Tricky Integral involving radicals

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I am trying to evaluate the following definite integral (for $a>0$):

$$I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx$$
Neither the substitution $u=left( 1-x^a right)^frac1a$ nor $u=left( 1-x^a right)^frac1a-x$ are appropriate.I have also tried Feynman’s trick (differentiated with respect to a) but I didn’t get any success. Thanks in advance.

calculus integration radicals

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edited Jul 16 at 15:12

asked Jul 16 at 15:03

user577488

1085

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How about $u=x^a$?
  – Chris2018
  Jul 16 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not sure let me try it
  – user577488
  Jul 16 at 15:09
  

  
  
  
  

add a comment | 

up vote
8
down vote

favorite

1

I am trying to evaluate the following definite integral (for $a>0$):

$$I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx$$
Neither the substitution $u=left( 1-x^a right)^frac1a$ nor $u=left( 1-x^a right)^frac1a-x$ are appropriate.I have also tried Feynman’s trick (differentiated with respect to a) but I didn’t get any success. Thanks in advance.

calculus integration radicals

share|cite|improve this question

edited Jul 16 at 15:12

asked Jul 16 at 15:03

user577488

1085

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How about $u=x^a$?
  – Chris2018
  Jul 16 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not sure let me try it
  – user577488
  Jul 16 at 15:09
  

  
  
  
  

add a comment | 

up vote
8
down vote

favorite

1

up vote
8
down vote

favorite

1

1

I am trying to evaluate the following definite integral (for $a>0$):

$$I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx$$
Neither the substitution $u=left( 1-x^a right)^frac1a$ nor $u=left( 1-x^a right)^frac1a-x$ are appropriate.I have also tried Feynman’s trick (differentiated with respect to a) but I didn’t get any success. Thanks in advance.

calculus integration radicals

share|cite|improve this question

edited Jul 16 at 15:12

asked Jul 16 at 15:03

user577488

1085

I am trying to evaluate the following definite integral (for $a>0$):

$$I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx$$
Neither the substitution $u=left( 1-x^a right)^frac1a$ nor $u=left( 1-x^a right)^frac1a-x$ are appropriate.I have also tried Feynman’s trick (differentiated with respect to a) but I didn’t get any success. Thanks in advance.

calculus integration radicals

share|cite|improve this question

edited Jul 16 at 15:12

asked Jul 16 at 15:03

user577488

1085

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 15:12

edited Jul 16 at 15:12

edited Jul 16 at 15:12

asked Jul 16 at 15:03

user577488

1085

asked Jul 16 at 15:03

user577488

1085

asked Jul 16 at 15:03

user577488

1085

1085

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How about $u=x^a$?
  – Chris2018
  Jul 16 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not sure let me try it
  – user577488
  Jul 16 at 15:09
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  How about $u=x^a$?
  – Chris2018
  Jul 16 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  I am not sure let me try it
  – user577488
  Jul 16 at 15:09
  

  
  
  
  

1

1

How about $u=x^a$?
– Chris2018
Jul 16 at 15:06

How about $u=x^a$?
– Chris2018
Jul 16 at 15:06

I am not sure let me try it
– user577488
Jul 16 at 15:09

I am not sure let me try it
– user577488
Jul 16 at 15:09

add a comment | 

2 Answers
2

active

oldest

votes

up vote
8
down vote



accepted

Using the substitution $u=left( 1-x^a right)^frac1a$
$$I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx=int_0^1u^a-1left( 1-u^a right)^frac1a-1left( u-left( 1-u^a right)^frac1a right)^2du$$
Since both $x and u$ are dummy variables
$$beginalign
& I=frac12int_0^1left( left( 1-x^a right)^frac1a-x right)^2+x^a-1left( 1-x^a right)^frac1a-1left( x-left( 1-x^a right)^frac1a right)^2dx \
& quad =frac12int_0^1left( left( 1-x^a right)^frac1a-x right)^2left( 1+x^a-1left( 1-x^a right)^frac1a-1 right)dx \
& quad =frac12int_0^1-frac13fracddxleft( left( 1-x^a right)^frac1a-x right)^3dx \
& quad =frac13 \
endalign$$

