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Prove that a definite integral is an infinite sum [duplicate]

How to find $ int_0^infty dfrac x1+e^x dx$Taylor Series of IntegralA definite integral in terms of Meijer G-functionConvergence of $int_0^infty fraclog x1+x^p dx$The easiest way to evaluate Gaussian integralProblem with evaluating the exact value of an integralTrig Integral With A Discontinuous Phase ShiftHow to evaluate this definite integral and find limit when upper limit of definite integral tends to infinityOn a simple integral involving Gompertz constantHow a Definite Integral can have the same value as just of one of the points of the integration interval?Calculate a definite integral involving sin and exp

3

2

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This question already has an answer here:

• 
  How to find $ int_0^infty dfrac x1+e^x dx$
  
  5 answers
  
  

I've been trying to solve this given equality involving an improper integral and an infinite sum without any substantial progress:
$$int_0^infty fracx1+e^xdx=sum_n=1^infty frac(-1)^n+1n^2$$

I tried various integration techniques such as change of variable ($1+e^x=t$ ; $e^x=t$), integration by parts and used Taylor expansion at every integral I arrived. However, I did not find any way of expressing the integrand as an infinite sums of functions that I could integrate term by term.

Any suggestions?

real-analysis integration sequences-and-series

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asked May 3 at 18:11

Javier Cuerva HerediaJavier Cuerva Heredia

191

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marked as duplicate by YuiTo Cheng, RRL real-analysis
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May 4 at 3:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that the sum in question is equal to $pi^2/12$. For more information, look up the Riemann zeta function.
  $endgroup$
  – Michael Seifert
  May 3 at 19:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of How to find $ int_0^infty dfrac x1+e^x dx$ (Your result is proved in the intermediate steps in Ron Gordan's answer; it also shows the sum is equal to $pi^2/12$ as mentioned by Michael Seifert)
  $endgroup$
  – YuiTo Cheng
  May 4 at 3:03
  
  

  
  
  
  

add a comment | 

3

2

$begingroup$

This question already has an answer here:

• 
  How to find $ int_0^infty dfrac x1+e^x dx$
  
  5 answers
  
  

I've been trying to solve this given equality involving an improper integral and an infinite sum without any substantial progress:
$$int_0^infty fracx1+e^xdx=sum_n=1^infty frac(-1)^n+1n^2$$

I tried various integration techniques such as change of variable ($1+e^x=t$ ; $e^x=t$), integration by parts and used Taylor expansion at every integral I arrived. However, I did not find any way of expressing the integrand as an infinite sums of functions that I could integrate term by term.

Any suggestions?

real-analysis integration sequences-and-series

share|cite|improve this question

asked May 3 at 18:11

Javier Cuerva HerediaJavier Cuerva Heredia

191

New contributor

Javier Cuerva Heredia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

marked as duplicate by YuiTo Cheng, RRL real-analysis
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May 4 at 3:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note that the sum in question is equal to $pi^2/12$. For more information, look up the Riemann zeta function.
  $endgroup$
  – Michael Seifert
  May 3 at 19:58
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of How to find $ int_0^infty dfrac x1+e^x dx$ (Your result is proved in the intermediate steps in Ron Gordan's answer; it also shows the sum is equal to $pi^2/12$ as mentioned by Michael Seifert)
  $endgroup$
  – YuiTo Cheng
  May 4 at 3:03
  
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  How to find $ int_0^infty dfrac x1+e^x dx$
  
  5 answers
  
  

I've been trying to solve this given equality involving an improper integral and an infinite sum without any substantial progress:
$$int_0^infty fracx1+e^xdx=sum_n=1^infty frac(-1)^n+1n^2$$

I tried various integration techniques such as change of variable ($1+e^x=t$ ; $e^x=t$), integration by parts and used Taylor expansion at every integral I arrived. However, I did not find any way of expressing the integrand as an infinite sums of functions that I could integrate term by term.

Any suggestions?

real-analysis integration sequences-and-series

share|cite|improve this question

asked May 3 at 18:11

Javier Cuerva HerediaJavier Cuerva Heredia

191

New contributor

Javier Cuerva Heredia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

asked May 3 at 18:11

Javier Cuerva HerediaJavier Cuerva Heredia

191

New contributor

Javier Cuerva Heredia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

asked May 3 at 18:11

Javier Cuerva HerediaJavier Cuerva Heredia

191

New contributor

Javier Cuerva Heredia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 3 at 18:11

Javier Cuerva HerediaJavier Cuerva Heredia

191

New contributor

Javier Cuerva Heredia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked May 3 at 18:11

