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
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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
$begingroup$
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
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.
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marked as duplicate by YuiTo Cheng, RRL real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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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.
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
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.
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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 |
$begingroup$
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
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.
$endgroup$
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?
This question already has an answer here:
How to find $ int_0^infty dfrac x1+e^x dx$
5 answers
real-analysis integration sequences-and-series
real-analysis integration sequences-and-series
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
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
asked May 3 at 18:11
Javier Cuerva HerediaJavier Cuerva Heredia
191
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.
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.
marked as duplicate by YuiTo Cheng, RRL real-analysis
Users with the real-analysis badge can single-handedly close real-analysis questions as duplicates and reopen them as needed.
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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 |
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
1
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
$begingroup$
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
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
$begingroup$
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 |
1 Answer
1
active
oldest
votes
$begingroup$
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
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 |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
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
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
answered May 3 at 18:14
Lord Shark the UnknownLord Shark the Unknown
110k1163137
$endgroup$
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
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
answered May 3 at 18:14
Lord Shark the UnknownLord Shark the Unknown
110k1163137
110k1163137
$begingroup$
(+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 |
$begingroup$
(+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
$begingroup$
(+1) That was nice!
$endgroup$
– José Carlos Santos
May 3 at 18:16
$begingroup$
(+1) That was nice!
$endgroup$
– José Carlos Santos
May 3 at 18:16
1
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
$begingroup$
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
$begingroup$
@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
$begingroup$
@Cryvate yes, I see this now. Thank you.
$endgroup$
– Matthieu
May 3 at 20:12
add a comment |
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