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Prove $int_0^2 fraclogxsqrt4-x^2textdx=0$ without integrating

3

3

$begingroup$

Inspired by this post:

Prove that $int_0^4 fracln xsqrt4x-x^2~dx=0$ (without trigonometric substitution)

I was wondering if there exists a similar pair of substitutions for the following integral:

$$I = int_0^2 fraclogxsqrt4-x^2textdx$$

More precisely:

Does there exist a linear and non-linear transformation (or possibly two non-linear transformations) $u(x), v(x)$ such that by applying them to the above integral, we can obtain $aI = bI$ where $mathbbR ni a neq b in mathbbR$

integration substitution

share|cite|improve this question

asked Mar 27 at 7:57

Jack LamJack Lam

2,0741826

$endgroup$

add a comment | 

3

3

$begingroup$

Inspired by this post:

Prove that $int_0^4 fracln xsqrt4x-x^2~dx=0$ (without trigonometric substitution)

I was wondering if there exists a similar pair of substitutions for the following integral:

$$I = int_0^2 fraclogxsqrt4-x^2textdx$$

More precisely:

Does there exist a linear and non-linear transformation (or possibly two non-linear transformations) $u(x), v(x)$ such that by applying them to the above integral, we can obtain $aI = bI$ where $mathbbR ni a neq b in mathbbR$

integration substitution

share|cite|improve this question

asked Mar 27 at 7:57

Jack LamJack Lam

2,0741826

$endgroup$

add a comment | 

$begingroup$

Inspired by this post:

Prove that $int_0^4 fracln xsqrt4x-x^2~dx=0$ (without trigonometric substitution)

I was wondering if there exists a similar pair of substitutions for the following integral:

$$I = int_0^2 fraclogxsqrt4-x^2textdx$$

More precisely:

Does there exist a linear and non-linear transformation (or possibly two non-linear transformations) $u(x), v(x)$ such that by applying them to the above integral, we can obtain $aI = bI$ where $mathbbR ni a neq b in mathbbR$

integration substitution

share|cite|improve this question

asked Mar 27 at 7:57

Jack LamJack Lam

2,0741826

$endgroup$

share|cite|improve this question

asked Mar 27 at 7:57

Jack LamJack Lam

2,0741826

share|cite|improve this question

asked Mar 27 at 7:57

Jack LamJack Lam

2,0741826

asked Mar 27 at 7:57

Jack LamJack Lam

2,0741826

asked Mar 27 at 7:57

Jack LamJack Lam

2,0741826

2 Answers
2

active

oldest

votes

5

$begingroup$

I have clearly overthought this.

Substitute $x mapsto x^2$ to transform $I$ into a quarter of the integral linked.

Nesting the transformations, we find the appropriate non-linear transformations for this integral are:

$$umapsto sqrt4-x^2, u mapsto x^2 - 2$$

Resulting in:

$$I=frac12int_0^2fraclog(4-u^2)sqrt4-u^2textdu$$

$$I=frac14int_-2^2fraclog(2+u)sqrt4-u^2textdu=frac14int_0^2fraclog(4-u^2)sqrt4-u^2textdu$$

$$I = fracmathcalI2 = fracmathcalI4$$

share|cite|improve this answer

answered Mar 27 at 10:47

Jack LamJack Lam

2,0741826

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice approach (+1)
  $endgroup$
  – Robert Z
  Mar 27 at 11:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice gymnastics ! and $to +1$ for sure.
  $endgroup$
  – Claude Leibovici
  Mar 27 at 11:34
  

  
  
  
  

add a comment | 

3

$begingroup$

I found a way of proving that $I=2I$ without integrating, although it is not similar to the one explained in the linked post.