Javier Cuerva HerediaJavier Cuerva Heredia

191

1 Answer
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14

$begingroup$

It's really just a geometric series:
beginalign
int_0^inftyfracx,dx1+e^x&=int_0^infty xe^-x(1+e^-x)^-1
=int_0^infty xsum_n=1^infty(-1)^n-1e^-nx,dx\
&=sum_n=1^infty(-1)^n-1int_0^infty xe^-nx,dx
=sum_n=1^inftyfrac(-1)^n-1n^2.
endalign

share|cite|improve this answer

answered May 3 at 18:14

Lord Shark the UnknownLord Shark the Unknown

110k1163137

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) That was nice!
  $endgroup$
  – José Carlos Santos
  May 3 at 18:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How did you make the following step? : $int_0^infty xe^-x(1+e^-x)^-1 =int_0^infty xsum_n=1^infty(-1)^n-1e^-nx,dx$. Thanks!
  $endgroup$
  – Matthieu
  May 3 at 19:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Matthieu the only reasonable explanation is a power series expansion, because it is $e^-nx$ suggests it is in the variable $e^x$ which seems about right to me: that sum could be the power series expansion (in the variable $e^x$) of $e^-x(1 + e^-x)^-1$.
  $endgroup$
  – Cryvate
  May 3 at 20:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Cryvate yes, I see this now. Thank you.
  $endgroup$
  – Matthieu
  May 3 at 20:12
  

  
  
  
  

add a comment | 

14

$begingroup$

It's really just a geometric series:
beginalign
int_0^inftyfracx,dx1+e^x&=int_0^infty xe^-x(1+e^-x)^-1
=int_0^infty xsum_n=1^infty(-1)^n-1e^-nx,dx\
&=sum_n=1^infty(-1)^n-1int_0^infty xe^-nx,dx
=sum_n=1^inftyfrac(-1)^n-1n^2.
endalign

share|cite|improve this answer

answered May 3 at 18:14

Lord Shark the UnknownLord Shark the Unknown

110k1163137

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) That was nice!
  $endgroup$
  – José Carlos Santos
  May 3 at 18:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How did you make the following step? : $int_0^infty xe^-x(1+e^-x)^-1 =int_0^infty xsum_n=1^infty(-1)^n-1e^-nx,dx$. Thanks!
  $endgroup$
  – Matthieu
  May 3 at 19:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Matthieu the only reasonable explanation is a power series expansion, because it is $e^-nx$ suggests it is in the variable $e^x$ which seems about right to me: that sum could be the power series expansion (in the variable $e^x$) of $e^-x(1 + e^-x)^-1$.
  $endgroup$
  – Cryvate
  May 3 at 20:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Cryvate yes, I see this now. Thank you.
  $endgroup$
  – Matthieu
  May 3 at 20:12
  

  
  
  
  

add a comment | 

14

$begingroup$

It's really just a geometric series:
beginalign
int_0^inftyfracx,dx1+e^x&=int_0^infty xe^-x(1+e^-x)^-1
=int_0^infty xsum_n=1^infty(-1)^n-1e^-nx,dx\
&=sum_n=1^infty(-1)^n-1int_0^infty xe^-nx,dx
=sum_n=1^inftyfrac(-1)^n-1n^2.
endalign

share|cite|improve this answer

answered May 3 at 18:14

Lord Shark the UnknownLord Shark the Unknown

110k1163137

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (+1) That was nice!
  $endgroup$
  – José Carlos Santos
  May 3 at 18:16
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How did you make the following step? : $int_0^infty xe^-x(1+e^-x)^-1 =int_0^infty xsum_n=1^infty(-1)^n-1e^-nx,dx$. Thanks!
  $endgroup$
  – Matthieu
  May 3 at 19:56
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Matthieu the only reasonable explanation is a power series expansion, because it is $e^-nx$ suggests it is in the variable $e^x$ which seems about right to me: that sum could be the power series expansion (in the variable $e^x$) of $e^-x(1 + e^-x)^-1$.
  $endgroup$
  – Cryvate
  May 3 at 20:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Cryvate yes, I see this now. Thank you.
  $endgroup$
  – Matthieu
  May 3 at 20:12
  

  
  
  
  

add a comment | 

$begingroup$

It's really just a geometric series:
beginalign
int_0^inftyfracx,dx1+e^x&=int_0^infty xe^-x(1+e^-x)^-1
=int_0^infty xsum_n=1^infty(-1)^n-1e^-nx,dx\
&=sum_n=1^infty(-1)^n-1int_0^infty xe^-nx,dx
=sum_n=1^inftyfrac(-1)^n-1n^2.
endalign

share|cite|improve this answer

answered May 3 at 18:14

Lord Shark the UnknownLord Shark the Unknown

110k1163137

$endgroup$

share|cite|improve this answer

answered May 3 at 18:14

Lord Shark the UnknownLord Shark the Unknown

110k1163137

answered May 3 at 18:14

Lord Shark the UnknownLord Shark the Unknown

110k1163137

answered May 3 at 18:14

Lord Shark the UnknownLord Shark the Unknown

110k1163137