Does $int_{0}^{infty}frac{r^2}{1+alpha r^4}dr$ converge?

1

$begingroup$

I want to check that $int_{mathbb R^3}frac{1}{1+x^4+y^4+z^4}dxdydz$ converges.

I moved to spherical coordinates, the integral became $int_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+r^4f(theta,phi)}dphi dtheta dr leqint_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+alpha r^4}dphi dtheta dr = 4pi int_{0}^{infty}frac{r^2}{1+alpha r^4}dr $

I don't know how to show this converges.

Edit: $f(theta,phi)$ is the usual spherical coordinates transform, but notice that it's bounded from below by a positive number, let that number be $alpha > 0$.

calculus integration improper-integrals

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asked Jan 20 at 21:00

Rick JokerRick Joker

399111

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1

$begingroup$

I want to check that $int_{mathbb R^3}frac{1}{1+x^4+y^4+z^4}dxdydz$ converges.

I moved to spherical coordinates, the integral became $int_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+r^4f(theta,phi)}dphi dtheta dr leqint_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+alpha r^4}dphi dtheta dr = 4pi int_{0}^{infty}frac{r^2}{1+alpha r^4}dr $

I don't know how to show this converges.

Edit: $f(theta,phi)$ is the usual spherical coordinates transform, but notice that it's bounded from below by a positive number, let that number be $alpha > 0$.

calculus integration improper-integrals

share|cite|improve this question

asked Jan 20 at 21:00

Rick JokerRick Joker

399111

$endgroup$

add a comment | 

1

1

1

$begingroup$

I want to check that $int_{mathbb R^3}frac{1}{1+x^4+y^4+z^4}dxdydz$ converges.

I moved to spherical coordinates, the integral became $int_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+r^4f(theta,phi)}dphi dtheta dr leqint_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+alpha r^4}dphi dtheta dr = 4pi int_{0}^{infty}frac{r^2}{1+alpha r^4}dr $

I don't know how to show this converges.

Edit: $f(theta,phi)$ is the usual spherical coordinates transform, but notice that it's bounded from below by a positive number, let that number be $alpha > 0$.

calculus integration improper-integrals

share|cite|improve this question

asked Jan 20 at 21:00

Rick JokerRick Joker

399111

$endgroup$

I want to check that $int_{mathbb R^3}frac{1}{1+x^4+y^4+z^4}dxdydz$ converges.

I moved to spherical coordinates, the integral became $int_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+r^4f(theta,phi)}dphi dtheta dr leqint_{0}^{infty}int_{0}^{pi}int_{0}^{2pi}frac{r^2sin(theta)}{1+alpha r^4}dphi dtheta dr = 4pi int_{0}^{infty}frac{r^2}{1+alpha r^4}dr $

I don't know how to show this converges.

Edit: $f(theta,phi)$ is the usual spherical coordinates transform, but notice that it's bounded from below by a positive number, let that number be $alpha > 0$.

calculus integration improper-integrals

calculus integration improper-integrals

share|cite|improve this question

asked Jan 20 at 21:00

Rick JokerRick Joker

399111

share|cite|improve this question

asked Jan 20 at 21:00

Rick JokerRick Joker

399111

share|cite|improve this question

share|cite|improve this question

asked Jan 20 at 21:00

Rick JokerRick Joker

399111

asked Jan 20 at 21:00

Rick JokerRick Joker

399111

asked Jan 20 at 21:00

Rick JokerRick Joker

399111

399111

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3 Answers
3

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1

$begingroup$

Hint: Split the integral
begin{align}
int^infty_0 frac{r^2}{1+alpha r^4} dr = int^infty_1frac{r^2}{1+alpha r^4} dr +int^1_0 frac{r^2}{1+alpha r^4} dr
end{align}

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answered Jan 20 at 21:04

Jacky ChongJacky Chong

19.1k21129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
  $endgroup$
  – Rick Joker
  Jan 20 at 21:05
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks anyways Jacky.
  $endgroup$
  – Rick Joker
  Jan 20 at 21:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RickJoker Yes.
  $endgroup$
  – Jacky Chong
  Jan 20 at 21:06
  

  
  
  
  

add a comment | 

2

$begingroup$

At 0, this is continuous. At $infty$ it behaves like $x mapsto 1/x^2$, which integrates finitely there. The integral converges.

