Fdtxjr

Are Captain Marvel's powers affected by Thanos breaking the Tesseract and claiming the stone?

Do you waste sorcery points if you try to apply metamagic to a spell from a scroll but fail to cast it?

Grepping string, but include all non-blank lines following each grep match

How much do grades matter for a future academia position?

Does the Crossbow Expert feat's extra crossbow attack work with the reaction attack from a Hunter ranger's Giant Killer feature?

How do I Interface a PS/2 Keyboard without Modern Techniques?

How do you justify more code being written by following clean code practices?

Do I have to take mana from my deck or hand when tapping a dual land?

Usage of an old photo with expired copyright

What (the heck) is a Super Worm Equinox Moon?

Why would five hundred and five be same as one?

Why didn't Voldemort know what Grindelwald looked like?

Do people actually use the word "kaputt" in conversation?

Is there a reason to prefer HFS+ over APFS for disk images in High Sierra and/or Mojave?

How can I, as DM, avoid the Conga Line of Death occurring when implementing some form of flanking rule?

If the only attacker is removed from combat, is a creature still counted as having attacked this turn?

Why is the principal energy of an electron lower for excited electrons in a higher energy state?

Showing mass murder in a kid's book

Check if object is null and return null

Echo with obfuscation

Would this string work as string?

How would you translate "more" for use as an interface button?

How do I prevent inappropriate ads from appearing in my game?

How to leave product feedback on macOS?

Show that $lim_nrightarrowinftyint_0^nfracsqrt xln x(1+x)^2,dx=pi$

Computation of $lim_n rightarrow infty n(int_0^infty frac11+x^4+x^n mathrmdm(x)-C)$Show that $lim_nrightarrow infty int_0^pi/2 2^n sqrtn sin^n(x) cos^n-2(x) ; dx = sqrt2pi$The value of $lim_nto inftyint_-infty^inftyf(x)cos^2 nx dx.$Show that $lim_t rightarrow infty tF(t) = 0.$Calculate $lim_nrightarrow infty int_-infty^inftyfrac11+fracx^4ndx$ if it exists.Calculate $lim_epsilon rightarrow 01over epsilon^2cdotbiggl( 1-1over2int_-1^1sqrtdt biggl)$Show that $lim_nrightarrow inftyint_a^pi/2sqrtn cos^n(x)dx=0$ for any $ain left(0,fracpi2right)$Find $lim_nrightarrowinftysum_k=1^nfracnk(2n-k+1)$ and $lim_nrightarrowinftysum_k=1^nfracnk(2n-k+1)-frac12ln(n)$.Find $lim_nrightarrowinftyint_2npi^2(n+1)pixln xcos x,dx$Show that $lim_nrightarrowinfty int_n^n+1fleft(fracx(n!)^frac1nright)dx=f(e)$

5

1

$begingroup$

Show that $lim_nrightarrowinftyint_0^nfracsqrt xln x(1+x)^2,dx=pi$. This limit puzzled me as I never worked on this kind before and I am inclined to think that, as the $n$ is in the integral's boundary, there is a theorem involved like the dominant convergence theorem. I tried integrating by parts, but the $0$ of the integral prevented me from applying $log0$. How should I handle it?

calculus integration limits

share|cite|improve this question

edited Mar 14 at 9:06

Bernard

123k741117

asked Mar 14 at 8:51

user651692

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The integral converges at $infty$. I mean, you want to compute: $int_0^inftyfracsqrt xln x(1+x)^2dx$
  $endgroup$
  – FDP
  Mar 14 at 9:03
  
  

  
  
  
  

add a comment | 

5

1

$begingroup$

Show that $lim_nrightarrowinftyint_0^nfracsqrt xln x(1+x)^2,dx=pi$. This limit puzzled me as I never worked on this kind before and I am inclined to think that, as the $n$ is in the integral's boundary, there is a theorem involved like the dominant convergence theorem. I tried integrating by parts, but the $0$ of the integral prevented me from applying $log0$. How should I handle it?

calculus integration limits

share|cite|improve this question

edited Mar 14 at 9:06

Bernard

123k741117

asked Mar 14 at 8:51

user651692

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The integral converges at $infty$. I mean, you want to compute: $int_0^inftyfracsqrt xln x(1+x)^2dx$
  $endgroup$
  – FDP
  Mar 14 at 9:03
  
  

  
  
