to the same optimal result, note that, by concavity of the logarithm, for every $1<x<e$, $ln x<e^{-1}x$ hence $$I_n<e^{-n}int_1^ex^ndx<e^{-n}int_0^ex^ndx=frac e{n+1}$$
    $endgroup$
    – Did
    Dec 10 '18 at 23:39














  • 1




    $begingroup$
    Your mistake is $-1+ecdot e^{-1} = 1$. It should be $0$. Actually, by dominated convergence theorem, $int_1^e (log x)^n dx to 0$. In the same spirit, we can show $int_1^A (log x)^n dx = mathcal{o}((log A)^n)$.
    $endgroup$
    – Song
    Dec 10 '18 at 23:22












  • $begingroup$
    Hi @Song - Yes, indeed- Thank you very much. I will edit my post.
    $endgroup$
    – DavidG
    Dec 10 '18 at 23:23






  • 2




    $begingroup$
    For a more direct approach leading to the same optimal result, note that, by concavity of the logarithm, for every $1<x<e$, $ln x<e^{-1}x$ hence $$I_n<e^{-n}int_1^ex^ndx<e^{-n}int_0^ex^ndx=frac e{n+1}$$
    $endgroup$
    – Did
    Dec 10 '18 at 23:39








1




1




$begingroup$
Your mistake is $-1+ecdot e^{-1} = 1$. It should be $0$. Actually, by dominated convergence theorem, $int_1^e (log x)^n dx to 0$. In the same spirit, we can show $int_1^A (log x)^n dx = mathcal{o}((log A)^n)$.
$endgroup$
– Song
Dec 10 '18 at 23:22






$begingroup$
Your mistake is $-1+ecdot e^{-1} = 1$. It should be $0$. Actually, by dominated convergence theorem, $int_1^e (log x)^n dx to 0$. In the same spirit, we can show $int_1^A (log x)^n dx = mathcal{o}((log A)^n)$.
$endgroup$
– Song
Dec 10 '18 at 23:22














$begingroup$
Hi @Song - Yes, indeed- Thank you very much. I will edit my post.
$endgroup$
– DavidG
Dec 10 '18 at 23:23




$begingroup$
Hi @Song - Yes, indeed- Thank you very much. I will edit my post.
$endgroup$
– DavidG
Dec 10 '18 at 23:23




2




2




$begingroup$
For a more direct approach leading to the same optimal result, note that, by concavity of the logarithm, for every $1<x<e$, $ln x<e^{-1}x$ hence $$I_n<e^{-n}int_1^ex^ndx<e^{-n}int_0^ex^ndx=frac e{n+1}$$
$endgroup$
– Did
Dec 10 '18 at 23:39




$begingroup$
For a more direct approach leading to the same optimal result, note that, by concavity of the logarithm, for every $1<x<e$, $ln x<e^{-1}x$ hence $$I_n<e^{-n}int_1^ex^ndx<e^{-n}int_0^ex^ndx=frac e{n+1}$$
$endgroup$
– Did
Dec 10 '18 at 23:39










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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3034629%2fshow-lim-n-rightarrow-infty-left-int-1e-left-lnx-rightn-dx-r%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
















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%2f3034629%2fshow-lim-n-rightarrow-infty-left-int-1e-left-lnx-rightn-dx-r%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