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answered Jan 20 at 21:10

ncmathsadistncmathsadist

42.9k260103

$endgroup$

add a comment | 

1

$begingroup$

There is a problem only at $infty$. You can user equivalence:

$$frac{r^2}{1+alpha r^4}sim_inftyfrac 1{alpha r^2},$$
and $;displaystyleint_1^inftyfrac 1{alpha r^2},mathrm d r;$ converges.

share|cite|improve this answer

answered Jan 20 at 21:10

BernardBernard

122k740116

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

1

$begingroup$

Hint: Split the integral
begin{align}
int^infty_0 frac{r^2}{1+alpha r^4} dr = int^infty_1frac{r^2}{1+alpha r^4} dr +int^1_0 frac{r^2}{1+alpha r^4} dr
end{align}

share|cite|improve this answer

answered Jan 20 at 21:04

Jacky ChongJacky Chong

19.1k21129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
  $endgroup$
  – Rick Joker
  Jan 20 at 21:05
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks anyways Jacky.
  $endgroup$
  – Rick Joker
  Jan 20 at 21:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RickJoker Yes.
  $endgroup$
  – Jacky Chong
  Jan 20 at 21:06
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint: Split the integral
begin{align}
int^infty_0 frac{r^2}{1+alpha r^4} dr = int^infty_1frac{r^2}{1+alpha r^4} dr +int^1_0 frac{r^2}{1+alpha r^4} dr
end{align}

share|cite|improve this answer

answered Jan 20 at 21:04

Jacky ChongJacky Chong

19.1k21129

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
  $endgroup$
  – Rick Joker
  Jan 20 at 21:05
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks anyways Jacky.
  $endgroup$
  – Rick Joker
  Jan 20 at 21:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RickJoker Yes.
  $endgroup$
  – Jacky Chong
  Jan 20 at 21:06
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

Hint: Split the integral
begin{align}
int^infty_0 frac{r^2}{1+alpha r^4} dr = int^infty_1frac{r^2}{1+alpha r^4} dr +int^1_0 frac{r^2}{1+alpha r^4} dr
end{align}

share|cite|improve this answer

answered Jan 20 at 21:04

Jacky ChongJacky Chong

19.1k21129

$endgroup$

Hint: Split the integral
begin{align}
int^infty_0 frac{r^2}{1+alpha r^4} dr = int^infty_1frac{r^2}{1+alpha r^4} dr +int^1_0 frac{r^2}{1+alpha r^4} dr
end{align}

share|cite|improve this answer

answered Jan 20 at 21:04

Jacky ChongJacky Chong

19.1k21129

share|cite|improve this answer

share|cite|improve this answer

answered Jan 20 at 21:04

Jacky ChongJacky Chong

19.1k21129

answered Jan 20 at 21:04

Jacky ChongJacky Chong

19.1k21129

answered Jan 20 at 21:04

Jacky ChongJacky Chong

19.1k21129

19.1k21129

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
  $endgroup$
  – Rick Joker
  Jan 20 at 21:05
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks anyways Jacky.
  $endgroup$
  – Rick Joker
  Jan 20 at 21:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RickJoker Yes.
  $endgroup$
  – Jacky Chong
  Jan 20 at 21:06
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
  $endgroup$
  – Rick Joker
  Jan 20 at 21:05
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks anyways Jacky.
  $endgroup$
  – Rick Joker
  Jan 20 at 21:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RickJoker Yes.
  $endgroup$
  – Jacky Chong
  Jan 20 at 21:06
  

  
  
  
  

$begingroup$
And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
$endgroup$
– Rick Joker
Jan 20 at 21:05