Let $x=2sin(t)$ then
$$beginalignI&=int_0^pi/2log(2sin(t)),dt\
&=int_0^pi/2log(4sin(t/2)cos(t/2)),dt\
&=int_0^pi/2log(2sin(t/2)),dt+int_0^pi/2log(2cos(t/2)),dt\
&=int_0^pi/2log(2sin(t/2)),dt+int_pi/2^pilog(2sin(s/2)),ds\
&=int_0^pilog(2sin(s/2)),ds=2int_0^pi/2log(2sin(u)),du=2I
endalign$$
where $s=pi-t$ and $u=s/2$. Hence $I=0$.

share|cite|improve this answer

edited Mar 27 at 11:31

answered Mar 27 at 9:12

Robert ZRobert Z

102k1072145

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  While this is a quite neat way of proving the given equality, the OP asks for a similar pair of substitutions $($similiar to the linked post$)$, namely a pair of linear and non-linear transformation in order to achieve the result.
  $endgroup$
  – mrtaurho
  Mar 27 at 9:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I missed the day when you reached the $100k$ reputation. Congratulations and thanks for posting so good answers so often. Cheers :-)
  $endgroup$
  – Claude Leibovici
  Mar 27 at 9:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ClaudeLeibovici Thanks for your generous comments!
  $endgroup$
  – Robert Z
  Mar 27 at 11:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mrtaurho It seems that OP found the appropriate non-linear transformations
  $endgroup$
  – Robert Z
  Mar 27 at 11:27
  

  
  
  
  

add a comment | 

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5

$begingroup$

I have clearly overthought this.

Substitute $x mapsto x^2$ to transform $I$ into a quarter of the integral linked.

Nesting the transformations, we find the appropriate non-linear transformations for this integral are:

$$umapsto sqrt4-x^2, u mapsto x^2 - 2$$

Resulting in:

$$I=frac12int_0^2fraclog(4-u^2)sqrt4-u^2textdu$$

$$I=frac14int_-2^2fraclog(2+u)sqrt4-u^2textdu=frac14int_0^2fraclog(4-u^2)sqrt4-u^2textdu$$

$$I = fracmathcalI2 = fracmathcalI4$$

share|cite|improve this answer

answered Mar 27 at 10:47

Jack LamJack Lam

2,0741826

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice approach (+1)
  $endgroup$
  – Robert Z
  Mar 27 at 11:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice gymnastics ! and $to +1$ for sure.
  $endgroup$
  – Claude Leibovici
  Mar 27 at 11:34
  

  
  
  
  

add a comment | 

5

$begingroup$

I have clearly overthought this.

Substitute $x mapsto x^2$ to transform $I$ into a quarter of the integral linked.

Nesting the transformations, we find the appropriate non-linear transformations for this integral are:

$$umapsto sqrt4-x^2, u mapsto x^2 - 2$$

Resulting in:

$$I=frac12int_0^2fraclog(4-u^2)sqrt4-u^2textdu$$

$$I=frac14int_-2^2fraclog(2+u)sqrt4-u^2textdu=frac14int_0^2fraclog(4-u^2)sqrt4-u^2textdu$$

$$I = fracmathcalI2 = fracmathcalI4$$

share|cite|improve this answer

answered Mar 27 at 10:47

Jack LamJack Lam

2,0741826

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice approach (+1)
  $endgroup$
  – Robert Z
  Mar 27 at 11:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice gymnastics ! and $to +1$ for sure.
  $endgroup$
  – Claude Leibovici
  Mar 27 at 11:34
  

  
  
  
  

add a comment | 

$begingroup$

I have clearly overthought this.

Substitute $x mapsto x^2$ to transform $I$ into a quarter of the integral linked.

Nesting the transformations, we find the appropriate non-linear transformations for this integral are:

$$umapsto sqrt4-x^2, u mapsto x^2 - 2$$

Resulting in:

$$I=frac12int_0^2fraclog(4-u^2)sqrt4-u^2textdu$$

$$I=frac14int_-2^2fraclog(2+u)sqrt4-u^2textdu=frac14int_0^2fraclog(4-u^2)sqrt4-u^2textdu$$

$$I = fracmathcalI2 = fracmathcalI4$$

share|cite|improve this answer

answered Mar 27 at 10:47

Jack LamJack Lam

2,0741826

$endgroup$

share|cite|improve this answer

answered Mar 27 at 10:47

Jack LamJack Lam

2,0741826

answered Mar 27 at 10:47

Jack LamJack Lam

2,0741826

answered Mar 27 at 10:47

Jack LamJack Lam

2,0741826

3

$begingroup$

I found a way of proving that $I=2I$ without integrating, although it is not similar to the one explained in the linked post.