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answered Jul 16 at 15:23

Vincent Law

1,467212

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up vote
6
down vote

Hint. By the change of variable
$$
u=x^a,qquad x=u^1/a,qquad dx=frac1au^1/a-1du,
$$ one gets
$$
I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx=frac1aint_0^1left( left( 1-uright)^frac1a-u^1/a right)^2u^1/a-1du
$$ then by expanding the square one is led to apply the standard Euler beta evaluation:
$$
int_0^1(1-u)^s-1 u^t-1,du = fracGamma(s)Gamma(t)Gamma(s+t),quad operatornameRe(s)>0,,operatornameRe(t)>0.
$$

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answered Jul 16 at 15:14

Olivier Oloa

106k17173292

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
8
down vote



accepted

Using the substitution $u=left( 1-x^a right)^frac1a$
$$I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx=int_0^1u^a-1left( 1-u^a right)^frac1a-1left( u-left( 1-u^a right)^frac1a right)^2du$$
Since both $x and u$ are dummy variables
$$beginalign
& I=frac12int_0^1left( left( 1-x^a right)^frac1a-x right)^2+x^a-1left( 1-x^a right)^frac1a-1left( x-left( 1-x^a right)^frac1a right)^2dx \
& quad =frac12int_0^1left( left( 1-x^a right)^frac1a-x right)^2left( 1+x^a-1left( 1-x^a right)^frac1a-1 right)dx \
& quad =frac12int_0^1-frac13fracddxleft( left( 1-x^a right)^frac1a-x right)^3dx \
& quad =frac13 \
endalign$$

share|cite|improve this answer

answered Jul 16 at 15:23

Vincent Law

1,467212

add a comment | 

up vote
8
down vote



accepted

Using the substitution $u=left( 1-x^a right)^frac1a$
$$I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx=int_0^1u^a-1left( 1-u^a right)^frac1a-1left( u-left( 1-u^a right)^frac1a right)^2du$$
Since both $x and u$ are dummy variables
$$beginalign
& I=frac12int_0^1left( left( 1-x^a right)^frac1a-x right)^2+x^a-1left( 1-x^a right)^frac1a-1left( x-left( 1-x^a right)^frac1a right)^2dx \
& quad =frac12int_0^1left( left( 1-x^a right)^frac1a-x right)^2left( 1+x^a-1left( 1-x^a right)^frac1a-1 right)dx \
& quad =frac12int_0^1-frac13fracddxleft( left( 1-x^a right)^frac1a-x right)^3dx \
& quad =frac13 \
endalign$$

share|cite|improve this answer

answered Jul 16 at 15:23

Vincent Law

1,467212

add a comment | 

up vote
8
down vote



accepted

up vote
8
down vote



accepted

Using the substitution $u=left( 1-x^a right)^frac1a$
$$I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx=int_0^1u^a-1left( 1-u^a right)^frac1a-1left( u-left( 1-u^a right)^frac1a right)^2du$$
Since both $x and u$ are dummy variables
$$beginalign
& I=frac12int_0^1left( left( 1-x^a right)^frac1a-x right)^2+x^a-1left( 1-x^a right)^frac1a-1left( x-left( 1-x^a right)^frac1a right)^2dx \
& quad =frac12int_0^1left( left( 1-x^a right)^frac1a-x right)^2left( 1+x^a-1left( 1-x^a right)^frac1a-1 right)dx \
& quad =frac12int_0^1-frac13fracddxleft( left( 1-x^a right)^frac1a-x right)^3dx \
& quad =frac13 \
endalign$$

