Double fractional part integral

Clash Royale CLAN TAG#URR8PPP

up vote
3
down vote

favorite

Let $$ denote the fractional part, does the following integral have a closed form ?
$$int_0^1int_0^1biggfrac1x,ybigg^2dx,dy$$

integration sequences-and-series definite-integrals fractional-part

share|cite|improve this question

asked Jul 19 at 6:35

Kays Tomy

934

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Probably duplicate: math.stackexchange.com/questions/875076/…
  – Mariusz Iwaniuk
  Jul 19 at 15:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk thank you for the update. I do not know that existed before already.
  – Kays Tomy
  Jul 19 at 16:30
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

Let $$ denote the fractional part, does the following integral have a closed form ?
$$int_0^1int_0^1biggfrac1x,ybigg^2dx,dy$$

integration sequences-and-series definite-integrals fractional-part

share|cite|improve this question

asked Jul 19 at 6:35

Kays Tomy

934

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Probably duplicate: math.stackexchange.com/questions/875076/…
  – Mariusz Iwaniuk
  Jul 19 at 15:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk thank you for the update. I do not know that existed before already.
  – Kays Tomy
  Jul 19 at 16:30
  

  
  
  
  

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

Let $$ denote the fractional part, does the following integral have a closed form ?
$$int_0^1int_0^1biggfrac1x,ybigg^2dx,dy$$

integration sequences-and-series definite-integrals fractional-part

share|cite|improve this question

asked Jul 19 at 6:35

Kays Tomy

934

Let $$ denote the fractional part, does the following integral have a closed form ?
$$int_0^1int_0^1biggfrac1x,ybigg^2dx,dy$$

integration sequences-and-series definite-integrals fractional-part

share|cite|improve this question

asked Jul 19 at 6:35

Kays Tomy

934

share|cite|improve this question

share|cite|improve this question

asked Jul 19 at 6:35

Kays Tomy

934

asked Jul 19 at 6:35

Kays Tomy

934

asked Jul 19 at 6:35

Kays Tomy

934

934

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Probably duplicate: math.stackexchange.com/questions/875076/…
  – Mariusz Iwaniuk
  Jul 19 at 15:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk thank you for the update. I do not know that existed before already.
  – Kays Tomy
  Jul 19 at 16:30
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Probably duplicate: math.stackexchange.com/questions/875076/…
  – Mariusz Iwaniuk
  Jul 19 at 15:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk thank you for the update. I do not know that existed before already.
  – Kays Tomy
  Jul 19 at 16:30
  

  
  
  
  

Probably duplicate: math.stackexchange.com/questions/875076/…
– Mariusz Iwaniuk
Jul 19 at 15:17

Probably duplicate: math.stackexchange.com/questions/875076/…
– Mariusz Iwaniuk
Jul 19 at 15:17

@MariuszIwaniuk thank you for the update. I do not know that existed before already.
– Kays Tomy
Jul 19 at 16:30

@MariuszIwaniuk thank you for the update. I do not know that existed before already.
– Kays Tomy
Jul 19 at 16:30

add a comment | 

1 Answer
1

active

oldest

votes

up vote
1
down vote

This is an incomplete answer that only addresses the 1-dimensional case.

We split the integral into continuous pieces:
$$beginalign
int_0^1 left frac1x right^2 mathrmdx
&= sum_n=1^infty int_frac1n+1^frac1n left frac1x right^2 mathrmdx \\
&= sum_n=1^infty int_frac1n+1^frac1n left( frac1x - nright)^2 mathrmdx \\
&= sum_n=1^infty left( n^2x - 2n ln |x| - frac1x right)biggrrvert_frac1n+1^frac1n \\
&= sum_n=1^infty left( n - 2n ln frac1n - n - fracn^2n+1 + 2n ln frac1n+1 + n + 1 right) \\
&= sum_n=1^infty left( 2n ln fracnn+1 + frac2n + 1n+1 right)
endalign$$

With the aid of computer algebra, we obtain that this series converges to $$ln (2pi) - gamma - 1$$ where $gamma$ is the Euler-Mascheroni constant.

share|cite|improve this answer

edited Jul 19 at 15:47

answered Jul 19 at 14:04

Fytch

159117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) and welcome to Math SE!
  – Szeto
  Jul 19 at 14:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  But notice the OP’s integral has a square there.
  – Szeto
  Jul 19 at 14:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Szeto Thanks, I just noticed and I'm working on a solution.
  – Fytch
  Jul 19 at 14:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $int_0^1 left frac1x right^2 mathrmdx=-1-gamma +ln (2)+ln (pi )$ :)
  – Mariusz Iwaniuk
  Jul 19 at 14:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk How did you arrive at that? Did you evaluate the emergent series using complex analysis?
  – Fytch
  Jul 19 at 15:16
  

  
  
  
  

 | 
show 4 more comments

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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2856330%2fdouble-fractional-part-integral%23new-answer', 'question_page');

);

Post as a guest

Name Email

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote

This is an incomplete answer that only addresses the 1-dimensional case.