Let $x=2sin(t)$ then
$$beginalignI&=int_0^pi/2log(2sin(t)),dt\
&=int_0^pi/2log(4sin(t/2)cos(t/2)),dt\
&=int_0^pi/2log(2sin(t/2)),dt+int_0^pi/2log(2cos(t/2)),dt\
&=int_0^pi/2log(2sin(t/2)),dt+int_pi/2^pilog(2sin(s/2)),ds\
&=int_0^pilog(2sin(s/2)),ds=2int_0^pi/2log(2sin(u)),du=2I
endalign$$
where $s=pi-t$ and $u=s/2$. Hence $I=0$.

share|cite|improve this answer

edited Mar 27 at 11:31

answered Mar 27 at 9:12

Robert ZRobert Z

102k1072145

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  While this is a quite neat way of proving the given equality, the OP asks for a similar pair of substitutions $($similiar to the linked post$)$, namely a pair of linear and non-linear transformation in order to achieve the result.
  $endgroup$
  – mrtaurho
  Mar 27 at 9:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I missed the day when you reached the $100k$ reputation. Congratulations and thanks for posting so good answers so often. Cheers :-)
  $endgroup$
  – Claude Leibovici
  Mar 27 at 9:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ClaudeLeibovici Thanks for your generous comments!
  $endgroup$
  – Robert Z
  Mar 27 at 11:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mrtaurho It seems that OP found the appropriate non-linear transformations
  $endgroup$
  – Robert Z
  Mar 27 at 11:27
  

  
  
  
  

add a comment | 

3

$begingroup$

I found a way of proving that $I=2I$ without integrating, although it is not similar to the one explained in the linked post.

Let $x=2sin(t)$ then
$$beginalignI&=int_0^pi/2log(2sin(t)),dt\
&=int_0^pi/2log(4sin(t/2)cos(t/2)),dt\
&=int_0^pi/2log(2sin(t/2)),dt+int_0^pi/2log(2cos(t/2)),dt\
&=int_0^pi/2log(2sin(t/2)),dt+int_pi/2^pilog(2sin(s/2)),ds\
&=int_0^pilog(2sin(s/2)),ds=2int_0^pi/2log(2sin(u)),du=2I
endalign$$
where $s=pi-t$ and $u=s/2$. Hence $I=0$.

share|cite|improve this answer

edited Mar 27 at 11:31

answered Mar 27 at 9:12

Robert ZRobert Z

102k1072145

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  While this is a quite neat way of proving the given equality, the OP asks for a similar pair of substitutions $($similiar to the linked post$)$, namely a pair of linear and non-linear transformation in order to achieve the result.
  $endgroup$
  – mrtaurho
  Mar 27 at 9:22
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I missed the day when you reached the $100k$ reputation. Congratulations and thanks for posting so good answers so often. Cheers :-)
  $endgroup$
  – Claude Leibovici
  Mar 27 at 9:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ClaudeLeibovici Thanks for your generous comments!
  $endgroup$
  – Robert Z
  Mar 27 at 11:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mrtaurho It seems that OP found the appropriate non-linear transformations
  $endgroup$
  – Robert Z
  Mar 27 at 11:27
  

  
  
  
  

add a comment | 

$begingroup$

I found a way of proving that $I=2I$ without integrating, although it is not similar to the one explained in the linked post.

Let $x=2sin(t)$ then
$$beginalignI&=int_0^pi/2log(2sin(t)),dt\
&=int_0^pi/2log(4sin(t/2)cos(t/2)),dt\
&=int_0^pi/2log(2sin(t/2)),dt+int_0^pi/2log(2cos(t/2)),dt\
&=int_0^pi/2log(2sin(t/2)),dt+int_pi/2^pilog(2sin(s/2)),ds\
&=int_0^pilog(2sin(s/2)),ds=2int_0^pi/2log(2sin(u)),du=2I
endalign$$
where $s=pi-t$ and $u=s/2$. Hence $I=0$.

share|cite|improve this answer

edited Mar 27 at 11:31

answered Mar 27 at 9:12

Robert ZRobert Z

102k1072145

$endgroup$

share|cite|improve this answer

edited Mar 27 at 11:31

answered Mar 27 at 9:12

Robert ZRobert Z

102k1072145

answered Mar 27 at 9:12

Robert ZRobert Z

102k1072145

answered Mar 27 at 9:12

Robert ZRobert Z

102k1072145