Fractional part of the floor function Integral [closed]

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Let $lfloorrfloor $ and $$ denote the floor function and the fractional part funtion, respectively. Then calculate in closed-form the following integral

$$int_0^1biggfrac1xbigglfloorfrac1xbiggrfloorbiggdx$$

integration sequences-and-series riemann-zeta eulers-constant fractional-part

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edited Jul 21 at 4:19

Abcd

2,3761624

asked Jul 20 at 15:15

Kays Tomy

934

closed as off-topic by amWhy, Adrian Keister, choco_addicted, Strants, Parcly Taxel Jul 24 at 15:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Adrian Keister, choco_addicted, Strants, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Anything you want to add to the question?
  – klirk
  Jul 20 at 15:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @klirk : What would OP add? Curly braces refer to fractional part of enclosed value, if that's what you mean. Fairly standard.
  – MPW
  Jul 20 at 15:23
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  He could add why he is interested in the solution, what he tried so far etc...
  – klirk
  Jul 20 at 15:36
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @klirk : Okay, fair enough. +1.
  – MPW
  Jul 20 at 15:42
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

Let $lfloorrfloor $ and $$ denote the floor function and the fractional part funtion, respectively. Then calculate in closed-form the following integral

$$int_0^1biggfrac1xbigglfloorfrac1xbiggrfloorbiggdx$$

integration sequences-and-series riemann-zeta eulers-constant fractional-part

share|cite|improve this question

edited Jul 21 at 4:19

Abcd

2,3761624

asked Jul 20 at 15:15

Kays Tomy

934

closed as off-topic by amWhy, Adrian Keister, choco_addicted, Strants, Parcly Taxel Jul 24 at 15:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Adrian Keister, choco_addicted, Strants, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Anything you want to add to the question?
  – klirk
  Jul 20 at 15:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @klirk : What would OP add? Curly braces refer to fractional part of enclosed value, if that's what you mean. Fairly standard.
  – MPW
  Jul 20 at 15:23
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  He could add why he is interested in the solution, what he tried so far etc...
  – klirk
  Jul 20 at 15:36
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @klirk : Okay, fair enough. +1.
  – MPW
  Jul 20 at 15:42
  

  
  
  
  

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

Let $lfloorrfloor $ and $$ denote the floor function and the fractional part funtion, respectively. Then calculate in closed-form the following integral

$$int_0^1biggfrac1xbigglfloorfrac1xbiggrfloorbiggdx$$

integration sequences-and-series riemann-zeta eulers-constant fractional-part

share|cite|improve this question

edited Jul 21 at 4:19

Abcd

2,3761624

asked Jul 20 at 15:15

Kays Tomy

934

Let $lfloorrfloor $ and $$ denote the floor function and the fractional part funtion, respectively. Then calculate in closed-form the following integral

$$int_0^1biggfrac1xbigglfloorfrac1xbiggrfloorbiggdx$$

integration sequences-and-series riemann-zeta eulers-constant fractional-part

share|cite|improve this question

edited Jul 21 at 4:19

Abcd

2,3761624

asked Jul 20 at 15:15

Kays Tomy

934

share|cite|improve this question

share|cite|improve this question

edited Jul 21 at 4:19

Abcd

2,3761624

edited Jul 21 at 4:19

Abcd

2,3761624

edited Jul 21 at 4:19

Abcd

2,3761624

2,3761624

asked Jul 20 at 15:15

Kays Tomy

934

asked Jul 20 at 15:15

Kays Tomy

934

asked Jul 20 at 15:15

Kays Tomy

934

934

closed as off-topic by amWhy, Adrian Keister, choco_addicted, Strants, Parcly Taxel Jul 24 at 15:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Adrian Keister, choco_addicted, Strants, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Adrian Keister, choco_addicted, Strants, Parcly Taxel Jul 24 at 15:32

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Adrian Keister, choco_addicted, Strants, Parcly Taxel

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Anything you want to add to the question?
  – klirk
  Jul 20 at 15:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @klirk : What would OP add? Curly braces refer to fractional part of enclosed value, if that's what you mean. Fairly standard.
  – MPW
  Jul 20 at 15:23
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  He could add why he is interested in the solution, what he tried so far etc...
  – klirk
  Jul 20 at 15:36
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @klirk : Okay, fair enough. +1.
  – MPW
  Jul 20 at 15:42
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Anything you want to add to the question?
  – klirk
  Jul 20 at 15:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @klirk : What would OP add? Curly braces refer to fractional part of enclosed value, if that's what you mean. Fairly standard.
  – MPW
  Jul 20 at 15:23
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  He could add why he is interested in the solution, what he tried so far etc...
  – klirk
  Jul 20 at 15:36
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @klirk : Okay, fair enough. +1.
  – MPW
  Jul 20 at 15:42
  