We split the integral into continuous pieces:
$$beginalign
int_0^1 left frac1x right^2 mathrmdx
&= sum_n=1^infty int_frac1n+1^frac1n left frac1x right^2 mathrmdx \\
&= sum_n=1^infty int_frac1n+1^frac1n left( frac1x - nright)^2 mathrmdx \\
&= sum_n=1^infty left( n^2x - 2n ln |x| - frac1x right)biggrrvert_frac1n+1^frac1n \\
&= sum_n=1^infty left( n - 2n ln frac1n - n - fracn^2n+1 + 2n ln frac1n+1 + n + 1 right) \\
&= sum_n=1^infty left( 2n ln fracnn+1 + frac2n + 1n+1 right)
endalign$$

With the aid of computer algebra, we obtain that this series converges to $$ln (2pi) - gamma - 1$$ where $gamma$ is the Euler-Mascheroni constant.

share|cite|improve this answer

edited Jul 19 at 15:47

answered Jul 19 at 14:04

Fytch

159117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) and welcome to Math SE!
  – Szeto
  Jul 19 at 14:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  But notice the OP’s integral has a square there.
  – Szeto
  Jul 19 at 14:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Szeto Thanks, I just noticed and I'm working on a solution.
  – Fytch
  Jul 19 at 14:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $int_0^1 left frac1x right^2 mathrmdx=-1-gamma +ln (2)+ln (pi )$ :)
  – Mariusz Iwaniuk
  Jul 19 at 14:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk How did you arrive at that? Did you evaluate the emergent series using complex analysis?
  – Fytch
  Jul 19 at 15:16
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote

This is an incomplete answer that only addresses the 1-dimensional case.

We split the integral into continuous pieces:
$$beginalign
int_0^1 left frac1x right^2 mathrmdx
&= sum_n=1^infty int_frac1n+1^frac1n left frac1x right^2 mathrmdx \\
&= sum_n=1^infty int_frac1n+1^frac1n left( frac1x - nright)^2 mathrmdx \\
&= sum_n=1^infty left( n^2x - 2n ln |x| - frac1x right)biggrrvert_frac1n+1^frac1n \\
&= sum_n=1^infty left( n - 2n ln frac1n - n - fracn^2n+1 + 2n ln frac1n+1 + n + 1 right) \\
&= sum_n=1^infty left( 2n ln fracnn+1 + frac2n + 1n+1 right)
endalign$$

With the aid of computer algebra, we obtain that this series converges to $$ln (2pi) - gamma - 1$$ where $gamma$ is the Euler-Mascheroni constant.

share|cite|improve this answer

edited Jul 19 at 15:47

answered Jul 19 at 14:04

Fytch

159117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) and welcome to Math SE!
  – Szeto
  Jul 19 at 14:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  But notice the OP’s integral has a square there.
  – Szeto
  Jul 19 at 14:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Szeto Thanks, I just noticed and I'm working on a solution.
  – Fytch
  Jul 19 at 14:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $int_0^1 left frac1x right^2 mathrmdx=-1-gamma +ln (2)+ln (pi )$ :)
  – Mariusz Iwaniuk
  Jul 19 at 14:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk How did you arrive at that? Did you evaluate the emergent series using complex analysis?
  – Fytch
  Jul 19 at 15:16
  

  
  
  
  

 | 
show 4 more comments

up vote
1
down vote

up vote
1
down vote

This is an incomplete answer that only addresses the 1-dimensional case.

We split the integral into continuous pieces:
$$beginalign
int_0^1 left frac1x right^2 mathrmdx
&= sum_n=1^infty int_frac1n+1^frac1n left frac1x right^2 mathrmdx \\
&= sum_n=1^infty int_frac1n+1^frac1n left( frac1x - nright)^2 mathrmdx \\
&= sum_n=1^infty left( n^2x - 2n ln |x| - frac1x right)biggrrvert_frac1n+1^frac1n \\
&= sum_n=1^infty left( n - 2n ln frac1n - n - fracn^2n+1 + 2n ln frac1n+1 + n + 1 right) \\
&= sum_n=1^infty left( 2n ln fracnn+1 + frac2n + 1n+1 right)
endalign$$

With the aid of computer algebra, we obtain that this series converges to $$ln (2pi) - gamma - 1$$ where $gamma$ is the Euler-Mascheroni constant.

share|cite|improve this answer

edited Jul 19 at 15:47

answered Jul 19 at 14:04

Fytch

159117

This is an incomplete answer that only addresses the 1-dimensional case.