  
  

add a comment | 

$begingroup$

Show that $lim_nrightarrowinftyint_0^nfracsqrt xln x(1+x)^2,dx=pi$. This limit puzzled me as I never worked on this kind before and I am inclined to think that, as the $n$ is in the integral's boundary, there is a theorem involved like the dominant convergence theorem. I tried integrating by parts, but the $0$ of the integral prevented me from applying $log0$. How should I handle it?

calculus integration limits

share|cite|improve this question

edited Mar 14 at 9:06

Bernard

123k741117

asked Mar 14 at 8:51

user651692

$endgroup$

share|cite|improve this question

edited Mar 14 at 9:06

Bernard

123k741117

asked Mar 14 at 8:51

user651692

share|cite|improve this question

edited Mar 14 at 9:06

Bernard

123k741117

asked Mar 14 at 8:51

user651692

edited Mar 14 at 9:06

Bernard

123k741117

edited Mar 14 at 9:06

Bernard

123k741117

4 Answers
4

active

oldest

votes

5

$begingroup$

The integrand is positive for $x>1$, so normally we'd just state the problem as $$int_0^inftyfracsqrtxln x(1+x)^2dx=pi.$$Let's first note that the substitution $x=tan^2 t$ allows us to solve a seemingly unrelated problem, $$int_0^inftyfracx^k-1(1+x)^2dx=int_0^pi/22sin^2k-1tcos^3-2k tdt=operatornameB(k,,2-k)\=Gamma(k)Gamma(1-k)=pi(1-k)cscpi k.$$(Look up Beta and Gamma functions if you don't know them well.) But this problem is not unrelated! Let's differentiate with respect to $k$: $$int_0^inftyfracx^k-1ln x(1+x)^2dx=-picscpi k[1+(1-k)cotpi k].$$Finally, substituting $k=frac32$ gives $$int_0^inftyfracsqrtxln x(1+x)^2dx=pi.$$

share|cite|improve this answer

edited Mar 14 at 9:11

answered Mar 14 at 9:01

J.G.J.G.

30.9k23149

$endgroup$

add a comment | 

5

$begingroup$

beginalignJ&=int_0^infty fracsqrtxln x(1+x)^2,dxendalign
Perform the change of variable $y=sqrtx$

beginalignJ&=int_0^infty frac4x^2ln x(1+x^2)^2,dx\
&=left[-frac12(1+x^2)times 4xln xright]_0^infty+int_0^infty frac2ln x1+x^2,dx+int_0^infty frac21+x^2,dx\
&=int_0^infty frac21+x^2,dx\
&=2Big[arctan xBig]_0^infty\
&=pi
endalign

NB:
beginalignint_0^infty fracln x1+x^2,dx=0endalign

(perform the change of variable $y=frac1x$ )

share|cite|improve this answer

answered Mar 14 at 9:35

FDPFDP

6,17211829

$endgroup$

add a comment | 

3

$begingroup$

Hint:

You can use the substitution $sqrt x=tiff x=t^2,;t>0$ and obtain for the indefinite integral:
$$int fracsqrt xln x(1+x)^2,dx=intfrac2tln t(1+t^2)^2,fracmathrm dtt=2intfracln t(1+t^2)^2,mathrm dt.$$
Now you get rid of the log with an integration by parts, setting
$$u=ln t,quadmathrm dv=fracmathrm d t(1+t^2)^2,enspacetextwhencequadmathrm d u=fracmathrm dtt,quad v= intfracmathrm d t(1+t^2)^2.$$
This last integral is classically obtained from
$$arctan t=intfracmathrm d t1+t^2$$
integrating the latter by parts (again!).

share|cite|improve this answer

edited Mar 14 at 10:11

answered Mar 14 at 9:21

BernardBernard

123k741117

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BarryCipra: Fixed. Thank you for pointing it!
  $endgroup$
  – Bernard
  Mar 14 at 10:12
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $x=tan^2theta$ and then
begineqnarray*
I&=&int_0^inftysqrt xln xover(1+x)^2,dx\
&=&int_0^fracpi2frac2tanthetaln(tantheta)sec^4(theta)2tanthetasec^2theta,dtheta\
&=&4int_0^fracpi2sin^2thetaln(tantheta),dtheta\
&=&2int_0^fracpi2[1-cos(2theta)]ln(tantheta),dtheta\
&=&-2int_0^fracpi2cos(2theta)ln(tantheta),dtheta.
endeqnarray*
Here
$$ int_0^fracpi2ln(tantheta),dtheta=0 $$
is used. Integration by parts gives
$$ I=-2int_0^fracpi2cos(2theta)ln(tantheta),dtheta=-2(-theta+frac12sin(2theta)ln(tantheta))bigg|_0^fracpi2=pi. $$

share|cite|improve this answer

edited Mar 14 at 20:21

answered Mar 14 at 14:52

xpaulxpaul

23.3k24655

$endgroup$

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3147735%2fshow-that-lim-n-rightarrow-infty-int-0n-frac-sqrt-x-ln-x1x2-dx%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