  
  
  
  

Anything you want to add to the question?
– klirk
Jul 20 at 15:20

Anything you want to add to the question?
– klirk
Jul 20 at 15:20

@klirk : What would OP add? Curly braces refer to fractional part of enclosed value, if that's what you mean. Fairly standard.
– MPW
Jul 20 at 15:23

@klirk : What would OP add? Curly braces refer to fractional part of enclosed value, if that's what you mean. Fairly standard.
– MPW
Jul 20 at 15:23

3

3

He could add why he is interested in the solution, what he tried so far etc...
– klirk
Jul 20 at 15:36

He could add why he is interested in the solution, what he tried so far etc...
– klirk
Jul 20 at 15:36

1

1

@klirk : Okay, fair enough. +1.
– MPW
Jul 20 at 15:42

@klirk : Okay, fair enough. +1.
– MPW
Jul 20 at 15:42

add a comment | 

2 Answers
2

active

oldest

votes

up vote
4
down vote

Hint: Can you compute the value of
$$I_nequivintlimits_1/n^1/(n+1)leftfrac1x leftlfloor frac 1xrightrfloorright; dx$$
and consider $$sum_n=1^inftyI_n$$?

share|cite|improve this answer

answered Jul 20 at 15:22

MPW

28.4k11853

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I hate constructive hints. +1
  – Kugelblitz
  Jul 20 at 15:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It can be shown that $I_n = - n logleft (n right ) + n logleft (n + 1 right ) - frac1n + 1 + sum_k=0^n - 2 fracnk + n - left(n + 1right)^2 + 2 $. Doing the final summation seems painful.
  – Donny Frank
  Jul 22 at 6:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sum _k=0^n-2 fracnk+n-(n+1)^2+2=n H_n^2-n H_-1+n+n^2$ where $H_n$ is harmonic number.
  – Mariusz Iwaniuk
  Jul 22 at 19:58
  

  
  
  
  

add a comment | 

up vote
2
down vote

$$int_0^1biggfrac1xbigglfloorfrac1xbiggrfloorbiggdx=sum_k=1^inftyint_frac1k+1^frac1kbiggfrac1xbigglfloorfrac1xbiggrfloorbiggdx=sum_k=1^inftyint_frac1k+1^frac1kbiggfrackxbiggdx$$also$$int_frac1k+1^frac1kbiggfrackxbiggdx=sum_l=0^k-1int_frackk^2+l+1^frackk^2+ldfrackx-k^2-ldx=sum_l=0^k-1klndfrack^2+l+1k^2+l-(k^2+l)dfrack(k^2+l)(k^2+l+1)=kln(1+dfrac1k)-kH_k^2+k+kH_k^2$$so the integral would be$$Large I=sum_k=1^inftykln(1+dfrac1k)-kH_k^2+k+kH_k^2approx0.4350$$here is a figure that how the integral converges:

share|cite|improve this answer

edited Jul 24 at 18:47

answered Jul 24 at 14:49

Mostafa Ayaz

8,5773630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sure it does...There was a typo :)
  – Mostafa Ayaz
  Jul 24 at 18:49
  

  
  
  
  

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

Hint: Can you compute the value of
$$I_nequivintlimits_1/n^1/(n+1)leftfrac1x leftlfloor frac 1xrightrfloorright; dx$$
and consider $$sum_n=1^inftyI_n$$?

share|cite|improve this answer

answered Jul 20 at 15:22

MPW

28.4k11853

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I hate constructive hints. +1
  – Kugelblitz
  Jul 20 at 15:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It can be shown that $I_n = - n logleft (n right ) + n logleft (n + 1 right ) - frac1n + 1 + sum_k=0^n - 2 fracnk + n - left(n + 1right)^2 + 2 $. Doing the final summation seems painful.
  – Donny Frank
  Jul 22 at 6:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sum _k=0^n-2 fracnk+n-(n+1)^2+2=n H_n^2-n H_-1+n+n^2$ where $H_n$ is harmonic number.
  – Mariusz Iwaniuk
  Jul 22 at 19:58
  

  
  
  
  

add a comment | 

up vote
4
down vote

Hint: Can you compute the value of
$$I_nequivintlimits_1/n^1/(n+1)leftfrac1x leftlfloor frac 1xrightrfloorright; dx$$
and consider $$sum_n=1^inftyI_n$$?

