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improper integral convergance

(Homework) Trouble calculating improper integral by parts.Improper integral $int_0^3 fracdx(x-1)^2/3$Determine the values of $p>0$ for which the improper integral $ int_0^inftyfracdxx^p sqrt1+x^p $ converges.Help with Infinite series + Integral Test + Improper IntegralsImproper integral: $int_0^infty fracsin^4xx^2dx$Proving $f(x) = fracsin xx$ converges by improper integral test?Unsure why this improper integral converges at 2.Determine whether the integral converges or diverges $int_1^infty fraclog x +sin xsqrtx dx$How can I determine all the values of p where an integral is improper?Prove integral diverges

0

$begingroup$

How can I determine if the following improper integral is converging or diverging

$$ int_1^+infty Bigg( fracx^24x^4+3x^3+2x^2+x Bigg) dx $$
$$ int_0^1 Bigg( fracx^24x^4+3x^3+2x^2+x Bigg) dx $$

So I see the two improper integral are the same, with differnet bounds. My understanding is to integrate it and then compute the bounds. If it approaches infinity or negative infinity it much diverge. If it approaches a finite number it converges. However i'm having trouble integrating this.

EDIT: I mistakengly put a p as an exponent

calculus integration convergence

share|cite|improve this question

edited Mar 22 at 0:52

mushi

asked Mar 22 at 0:42

mushimushi

63

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your integrand is comparable to (and in fact smaller than) $1/x^2$ as $x to infty$. Then you can apply the p-test and set $2p>1$.
  $endgroup$
  – Andrew Li
  Mar 22 at 0:46
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As a first step, divide the numerator and denominator by $x$
  $endgroup$
  – Mark Viola
  Mar 22 at 0:59
  

  
  
  
  

add a comment | 

0

$begingroup$

How can I determine if the following improper integral is converging or diverging

$$ int_1^+infty Bigg( fracx^24x^4+3x^3+2x^2+x Bigg) dx $$
$$ int_0^1 Bigg( fracx^24x^4+3x^3+2x^2+x Bigg) dx $$

So I see the two improper integral are the same, with differnet bounds. My understanding is to integrate it and then compute the bounds. If it approaches infinity or negative infinity it much diverge. If it approaches a finite number it converges. However i'm having trouble integrating this.

EDIT: I mistakengly put a p as an exponent

calculus integration convergence

share|cite|improve this question

edited Mar 22 at 0:52

mushi

asked Mar 22 at 0:42

mushimushi

63

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your integrand is comparable to (and in fact smaller than) $1/x^2$ as $x to infty$. Then you can apply the p-test and set $2p>1$.
  $endgroup$
  – Andrew Li
  Mar 22 at 0:46
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As a first step, divide the numerator and denominator by $x$
  $endgroup$
  – Mark Viola
  Mar 22 at 0:59
  

  
  
  
  

add a comment | 

$begingroup$

How can I determine if the following improper integral is converging or diverging

$$ int_1^+infty Bigg( fracx^24x^4+3x^3+2x^2+x Bigg) dx $$
$$ int_0^1 Bigg( fracx^24x^4+3x^3+2x^2+x Bigg) dx $$

So I see the two improper integral are the same, with differnet bounds. My understanding is to integrate it and then compute the bounds. If it approaches infinity or negative infinity it much diverge. If it approaches a finite number it converges. However i'm having trouble integrating this.

EDIT: I mistakengly put a p as an exponent

calculus integration convergence

share|cite|improve this question

edited Mar 22 at 0:52

mushi

asked Mar 22 at 0:42

mushimushi

63

$endgroup$

share|cite|improve this question

edited Mar 22 at 0:52

mushi

asked Mar 22 at 0:42

mushimushi

63

share|cite|improve this question

edited Mar 22 at 0:52

mushi

asked Mar 22 at 0:42

mushimushi

63

asked Mar 22 at 0:42

mushimushi

63

asked Mar 22 at 0:42

mushimushi

63

2 Answers
2

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oldest

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1

$begingroup$

Hint

$$fracx^24x^4+3x^3+2x^2+x<x^2over 4x^4=1over 4x^2$$

share|cite|improve this answer

answered Mar 22 at 12:04

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

1

$begingroup$

The integrand common to these integrals is $O(x)$ for small $x$, finite and positive for $x>0$, and $O(x^-2)$ for $x<0$, so each integral converges. In particular, if $epsilon>0$ then $int_0^epsilon x^p dx$ converges for $p>-1$, while $int_epsilon^infty x^p dx$ converges for $p<-1$.

share|cite|improve this answer

answered Mar 22 at 12:57

J.G.J.G.

32.8k23250

$endgroup$

add a comment | 

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1

$begingroup$

Hint

$$fracx^24x^4+3x^3+2x^2+x<x^2over 4x^4=1over 4x^2$$

share|cite|improve this answer

answered Mar 22 at 12:04

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

1

$begingroup$

Hint

$$fracx^24x^4+3x^3+2x^2+x<x^2over 4x^4=1over 4x^2$$

share|cite|improve this answer

answered Mar 22 at 12:04

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

$begingroup$

Hint

$$fracx^24x^4+3x^3+2x^2+x<x^2over 4x^4=1over 4x^2$$

share|cite|improve this answer

answered Mar 22 at 12:04

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

share|cite|improve this answer

answered Mar 22 at 12:04

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 22 at 12:04

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 22 at 12:04

Mostafa AyazMostafa Ayaz

18.2k31040

1

$begingroup$

The integrand common to these integrals is $O(x)$ for small $x$, finite and positive for $x>0$, and $O(x^-2)$ for $x<0$, so each integral converges. In particular, if $epsilon>0$ then $int_0^epsilon x^p dx$ converges for $p>-1$, while $int_epsilon^infty x^p dx$ converges for $p<-1$.

share|cite|improve this answer

answered Mar 22 at 12:57

J.G.J.G.

32.8k23250

$endgroup$

add a comment | 

1

$begingroup$

The integrand common to these integrals is $O(x)$ for small $x$, finite and positive for $x>0$, and $O(x^-2)$ for $x<0$, so each integral converges. In particular, if $epsilon>0$ then $int_0^epsilon x^p dx$ converges for $p>-1$, while $int_epsilon^infty x^p dx$ converges for $p<-1$.

share|cite|improve this answer

answered Mar 22 at 12:57

J.G.J.G.

32.8k23250

$endgroup$

add a comment | 

$begingroup$

The integrand common to these integrals is $O(x)$ for small $x$, finite and positive for $x>0$, and $O(x^-2)$ for $x<0$, so each integral converges. In particular, if $epsilon>0$ then $int_0^epsilon x^p dx$ converges for $p>-1$, while $int_epsilon^infty x^p dx$ converges for $p<-1$.

share|cite|improve this answer

answered Mar 22 at 12:57

J.G.J.G.

32.8k23250

$endgroup$

share|cite|improve this answer

answered Mar 22 at 12:57

J.G.J.G.

32.8k23250

answered Mar 22 at 12:57

J.G.J.G.

32.8k23250

answered Mar 22 at 12:57

J.G.J.G.

32.8k23250