share|cite|improve this answer

answered Jul 16 at 15:23

Vincent Law

1,467212

Using the substitution $u=left( 1-x^a right)^frac1a$
$$I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx=int_0^1u^a-1left( 1-u^a right)^frac1a-1left( u-left( 1-u^a right)^frac1a right)^2du$$
Since both $x and u$ are dummy variables
$$beginalign
& I=frac12int_0^1left( left( 1-x^a right)^frac1a-x right)^2+x^a-1left( 1-x^a right)^frac1a-1left( x-left( 1-x^a right)^frac1a right)^2dx \
& quad =frac12int_0^1left( left( 1-x^a right)^frac1a-x right)^2left( 1+x^a-1left( 1-x^a right)^frac1a-1 right)dx \
& quad =frac12int_0^1-frac13fracddxleft( left( 1-x^a right)^frac1a-x right)^3dx \
& quad =frac13 \
endalign$$

share|cite|improve this answer

answered Jul 16 at 15:23

Vincent Law

1,467212

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 15:23

Vincent Law

1,467212

answered Jul 16 at 15:23

Vincent Law

1,467212

answered Jul 16 at 15:23

Vincent Law

1,467212

1,467212

add a comment | 

add a comment | 

up vote
6
down vote

Hint. By the change of variable
$$
u=x^a,qquad x=u^1/a,qquad dx=frac1au^1/a-1du,
$$ one gets
$$
I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx=frac1aint_0^1left( left( 1-uright)^frac1a-u^1/a right)^2u^1/a-1du
$$ then by expanding the square one is led to apply the standard Euler beta evaluation:
$$
int_0^1(1-u)^s-1 u^t-1,du = fracGamma(s)Gamma(t)Gamma(s+t),quad operatornameRe(s)>0,,operatornameRe(t)>0.
$$

share|cite|improve this answer

answered Jul 16 at 15:14

Olivier Oloa

106k17173292

add a comment | 

up vote
6
down vote

Hint. By the change of variable
$$
u=x^a,qquad x=u^1/a,qquad dx=frac1au^1/a-1du,
$$ one gets
$$
I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx=frac1aint_0^1left( left( 1-uright)^frac1a-u^1/a right)^2u^1/a-1du
$$ then by expanding the square one is led to apply the standard Euler beta evaluation:
$$
int_0^1(1-u)^s-1 u^t-1,du = fracGamma(s)Gamma(t)Gamma(s+t),quad operatornameRe(s)>0,,operatornameRe(t)>0.
$$

share|cite|improve this answer

answered Jul 16 at 15:14

Olivier Oloa

106k17173292

add a comment | 

up vote
6
down vote

up vote
6
down vote

Hint. By the change of variable
$$
u=x^a,qquad x=u^1/a,qquad dx=frac1au^1/a-1du,
$$ one gets
$$
I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx=frac1aint_0^1left( left( 1-uright)^frac1a-u^1/a right)^2u^1/a-1du
$$ then by expanding the square one is led to apply the standard Euler beta evaluation:
$$
int_0^1(1-u)^s-1 u^t-1,du = fracGamma(s)Gamma(t)Gamma(s+t),quad operatornameRe(s)>0,,operatornameRe(t)>0.
$$

share|cite|improve this answer

answered Jul 16 at 15:14

Olivier Oloa

106k17173292

Hint. By the change of variable
$$
u=x^a,qquad x=u^1/a,qquad dx=frac1au^1/a-1du,
$$ one gets
$$
I=int_0^1left( left( 1-x^a right)^frac1a-x right)^2dx=frac1aint_0^1left( left( 1-uright)^frac1a-u^1/a right)^2u^1/a-1du
$$ then by expanding the square one is led to apply the standard Euler beta evaluation:
$$
int_0^1(1-u)^s-1 u^t-1,du = fracGamma(s)Gamma(t)Gamma(s+t),quad operatornameRe(s)>0,,operatornameRe(t)>0.
$$

share|cite|improve this answer

answered Jul 16 at 15:14

Olivier Oloa

106k17173292

share|cite|improve this answer

share|cite|improve this answer

answered Jul 16 at 15:14

Olivier Oloa

106k17173292

answered Jul 16 at 15:14

Olivier Oloa

106k17173292

answered Jul 16 at 15:14

Olivier Oloa

106k17173292

106k17173292

add a comment | 

add a comment | 

 

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