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integrating $ln(x^2)/e^{x^2}$

How to show that this integral diverges or converges?Dirichelet Integral test limitsConvergence $I=int_0^infty frac{sin x}{x^s}dx$Improper Integral $int_0^1left(left{frac1xright}-frac12right)frac{log(x)}xdx$Is this series $sumlimits_{n=0}^infty frac{1+sin n}{10^n}$ divergent or convergent?Determine whether the integral is convergent or divergent. $int^{infty}_1 81frac{ln(x)}{x}dx$Determine that $Sigma^{infty}_{n=1}frac{n!}{n^n}$ is convergent.Convergence or Divergence integral questioncomparison test on $1/(e^x) vs 1/(e^x+1)$Testing convergence/divergence

2

1

$begingroup$

I am stuck when trying to find $$int_0^infty frac{ln(x^2)}{exp(x^2)} , dx $$
The context of the question is asking if the integral is convergent of divergent?

calculus integration convergence divergence

share|cite|improve this question

edited Mar 11 at 19:28

James odare

608

asked Mar 10 at 21:16

Suhas SuryaSuhas Surya

183

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that you do not have to actually compute anything to solve this problem! We are only interested in if the integral converges or not.
  $endgroup$
  – ItsJustVennDiagramsBro
  Mar 10 at 21:18
  

  
  
  
  

add a comment | 

2

1

$begingroup$

I am stuck when trying to find $$int_0^infty frac{ln(x^2)}{exp(x^2)} , dx $$
The context of the question is asking if the integral is convergent of divergent?

calculus integration convergence divergence

share|cite|improve this question

edited Mar 11 at 19:28

James odare

608

asked Mar 10 at 21:16

Suhas SuryaSuhas Surya

183

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that you do not have to actually compute anything to solve this problem! We are only interested in if the integral converges or not.
  $endgroup$
  – ItsJustVennDiagramsBro
  Mar 10 at 21:18
  

  
  
  
  

add a comment | 

$begingroup$

I am stuck when trying to find $$int_0^infty frac{ln(x^2)}{exp(x^2)} , dx $$
The context of the question is asking if the integral is convergent of divergent?

calculus integration convergence divergence

share|cite|improve this question

edited Mar 11 at 19:28

James odare

608

asked Mar 10 at 21:16

Suhas SuryaSuhas Surya

183

$endgroup$

share|cite|improve this question

edited Mar 11 at 19:28

James odare

608

asked Mar 10 at 21:16

Suhas SuryaSuhas Surya

183

share|cite|improve this question

edited Mar 11 at 19:28

James odare

608

asked Mar 10 at 21:16

Suhas SuryaSuhas Surya

183

edited Mar 11 at 19:28

James odare

608

edited Mar 11 at 19:28

James odare

608

asked Mar 10 at 21:16

Suhas SuryaSuhas Surya

183

asked Mar 10 at 21:16

Suhas SuryaSuhas Surya

183

3 Answers
3

active

oldest

votes

4

$begingroup$

If you want the explicit value, well,
$$ int_{0}^{+infty}2log(x)e^{-x^2},dx stackrel{xmapstosqrt{x}}{=}frac{1}{2}int_{0}^{+infty}x^{-1/2}log(x)e^{-x},dx $$
equals $frac{1}{2}Gamma'left(frac{1}{2}right)=frac{sqrt{pi}}{2}psileft(frac{1}{2}right)$ by the dominated convergence theorem and the integral definition of the $Gamma$ function, together with the well known $Gammaleft(frac{1}{2}right)=sqrt{pi}$. Since $psi(1)=Gamma'(1)=-gamma$ and
$$ sum_{ngeq 0}frac{1}{(n+a)(n+b)}=frac{psi(a)-psi(b)}{a-b},$$
by picking $a=frac{1}{2}$, $b=1$ and rearranging we get
$$ int_{0}^{+infty}log(x^2)e^{-x^2},dx = color{blue}{-frac{sqrt{pi}}{2}(gamma+2log 2)}$$
which is approximately $-frac{7}{4}$.