share|cite|improve this answer

answered Jul 20 at 15:22

MPW

28.4k11853

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I hate constructive hints. +1
  – Kugelblitz
  Jul 20 at 15:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It can be shown that $I_n = - n logleft (n right ) + n logleft (n + 1 right ) - frac1n + 1 + sum_k=0^n - 2 fracnk + n - left(n + 1right)^2 + 2 $. Doing the final summation seems painful.
  – Donny Frank
  Jul 22 at 6:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sum _k=0^n-2 fracnk+n-(n+1)^2+2=n H_n^2-n H_-1+n+n^2$ where $H_n$ is harmonic number.
  – Mariusz Iwaniuk
  Jul 22 at 19:58
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

Hint: Can you compute the value of
$$I_nequivintlimits_1/n^1/(n+1)leftfrac1x leftlfloor frac 1xrightrfloorright; dx$$
and consider $$sum_n=1^inftyI_n$$?

share|cite|improve this answer

answered Jul 20 at 15:22

MPW

28.4k11853

Hint: Can you compute the value of
$$I_nequivintlimits_1/n^1/(n+1)leftfrac1x leftlfloor frac 1xrightrfloorright; dx$$
and consider $$sum_n=1^inftyI_n$$?

share|cite|improve this answer

answered Jul 20 at 15:22

MPW

28.4k11853

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 15:22

MPW

28.4k11853

answered Jul 20 at 15:22

MPW

28.4k11853

answered Jul 20 at 15:22

MPW

28.4k11853

28.4k11853

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I hate constructive hints. +1
  – Kugelblitz
  Jul 20 at 15:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It can be shown that $I_n = - n logleft (n right ) + n logleft (n + 1 right ) - frac1n + 1 + sum_k=0^n - 2 fracnk + n - left(n + 1right)^2 + 2 $. Doing the final summation seems painful.
  – Donny Frank
  Jul 22 at 6:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sum _k=0^n-2 fracnk+n-(n+1)^2+2=n H_n^2-n H_-1+n+n^2$ where $H_n$ is harmonic number.
  – Mariusz Iwaniuk
  Jul 22 at 19:58
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I hate constructive hints. +1
  – Kugelblitz
  Jul 20 at 15:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  It can be shown that $I_n = - n logleft (n right ) + n logleft (n + 1 right ) - frac1n + 1 + sum_k=0^n - 2 fracnk + n - left(n + 1right)^2 + 2 $. Doing the final summation seems painful.
  – Donny Frank
  Jul 22 at 6:37
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $sum _k=0^n-2 fracnk+n-(n+1)^2+2=n H_n^2-n H_-1+n+n^2$ where $H_n$ is harmonic number.
  – Mariusz Iwaniuk
  Jul 22 at 19:58
  

  
  
  
  

I hate constructive hints. +1
– Kugelblitz
Jul 20 at 15:25

I hate constructive hints. +1
– Kugelblitz
Jul 20 at 15:25

1

1

It can be shown that $I_n = - n logleft (n right ) + n logleft (n + 1 right ) - frac1n + 1 + sum_k=0^n - 2 fracnk + n - left(n + 1right)^2 + 2 $. Doing the final summation seems painful.
– Donny Frank
Jul 22 at 6:37

It can be shown that $I_n = - n logleft (n right ) + n logleft (n + 1 right ) - frac1n + 1 + sum_k=0^n - 2 fracnk + n - left(n + 1right)^2 + 2 $. Doing the final summation seems painful.
– Donny Frank
Jul 22 at 6:37

1

1

$sum _k=0^n-2 fracnk+n-(n+1)^2+2=n H_n^2-n H_-1+n+n^2$ where $H_n$ is harmonic number.
– Mariusz Iwaniuk
Jul 22 at 19:58

$sum _k=0^n-2 fracnk+n-(n+1)^2+2=n H_n^2-n H_-1+n+n^2$ where $H_n$ is harmonic number.
– Mariusz Iwaniuk
Jul 22 at 19:58

add a comment | 

up vote
2
down vote

$$int_0^1biggfrac1xbigglfloorfrac1xbiggrfloorbiggdx=sum_k=1^inftyint_frac1k+1^frac1kbiggfrac1xbigglfloorfrac1xbiggrfloorbiggdx=sum_k=1^inftyint_frac1k+1^frac1kbiggfrackxbiggdx$$also$$int_frac1k+1^frac1kbiggfrackxbiggdx=sum_l=0^k-1int_frackk^2+l+1^frackk^2+ldfrackx-k^2-ldx=sum_l=0^k-1klndfrack^2+l+1k^2+l-(k^2+l)dfrack(k^2+l)(k^2+l+1)=kln(1+dfrac1k)-kH_k^2+k+kH_k^2$$so the integral would be$$Large I=sum_k=1^inftykln(1+dfrac1k)-kH_k^2+k+kH_k^2approx0.4350$$here is a figure that how the integral converges:

share|cite|improve this answer

edited Jul 24 at 18:47

answered Jul 24 at 14:49

Mostafa Ayaz

8,5773630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sure it does...There was a typo :)
  – Mostafa Ayaz
  Jul 24 at 18:49
  

