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Calculating the integral $int_0^infty fracarctan xx(ln(x+2))^2 dx$

0

1

$begingroup$

I'm trying to prove that the integral of the following function converges:
$$int_0^infty fracarctan xx(ln(x+2))^2 dx$$
it's easy to prove that for each integral starting from $1$ using Dirichlet but i didn't manage to proof for the integral starting at $0$

Can I have a hint?
Thank you

integration

share|cite|improve this question

edited Mar 25 at 9:19

mathreadler

15.5k72263

asked Mar 25 at 9:18

MyNickMyNick

214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The title is a bit misleading since calculating an indefinite integral is usually much more difficult than proving its convergence only.
  $endgroup$
  – Jack D'Aurizio
  Mar 25 at 15:29
  

  
  
  
  

add a comment | 

0

1

$begingroup$

I'm trying to prove that the integral of the following function converges:
$$int_0^infty fracarctan xx(ln(x+2))^2 dx$$
it's easy to prove that for each integral starting from $1$ using Dirichlet but i didn't manage to proof for the integral starting at $0$

Can I have a hint?
Thank you

integration

share|cite|improve this question

edited Mar 25 at 9:19

mathreadler

15.5k72263

asked Mar 25 at 9:18

MyNickMyNick

214

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The title is a bit misleading since calculating an indefinite integral is usually much more difficult than proving its convergence only.
  $endgroup$
  – Jack D'Aurizio
  Mar 25 at 15:29
  

  
  
  
  

add a comment | 

$begingroup$

I'm trying to prove that the integral of the following function converges:
$$int_0^infty fracarctan xx(ln(x+2))^2 dx$$
it's easy to prove that for each integral starting from $1$ using Dirichlet but i didn't manage to proof for the integral starting at $0$

Can I have a hint?
Thank you

integration

share|cite|improve this question

edited Mar 25 at 9:19

mathreadler

15.5k72263

asked Mar 25 at 9:18

MyNickMyNick

214

$endgroup$

share|cite|improve this question

edited Mar 25 at 9:19

mathreadler

15.5k72263

asked Mar 25 at 9:18

MyNickMyNick

214

share|cite|improve this question

edited Mar 25 at 9:19

mathreadler

15.5k72263

asked Mar 25 at 9:18

MyNickMyNick

214

edited Mar 25 at 9:19

mathreadler

15.5k72263

edited Mar 25 at 9:19

mathreadler

15.5k72263

asked Mar 25 at 9:18

MyNickMyNick

214

asked Mar 25 at 9:18

MyNickMyNick

214

3 Answers
3

active

oldest

votes

2

$begingroup$

$$lim_xto 0fracarctan(x)xln^2(x+2)=frac1ln^2(2)$$ and thus it's prolongeable by continuity at $0$.

share|cite|improve this answer

edited Mar 25 at 9:35

answered Mar 25 at 9:30

user657324user657324

59510

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you supposed to get 0/0, what am i missing?
  $endgroup$
  – MyNick
  Mar 25 at 10:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The very standard limit $lim_xto 0fracarctan(x)x=1.$ @MyNick
  $endgroup$
  – user657324
  Mar 25 at 10:04
  
  

  
  
  
  

add a comment | 

2

$begingroup$

$int_2^infty frac 1 x(log, x)^2, dx <infty$ as seen easily buy the substitution $y=log, x$. Now just use the fact that $arctan$ is bounded. Integrability near $0$ is clear.

share|cite|improve this answer

answered Mar 25 at 9:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i dont see how it's clear, can you give anotgher hint please?
  $endgroup$
  – MyNick
  Mar 25 at 10:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $frac arctan, x x to 1$ as $ x to 0$ (by L'Hopital's Rule) so the integrand is bounded on $(0,2)$. Is this the part where you had difficulty?
  $endgroup$
  – Kavi Rama Murthy
  Mar 25 at 10:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  didn't know that limit. Thank you :)
  $endgroup$
  – MyNick
  Mar 25 at 10:06
  

  
  
  
  

add a comment | 

1

$begingroup$

There is no problem around $x=0$ since, by Taylor expansion,
$$fractan ^-1(x)x log ^2(x+2)=frac1log ^2(2)-fracxlog ^3(2)+Oleft(x^2right)$$

I do not finish since Kavi Rama Murthy explained what is going on for infinite values of $x$.

share|cite|improve this answer

answered Mar 25 at 9:41

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

add a comment | 

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2

$begingroup$

$$lim_xto 0fracarctan(x)xln^2(x+2)=frac1ln^2(2)$$ and thus it's prolongeable by continuity at $0$.

share|cite|improve this answer

edited Mar 25 at 9:35

answered Mar 25 at 9:30

user657324user657324

59510

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you supposed to get 0/0, what am i missing?
  $endgroup$
  – MyNick
  Mar 25 at 10:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The very standard limit $lim_xto 0fracarctan(x)x=1.$ @MyNick
  $endgroup$
  – user657324
  Mar 25 at 10:04
  