We split the integral into continuous pieces:
$$beginalign
int_0^1 left frac1x right^2 mathrmdx
&= sum_n=1^infty int_frac1n+1^frac1n left frac1x right^2 mathrmdx \\
&= sum_n=1^infty int_frac1n+1^frac1n left( frac1x - nright)^2 mathrmdx \\
&= sum_n=1^infty left( n^2x - 2n ln |x| - frac1x right)biggrrvert_frac1n+1^frac1n \\
&= sum_n=1^infty left( n - 2n ln frac1n - n - fracn^2n+1 + 2n ln frac1n+1 + n + 1 right) \\
&= sum_n=1^infty left( 2n ln fracnn+1 + frac2n + 1n+1 right)
endalign$$

With the aid of computer algebra, we obtain that this series converges to $$ln (2pi) - gamma - 1$$ where $gamma$ is the Euler-Mascheroni constant.

share|cite|improve this answer

edited Jul 19 at 15:47

answered Jul 19 at 14:04

Fytch

159117

share|cite|improve this answer

share|cite|improve this answer

edited Jul 19 at 15:47

edited Jul 19 at 15:47

edited Jul 19 at 15:47

answered Jul 19 at 14:04

Fytch

159117

answered Jul 19 at 14:04

Fytch

159117

answered Jul 19 at 14:04

Fytch

159117

159117

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) and welcome to Math SE!
  – Szeto
  Jul 19 at 14:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  But notice the OP’s integral has a square there.
  – Szeto
  Jul 19 at 14:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Szeto Thanks, I just noticed and I'm working on a solution.
  – Fytch
  Jul 19 at 14:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $int_0^1 left frac1x right^2 mathrmdx=-1-gamma +ln (2)+ln (pi )$ :)
  – Mariusz Iwaniuk
  Jul 19 at 14:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk How did you arrive at that? Did you evaluate the emergent series using complex analysis?
  – Fytch
  Jul 19 at 15:16
  

  
  
  
  

 | 
show 4 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  (+1) and welcome to Math SE!
  – Szeto
  Jul 19 at 14:10
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  But notice the OP’s integral has a square there.
  – Szeto
  Jul 19 at 14:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Szeto Thanks, I just noticed and I'm working on a solution.
  – Fytch
  Jul 19 at 14:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $int_0^1 left frac1x right^2 mathrmdx=-1-gamma +ln (2)+ln (pi )$ :)
  – Mariusz Iwaniuk
  Jul 19 at 14:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @MariuszIwaniuk How did you arrive at that? Did you evaluate the emergent series using complex analysis?
  – Fytch
  Jul 19 at 15:16
  

  
  
  
  

(+1) and welcome to Math SE!
– Szeto
Jul 19 at 14:10

(+1) and welcome to Math SE!
– Szeto
Jul 19 at 14:10

1

1

But notice the OP’s integral has a square there.
– Szeto
Jul 19 at 14:11

But notice the OP’s integral has a square there.
– Szeto
Jul 19 at 14:11

@Szeto Thanks, I just noticed and I'm working on a solution.
– Fytch
Jul 19 at 14:19

@Szeto Thanks, I just noticed and I'm working on a solution.
– Fytch
Jul 19 at 14:19

$int_0^1 left frac1x right^2 mathrmdx=-1-gamma +ln (2)+ln (pi )$ :)
– Mariusz Iwaniuk
Jul 19 at 14:39

$int_0^1 left frac1x right^2 mathrmdx=-1-gamma +ln (2)+ln (pi )$ :)
– Mariusz Iwaniuk
Jul 19 at 14:39

@MariuszIwaniuk How did you arrive at that? Did you evaluate the emergent series using complex analysis?
– Fytch
Jul 19 at 15:16

@MariuszIwaniuk How did you arrive at that? Did you evaluate the emergent series using complex analysis?
– Fytch
Jul 19 at 15:16

 | 
show 4 more comments

 

draft saved

draft discarded

 

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

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2856330%2fdouble-fractional-part-integral%23new-answer', 'question_page');

);

Post as a guest

Name Email

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

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

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

Name Email

Name Email

Name

Email

Name Email

Name

Email