share|cite|improve this answer

answered Mar 11 at 0:51

Jack D'AurizioJack D'Aurizio

291k33284667

$endgroup$

add a comment | 

1

$begingroup$

The function is equivalent to $2ln{|x|}$ at $x$ goes to $0$ so is integrable there.
The function is $o(e^{-|x|})$ as $|x|$ goes to infinity, so is integrable there as well.

share|cite|improve this answer

answered Mar 10 at 21:22

MindlackMindlack

4,920211

$endgroup$

add a comment | 

1

$begingroup$

Near $0$, your function is equivalent to $ln(x^2)=2 ln(x)$ which is integrable.

Near $+infty$, your function is a $o left( frac{x^2}{e^{x^2}} right)$ which is integrable.

share|cite|improve this answer

answered Mar 10 at 21:23

TheSilverDoeTheSilverDoe

3,866112

$endgroup$

add a comment | 

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4

$begingroup$

If you want the explicit value, well,
$$ int_{0}^{+infty}2log(x)e^{-x^2},dx stackrel{xmapstosqrt{x}}{=}frac{1}{2}int_{0}^{+infty}x^{-1/2}log(x)e^{-x},dx $$
equals $frac{1}{2}Gamma'left(frac{1}{2}right)=frac{sqrt{pi}}{2}psileft(frac{1}{2}right)$ by the dominated convergence theorem and the integral definition of the $Gamma$ function, together with the well known $Gammaleft(frac{1}{2}right)=sqrt{pi}$. Since $psi(1)=Gamma'(1)=-gamma$ and
$$ sum_{ngeq 0}frac{1}{(n+a)(n+b)}=frac{psi(a)-psi(b)}{a-b},$$
by picking $a=frac{1}{2}$, $b=1$ and rearranging we get
$$ int_{0}^{+infty}log(x^2)e^{-x^2},dx = color{blue}{-frac{sqrt{pi}}{2}(gamma+2log 2)}$$
which is approximately $-frac{7}{4}$.

share|cite|improve this answer

answered Mar 11 at 0:51

Jack D'AurizioJack D'Aurizio

291k33284667

$endgroup$

add a comment | 

4

$begingroup$

If you want the explicit value, well,
$$ int_{0}^{+infty}2log(x)e^{-x^2},dx stackrel{xmapstosqrt{x}}{=}frac{1}{2}int_{0}^{+infty}x^{-1/2}log(x)e^{-x},dx $$
equals $frac{1}{2}Gamma'left(frac{1}{2}right)=frac{sqrt{pi}}{2}psileft(frac{1}{2}right)$ by the dominated convergence theorem and the integral definition of the $Gamma$ function, together with the well known $Gammaleft(frac{1}{2}right)=sqrt{pi}$. Since $psi(1)=Gamma'(1)=-gamma$ and
$$ sum_{ngeq 0}frac{1}{(n+a)(n+b)}=frac{psi(a)-psi(b)}{a-b},$$
by picking $a=frac{1}{2}$, $b=1$ and rearranging we get
$$ int_{0}^{+infty}log(x^2)e^{-x^2},dx = color{blue}{-frac{sqrt{pi}}{2}(gamma+2log 2)}$$
which is approximately $-frac{7}{4}$.