  

  
  
  
  

add a comment | 

2

$begingroup$

$$lim_xto 0fracarctan(x)xln^2(x+2)=frac1ln^2(2)$$ and thus it's prolongeable by continuity at $0$.

share|cite|improve this answer

edited Mar 25 at 9:35

answered Mar 25 at 9:30

user657324user657324

59510

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  you supposed to get 0/0, what am i missing?
  $endgroup$
  – MyNick
  Mar 25 at 10:01
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The very standard limit $lim_xto 0fracarctan(x)x=1.$ @MyNick
  $endgroup$
  – user657324
  Mar 25 at 10:04
  
  

  
  
  
  

add a comment | 

$begingroup$

$$lim_xto 0fracarctan(x)xln^2(x+2)=frac1ln^2(2)$$ and thus it's prolongeable by continuity at $0$.

share|cite|improve this answer

edited Mar 25 at 9:35

answered Mar 25 at 9:30

user657324user657324

59510

$endgroup$

share|cite|improve this answer

edited Mar 25 at 9:35

answered Mar 25 at 9:30

user657324user657324

59510

answered Mar 25 at 9:30

user657324user657324

59510

answered Mar 25 at 9:30

user657324user657324

59510

2

$begingroup$

$int_2^infty frac 1 x(log, x)^2, dx <infty$ as seen easily buy the substitution $y=log, x$. Now just use the fact that $arctan$ is bounded. Integrability near $0$ is clear.

share|cite|improve this answer

answered Mar 25 at 9:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i dont see how it's clear, can you give anotgher hint please?
  $endgroup$
  – MyNick
  Mar 25 at 10:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $frac arctan, x x to 1$ as $ x to 0$ (by L'Hopital's Rule) so the integrand is bounded on $(0,2)$. Is this the part where you had difficulty?
  $endgroup$
  – Kavi Rama Murthy
  Mar 25 at 10:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  didn't know that limit. Thank you :)
  $endgroup$
  – MyNick
  Mar 25 at 10:06
  

  
  
  
  

add a comment | 

2

$begingroup$

$int_2^infty frac 1 x(log, x)^2, dx <infty$ as seen easily buy the substitution $y=log, x$. Now just use the fact that $arctan$ is bounded. Integrability near $0$ is clear.

share|cite|improve this answer

answered Mar 25 at 9:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  i dont see how it's clear, can you give anotgher hint please?
  $endgroup$
  – MyNick
  Mar 25 at 10:02
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $frac arctan, x x to 1$ as $ x to 0$ (by L'Hopital's Rule) so the integrand is bounded on $(0,2)$. Is this the part where you had difficulty?
  $endgroup$
  – Kavi Rama Murthy
  Mar 25 at 10:05
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  didn't know that limit. Thank you :)
  $endgroup$
  – MyNick
  Mar 25 at 10:06
  

  
  
  
  

add a comment | 

$begingroup$

$int_2^infty frac 1 x(log, x)^2, dx <infty$ as seen easily buy the substitution $y=log, x$. Now just use the fact that $arctan$ is bounded. Integrability near $0$ is clear.

share|cite|improve this answer

answered Mar 25 at 9:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

$endgroup$

share|cite|improve this answer

answered Mar 25 at 9:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

answered Mar 25 at 9:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

answered Mar 25 at 9:36

Kavi Rama MurthyKavi Rama Murthy

74.6k53270

1

$begingroup$

There is no problem around $x=0$ since, by Taylor expansion,
$$fractan ^-1(x)x log ^2(x+2)=frac1log ^2(2)-fracxlog ^3(2)+Oleft(x^2right)$$

I do not finish since Kavi Rama Murthy explained what is going on for infinite values of $x$.

share|cite|improve this answer

answered Mar 25 at 9:41

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

add a comment | 

1

$begingroup$

There is no problem around $x=0$ since, by Taylor expansion,
$$fractan ^-1(x)x log ^2(x+2)=frac1log ^2(2)-fracxlog ^3(2)+Oleft(x^2right)$$

I do not finish since Kavi Rama Murthy explained what is going on for infinite values of $x$.

share|cite|improve this answer

answered Mar 25 at 9:41

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

add a comment | 

$begingroup$

There is no problem around $x=0$ since, by Taylor expansion,
$$fractan ^-1(x)x log ^2(x+2)=frac1log ^2(2)-fracxlog ^3(2)+Oleft(x^2right)$$

I do not finish since Kavi Rama Murthy explained what is going on for infinite values of $x$.

share|cite|improve this answer

answered Mar 25 at 9:41

Claude LeiboviciClaude Leibovici

125k1158135

$endgroup$

share|cite|improve this answer

answered Mar 25 at 9:41

Claude LeiboviciClaude Leibovici

125k1158135

answered Mar 25 at 9:41

Claude LeiboviciClaude Leibovici

125k1158135

answered Mar 25 at 9:41

Claude LeiboviciClaude Leibovici

125k1158135