5

$begingroup$

The integrand is positive for $x>1$, so normally we'd just state the problem as $$int_0^inftyfracsqrtxln x(1+x)^2dx=pi.$$Let's first note that the substitution $x=tan^2 t$ allows us to solve a seemingly unrelated problem, $$int_0^inftyfracx^k-1(1+x)^2dx=int_0^pi/22sin^2k-1tcos^3-2k tdt=operatornameB(k,,2-k)\=Gamma(k)Gamma(1-k)=pi(1-k)cscpi k.$$(Look up Beta and Gamma functions if you don't know them well.) But this problem is not unrelated! Let's differentiate with respect to $k$: $$int_0^inftyfracx^k-1ln x(1+x)^2dx=-picscpi k[1+(1-k)cotpi k].$$Finally, substituting $k=frac32$ gives $$int_0^inftyfracsqrtxln x(1+x)^2dx=pi.$$

share|cite|improve this answer

edited Mar 14 at 9:11

answered Mar 14 at 9:01

J.G.J.G.

30.9k23149

$endgroup$

add a comment | 

5

$begingroup$

The integrand is positive for $x>1$, so normally we'd just state the problem as $$int_0^inftyfracsqrtxln x(1+x)^2dx=pi.$$Let's first note that the substitution $x=tan^2 t$ allows us to solve a seemingly unrelated problem, $$int_0^inftyfracx^k-1(1+x)^2dx=int_0^pi/22sin^2k-1tcos^3-2k tdt=operatornameB(k,,2-k)\=Gamma(k)Gamma(1-k)=pi(1-k)cscpi k.$$(Look up Beta and Gamma functions if you don't know them well.) But this problem is not unrelated! Let's differentiate with respect to $k$: $$int_0^inftyfracx^k-1ln x(1+x)^2dx=-picscpi k[1+(1-k)cotpi k].$$Finally, substituting $k=frac32$ gives $$int_0^inftyfracsqrtxln x(1+x)^2dx=pi.$$

share|cite|improve this answer

edited Mar 14 at 9:11

answered Mar 14 at 9:01

J.G.J.G.

30.9k23149

$endgroup$

add a comment | 

$begingroup$

The integrand is positive for $x>1$, so normally we'd just state the problem as $$int_0^inftyfracsqrtxln x(1+x)^2dx=pi.$$Let's first note that the substitution $x=tan^2 t$ allows us to solve a seemingly unrelated problem, $$int_0^inftyfracx^k-1(1+x)^2dx=int_0^pi/22sin^2k-1tcos^3-2k tdt=operatornameB(k,,2-k)\=Gamma(k)Gamma(1-k)=pi(1-k)cscpi k.$$(Look up Beta and Gamma functions if you don't know them well.) But this problem is not unrelated! Let's differentiate with respect to $k$: $$int_0^inftyfracx^k-1ln x(1+x)^2dx=-picscpi k[1+(1-k)cotpi k].$$Finally, substituting $k=frac32$ gives $$int_0^inftyfracsqrtxln x(1+x)^2dx=pi.$$

share|cite|improve this answer

edited Mar 14 at 9:11

answered Mar 14 at 9:01

J.G.J.G.

30.9k23149

$endgroup$

share|cite|improve this answer

edited Mar 14 at 9:11

answered Mar 14 at 9:01

J.G.J.G.

30.9k23149

answered Mar 14 at 9:01

J.G.J.G.

30.9k23149

answered Mar 14 at 9:01

J.G.J.G.