share|cite|improve this answer

answered Mar 11 at 0:51

Jack D'AurizioJack D'Aurizio

291k33284667

$endgroup$

add a comment | 

$begingroup$

If you want the explicit value, well,
$$ int_{0}^{+infty}2log(x)e^{-x^2},dx stackrel{xmapstosqrt{x}}{=}frac{1}{2}int_{0}^{+infty}x^{-1/2}log(x)e^{-x},dx $$
equals $frac{1}{2}Gamma'left(frac{1}{2}right)=frac{sqrt{pi}}{2}psileft(frac{1}{2}right)$ by the dominated convergence theorem and the integral definition of the $Gamma$ function, together with the well known $Gammaleft(frac{1}{2}right)=sqrt{pi}$. Since $psi(1)=Gamma'(1)=-gamma$ and
$$ sum_{ngeq 0}frac{1}{(n+a)(n+b)}=frac{psi(a)-psi(b)}{a-b},$$
by picking $a=frac{1}{2}$, $b=1$ and rearranging we get
$$ int_{0}^{+infty}log(x^2)e^{-x^2},dx = color{blue}{-frac{sqrt{pi}}{2}(gamma+2log 2)}$$
which is approximately $-frac{7}{4}$.

share|cite|improve this answer

answered Mar 11 at 0:51

Jack D'AurizioJack D'Aurizio

291k33284667

$endgroup$

share|cite|improve this answer

answered Mar 11 at 0:51

Jack D'AurizioJack D'Aurizio

291k33284667

answered Mar 11 at 0:51

Jack D'AurizioJack D'Aurizio

291k33284667

answered Mar 11 at 0:51

Jack D'AurizioJack D'Aurizio

291k33284667

1

$begingroup$

The function is equivalent to $2ln{|x|}$ at $x$ goes to $0$ so is integrable there.
The function is $o(e^{-|x|})$ as $|x|$ goes to infinity, so is integrable there as well.

share|cite|improve this answer

answered Mar 10 at 21:22

MindlackMindlack

4,920211

$endgroup$

add a comment | 

1

$begingroup$

The function is equivalent to $2ln{|x|}$ at $x$ goes to $0$ so is integrable there.
The function is $o(e^{-|x|})$ as $|x|$ goes to infinity, so is integrable there as well.

share|cite|improve this answer

answered Mar 10 at 21:22

MindlackMindlack

4,920211

$endgroup$

add a comment | 

$begingroup$

The function is equivalent to $2ln{|x|}$ at $x$ goes to $0$ so is integrable there.
The function is $o(e^{-|x|})$ as $|x|$ goes to infinity, so is integrable there as well.

share|cite|improve this answer

answered Mar 10 at 21:22

MindlackMindlack

4,920211

$endgroup$

share|cite|improve this answer

answered Mar 10 at 21:22

MindlackMindlack

4,920211

answered Mar 10 at 21:22

MindlackMindlack

4,920211

answered Mar 10 at 21:22

MindlackMindlack

4,920211

1

$begingroup$

Near $0$, your function is equivalent to $ln(x^2)=2 ln(x)$ which is integrable.

Near $+infty$, your function is a $o left( frac{x^2}{e^{x^2}} right)$ which is integrable.

share|cite|improve this answer

answered Mar 10 at 21:23

TheSilverDoeTheSilverDoe

3,866112

$endgroup$

add a comment | 

1

$begingroup$

Near $0$, your function is equivalent to $ln(x^2)=2 ln(x)$ which is integrable.

Near $+infty$, your function is a $o left( frac{x^2}{e^{x^2}} right)$ which is integrable.

share|cite|improve this answer

answered Mar 10 at 21:23

TheSilverDoeTheSilverDoe

3,866112

$endgroup$

add a comment | 

$begingroup$

Near $0$, your function is equivalent to $ln(x^2)=2 ln(x)$ which is integrable.

Near $+infty$, your function is a $o left( frac{x^2}{e^{x^2}} right)$ which is integrable.

share|cite|improve this answer

answered Mar 10 at 21:23

TheSilverDoeTheSilverDoe

3,866112

$endgroup$

share|cite|improve this answer

answered Mar 10 at 21:23

TheSilverDoeTheSilverDoe

3,866112

answered Mar 10 at 21:23

TheSilverDoeTheSilverDoe

3,866112

answered Mar 10 at 21:23

TheSilverDoeTheSilverDoe

3,866112