  
  
  
  

add a comment | 

up vote
2
down vote

$$int_0^1biggfrac1xbigglfloorfrac1xbiggrfloorbiggdx=sum_k=1^inftyint_frac1k+1^frac1kbiggfrac1xbigglfloorfrac1xbiggrfloorbiggdx=sum_k=1^inftyint_frac1k+1^frac1kbiggfrackxbiggdx$$also$$int_frac1k+1^frac1kbiggfrackxbiggdx=sum_l=0^k-1int_frackk^2+l+1^frackk^2+ldfrackx-k^2-ldx=sum_l=0^k-1klndfrack^2+l+1k^2+l-(k^2+l)dfrack(k^2+l)(k^2+l+1)=kln(1+dfrac1k)-kH_k^2+k+kH_k^2$$so the integral would be$$Large I=sum_k=1^inftykln(1+dfrac1k)-kH_k^2+k+kH_k^2approx0.4350$$here is a figure that how the integral converges:

share|cite|improve this answer

edited Jul 24 at 18:47

answered Jul 24 at 14:49

Mostafa Ayaz

8,5773630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sure it does...There was a typo :)
  – Mostafa Ayaz
  Jul 24 at 18:49
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

$$int_0^1biggfrac1xbigglfloorfrac1xbiggrfloorbiggdx=sum_k=1^inftyint_frac1k+1^frac1kbiggfrac1xbigglfloorfrac1xbiggrfloorbiggdx=sum_k=1^inftyint_frac1k+1^frac1kbiggfrackxbiggdx$$also$$int_frac1k+1^frac1kbiggfrackxbiggdx=sum_l=0^k-1int_frackk^2+l+1^frackk^2+ldfrackx-k^2-ldx=sum_l=0^k-1klndfrack^2+l+1k^2+l-(k^2+l)dfrack(k^2+l)(k^2+l+1)=kln(1+dfrac1k)-kH_k^2+k+kH_k^2$$so the integral would be$$Large I=sum_k=1^inftykln(1+dfrac1k)-kH_k^2+k+kH_k^2approx0.4350$$here is a figure that how the integral converges:

share|cite|improve this answer

edited Jul 24 at 18:47

answered Jul 24 at 14:49

Mostafa Ayaz

8,5773630

$$int_0^1biggfrac1xbigglfloorfrac1xbiggrfloorbiggdx=sum_k=1^inftyint_frac1k+1^frac1kbiggfrac1xbigglfloorfrac1xbiggrfloorbiggdx=sum_k=1^inftyint_frac1k+1^frac1kbiggfrackxbiggdx$$also$$int_frac1k+1^frac1kbiggfrackxbiggdx=sum_l=0^k-1int_frackk^2+l+1^frackk^2+ldfrackx-k^2-ldx=sum_l=0^k-1klndfrack^2+l+1k^2+l-(k^2+l)dfrack(k^2+l)(k^2+l+1)=kln(1+dfrac1k)-kH_k^2+k+kH_k^2$$so the integral would be$$Large I=sum_k=1^inftykln(1+dfrac1k)-kH_k^2+k+kH_k^2approx0.4350$$here is a figure that how the integral converges:

share|cite|improve this answer

edited Jul 24 at 18:47

answered Jul 24 at 14:49

Mostafa Ayaz

8,5773630

share|cite|improve this answer

share|cite|improve this answer

edited Jul 24 at 18:47

edited Jul 24 at 18:47

edited Jul 24 at 18:47

answered Jul 24 at 14:49

Mostafa Ayaz

8,5773630

answered Jul 24 at 14:49

Mostafa Ayaz

8,5773630

answered Jul 24 at 14:49

Mostafa Ayaz

8,5773630

8,5773630

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sure it does...There was a typo :)
  – Mostafa Ayaz
  Jul 24 at 18:49
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Sure it does...There was a typo :)
  – Mostafa Ayaz
  Jul 24 at 18:49
  

  
  
  
  

Sure it does...There was a typo :)
– Mostafa Ayaz
Jul 24 at 18:49

Sure it does...There was a typo :)
– Mostafa Ayaz
Jul 24 at 18:49

add a comment |