$begingroup$
And $int_{1}^{infty} frac{r^2}{1 + alpha r^4} drleq int_{1}^{infty}frac{r^2}{alpha r^4}dr$. Literally just thought of that trick too.
$endgroup$
– Rick Joker
Jan 20 at 21:05

$begingroup$
Thanks anyways Jacky.
$endgroup$
– Rick Joker
Jan 20 at 21:06

$begingroup$
Thanks anyways Jacky.
$endgroup$
– Rick Joker
Jan 20 at 21:06

$begingroup$
@RickJoker Yes.
$endgroup$
– Jacky Chong
Jan 20 at 21:06

$begingroup$
@RickJoker Yes.
$endgroup$
– Jacky Chong
Jan 20 at 21:06

add a comment | 

2

$begingroup$

At 0, this is continuous. At $infty$ it behaves like $x mapsto 1/x^2$, which integrates finitely there. The integral converges.

share|cite|improve this answer

answered Jan 20 at 21:10

ncmathsadistncmathsadist

42.9k260103

$endgroup$

add a comment | 

2

$begingroup$

At 0, this is continuous. At $infty$ it behaves like $x mapsto 1/x^2$, which integrates finitely there. The integral converges.

share|cite|improve this answer

answered Jan 20 at 21:10

ncmathsadistncmathsadist

42.9k260103

$endgroup$

add a comment | 

2

2

2

$begingroup$

At 0, this is continuous. At $infty$ it behaves like $x mapsto 1/x^2$, which integrates finitely there. The integral converges.

share|cite|improve this answer

answered Jan 20 at 21:10

ncmathsadistncmathsadist

42.9k260103

$endgroup$

At 0, this is continuous. At $infty$ it behaves like $x mapsto 1/x^2$, which integrates finitely there. The integral converges.

share|cite|improve this answer

answered Jan 20 at 21:10

ncmathsadistncmathsadist

42.9k260103

share|cite|improve this answer

share|cite|improve this answer

answered Jan 20 at 21:10

ncmathsadistncmathsadist

42.9k260103

answered Jan 20 at 21:10

ncmathsadistncmathsadist

42.9k260103

answered Jan 20 at 21:10

ncmathsadistncmathsadist

42.9k260103

42.9k260103

add a comment | 

add a comment | 

1

$begingroup$

There is a problem only at $infty$. You can user equivalence:

$$frac{r^2}{1+alpha r^4}sim_inftyfrac 1{alpha r^2},$$
and $;displaystyleint_1^inftyfrac 1{alpha r^2},mathrm d r;$ converges.

share|cite|improve this answer

answered Jan 20 at 21:10

BernardBernard

122k740116

$endgroup$

add a comment | 

1

$begingroup$

There is a problem only at $infty$. You can user equivalence:

$$frac{r^2}{1+alpha r^4}sim_inftyfrac 1{alpha r^2},$$
and $;displaystyleint_1^inftyfrac 1{alpha r^2},mathrm d r;$ converges.

share|cite|improve this answer

answered Jan 20 at 21:10

BernardBernard

122k740116

$endgroup$

add a comment | 

1

1

1

$begingroup$

There is a problem only at $infty$. You can user equivalence:

$$frac{r^2}{1+alpha r^4}sim_inftyfrac 1{alpha r^2},$$
and $;displaystyleint_1^inftyfrac 1{alpha r^2},mathrm d r;$ converges.

share|cite|improve this answer

answered Jan 20 at 21:10

BernardBernard

122k740116

$endgroup$

There is a problem only at $infty$. You can user equivalence:

$$frac{r^2}{1+alpha r^4}sim_inftyfrac 1{alpha r^2},$$
and $;displaystyleint_1^inftyfrac 1{alpha r^2},mathrm d r;$ converges.

share|cite|improve this answer

answered Jan 20 at 21:10

BernardBernard

122k740116

share|cite|improve this answer

share|cite|improve this answer

answered Jan 20 at 21:10

BernardBernard

122k740116

answered Jan 20 at 21:10

BernardBernard

122k740116

answered Jan 20 at 21:10

BernardBernard

122k740116

122k740116

add a comment | 

add a comment | 

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