30.9k23149

5

$begingroup$

beginalignJ&=int_0^infty fracsqrtxln x(1+x)^2,dxendalign
Perform the change of variable $y=sqrtx$

beginalignJ&=int_0^infty frac4x^2ln x(1+x^2)^2,dx\
&=left[-frac12(1+x^2)times 4xln xright]_0^infty+int_0^infty frac2ln x1+x^2,dx+int_0^infty frac21+x^2,dx\
&=int_0^infty frac21+x^2,dx\
&=2Big[arctan xBig]_0^infty\
&=pi
endalign

NB:
beginalignint_0^infty fracln x1+x^2,dx=0endalign

(perform the change of variable $y=frac1x$ )

share|cite|improve this answer

answered Mar 14 at 9:35

FDPFDP

6,17211829

$endgroup$

add a comment | 

5

$begingroup$

beginalignJ&=int_0^infty fracsqrtxln x(1+x)^2,dxendalign
Perform the change of variable $y=sqrtx$

beginalignJ&=int_0^infty frac4x^2ln x(1+x^2)^2,dx\
&=left[-frac12(1+x^2)times 4xln xright]_0^infty+int_0^infty frac2ln x1+x^2,dx+int_0^infty frac21+x^2,dx\
&=int_0^infty frac21+x^2,dx\
&=2Big[arctan xBig]_0^infty\
&=pi
endalign

NB:
beginalignint_0^infty fracln x1+x^2,dx=0endalign

(perform the change of variable $y=frac1x$ )

share|cite|improve this answer

answered Mar 14 at 9:35

FDPFDP

6,17211829

$endgroup$

add a comment | 

$begingroup$

beginalignJ&=int_0^infty fracsqrtxln x(1+x)^2,dxendalign
Perform the change of variable $y=sqrtx$

beginalignJ&=int_0^infty frac4x^2ln x(1+x^2)^2,dx\
&=left[-frac12(1+x^2)times 4xln xright]_0^infty+int_0^infty frac2ln x1+x^2,dx+int_0^infty frac21+x^2,dx\
&=int_0^infty frac21+x^2,dx\
&=2Big[arctan xBig]_0^infty\
&=pi
endalign

NB:
beginalignint_0^infty fracln x1+x^2,dx=0endalign

(perform the change of variable $y=frac1x$ )

share|cite|improve this answer

answered Mar 14 at 9:35

FDPFDP

6,17211829

$endgroup$

share|cite|improve this answer

answered Mar 14 at 9:35

FDPFDP

6,17211829

answered Mar 14 at 9:35

FDPFDP

6,17211829

answered Mar 14 at 9:35

FDPFDP

6,17211829

3

$begingroup$

Hint:

You can use the substitution $sqrt x=tiff x=t^2,;t>0$ and obtain for the indefinite integral:
$$int fracsqrt xln x(1+x)^2,dx=intfrac2tln t(1+t^2)^2,fracmathrm dtt=2intfracln t(1+t^2)^2,mathrm dt.$$
Now you get rid of the log with an integration by parts, setting
$$u=ln t,quadmathrm dv=fracmathrm d t(1+t^2)^2,enspacetextwhencequadmathrm d u=fracmathrm dtt,quad v= intfracmathrm d t(1+t^2)^2.$$
This last integral is classically obtained from
$$arctan t=intfracmathrm d t1+t^2$$
integrating the latter by parts (again!).

share|cite|improve this answer

edited Mar 14 at 10:11

answered Mar 14 at 9:21

BernardBernard

123k741117

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BarryCipra: Fixed. Thank you for pointing it!
  $endgroup$
  – Bernard
  Mar 14 at 10:12
  

  
  
  
  

add a comment | 

3

$begingroup$

Hint:

You can use the substitution $sqrt x=tiff x=t^2,;t>0$ and obtain for the indefinite integral:
$$int fracsqrt xln x(1+x)^2,dx=intfrac2tln t(1+t^2)^2,fracmathrm dtt=2intfracln t(1+t^2)^2,mathrm dt.$$
Now you get rid of the log with an integration by parts, setting
$$u=ln t,quadmathrm dv=fracmathrm d t(1+t^2)^2,enspacetextwhencequadmathrm d u=fracmathrm dtt,quad v= intfracmathrm d t(1+t^2)^2.$$
This last integral is classically obtained from
$$arctan t=intfracmathrm d t1+t^2$$
integrating the latter by parts (again!).

share|cite|improve this answer

edited Mar 14 at 10:11

answered Mar 14 at 9:21

BernardBernard

123k741117

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @BarryCipra: Fixed. Thank you for pointing it!
  $endgroup$
  – Bernard
  Mar 14 at 10:12
  

  
  
  
  

add a comment | 

$begingroup$

Hint:

You can use the substitution $sqrt x=tiff x=t^2,;t>0$ and obtain for the indefinite integral:
$$int fracsqrt xln x(1+x)^2,dx=intfrac2tln t(1+t^2)^2,fracmathrm dtt=2intfracln t(1+t^2)^2,mathrm dt.$$
Now you get rid of the log with an integration by parts, setting
$$u=ln t,quadmathrm dv=fracmathrm d t(1+t^2)^2,enspacetextwhencequadmathrm d u=fracmathrm dtt,quad v= intfracmathrm d t(1+t^2)^2.$$
This last integral is classically obtained from
$$arctan t=intfracmathrm d t1+t^2$$
integrating the latter by parts (again!).

share|cite|improve this answer

edited Mar 14 at 10:11

answered Mar 14 at 9:21

BernardBernard

123k741117

$endgroup$

share|cite|improve this answer

edited Mar 14 at 10:11

answered Mar 14 at 9:21

BernardBernard

123k741117

answered Mar 14 at 9:21

BernardBernard

123k741117

answered Mar 14 at 9:21

BernardBernard

123k741117

2

$begingroup$

Let $x=tan^2theta$ and then
begineqnarray*
I&=&int_0^inftysqrt xln xover(1+x)^2,dx\
&=&int_0^fracpi2frac2tanthetaln(tantheta)sec^4(theta)2tanthetasec^2theta,dtheta\
&=&4int_0^fracpi2sin^2thetaln(tantheta),dtheta\
&=&2int_0^fracpi2[1-cos(2theta)]ln(tantheta),dtheta\
&=&-2int_0^fracpi2cos(2theta)ln(tantheta),dtheta.
endeqnarray*
Here
$$ int_0^fracpi2ln(tantheta),dtheta=0 $$
is used. Integration by parts gives
$$ I=-2int_0^fracpi2cos(2theta)ln(tantheta),dtheta=-2(-theta+frac12sin(2theta)ln(tantheta))bigg|_0^fracpi2=pi. $$

share|cite|improve this answer

edited Mar 14 at 20:21

answered Mar 14 at 14:52

xpaulxpaul

23.3k24655

$endgroup$

add a comment | 

2

$begingroup$

Let $x=tan^2theta$ and then
begineqnarray*
I&=&int_0^inftysqrt xln xover(1+x)^2,dx\
&=&int_0^fracpi2frac2tanthetaln(tantheta)sec^4(theta)2tanthetasec^2theta,dtheta\
&=&4int_0^fracpi2sin^2thetaln(tantheta),dtheta\
&=&2int_0^fracpi2[1-cos(2theta)]ln(tantheta),dtheta\
&=&-2int_0^fracpi2cos(2theta)ln(tantheta),dtheta.
endeqnarray*
Here
$$ int_0^fracpi2ln(tantheta),dtheta=0 $$
is used. Integration by parts gives
$$ I=-2int_0^fracpi2cos(2theta)ln(tantheta),dtheta=-2(-theta+frac12sin(2theta)ln(tantheta))bigg|_0^fracpi2=pi. $$

share|cite|improve this answer

edited Mar 14 at 20:21

answered Mar 14 at 14:52

xpaulxpaul

23.3k24655

$endgroup$

add a comment | 

$begingroup$

Let $x=tan^2theta$ and then
begineqnarray*
I&=&int_0^inftysqrt xln xover(1+x)^2,dx\
&=&int_0^fracpi2frac2tanthetaln(tantheta)sec^4(theta)2tanthetasec^2theta,dtheta\
&=&4int_0^fracpi2sin^2thetaln(tantheta),dtheta\
&=&2int_0^fracpi2[1-cos(2theta)]ln(tantheta),dtheta\
&=&-2int_0^fracpi2cos(2theta)ln(tantheta),dtheta.
endeqnarray*
Here
$$ int_0^fracpi2ln(tantheta),dtheta=0 $$
is used. Integration by parts gives
$$ I=-2int_0^fracpi2cos(2theta)ln(tantheta),dtheta=-2(-theta+frac12sin(2theta)ln(tantheta))bigg|_0^fracpi2=pi. $$

share|cite|improve this answer

edited Mar 14 at 20:21

answered Mar 14 at 14:52

xpaulxpaul

23.3k24655

$endgroup$

share|cite|improve this answer

edited Mar 14 at 20:21

answered Mar 14 at 14:52

xpaulxpaul

23.3k24655

answered Mar 14 at 14:52

xpaulxpaul

23.3k24655

answered Mar 14 at 14:52

xpaulxpaul

23.3k24655