Integral identity in statistical physics [closed]
$begingroup$
I found this identity in a statistical physics book, but I can't find a way to demonstrate it and I don't like to take thing as true without knowing why.
The identity is:
$$ |z|=frac{1}{pi} int_0^{pi}log(2cosh z - 2cos t)dt $$
Any help? Even just a reference is enough, if you know how this identity is called and/or where to find it.
integration definite-integrals
edited Jan 14 at 18:03
gt6989b
35k22557
asked Jan 14 at 18:01
Luca MartinoiaLuca Martinoia
33
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closed as off-topic by Namaste, Cesareo, Xander Henderson, Leucippus, mrtaurho Jan 15 at 5:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Cesareo, Xander Henderson, Leucippus, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I found this identity in a statistical physics book, but I can't find a way to demonstrate it and I don't like to take thing as true without knowing why.
The identity is:
$$ |z|=frac{1}{pi} int_0^{pi}log(2cosh z - 2cos t)dt $$
Any help? Even just a reference is enough, if you know how this identity is called and/or where to find it.
integration definite-integrals
edited Jan 14 at 18:03
gt6989b
35k22557
asked Jan 14 at 18:01
Luca MartinoiaLuca Martinoia
33
$endgroup$
closed as off-topic by Namaste, Cesareo, Xander Henderson, Leucippus, mrtaurho Jan 15 at 5:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Cesareo, Xander Henderson, Leucippus, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
I think this appeared here on MSE before, but it's quite hard to find it. Anyway something similar on AoPS: artofproblemsolving.com/community/c7h17769p120601
$endgroup$
– Zacky
Jan 14 at 18:09
$begingroup$
I tried to look for it, but I'm kinda new to the side and it was hard to find an integral like this. Sadly AoPS seems down now, but thanks for the help.
$endgroup$
– Luca Martinoia
Jan 14 at 18:15
add a comment |
$begingroup$
I found this identity in a statistical physics book, but I can't find a way to demonstrate it and I don't like to take thing as true without knowing why.
The identity is:
$$ |z|=frac{1}{pi} int_0^{pi}log(2cosh z - 2cos t)dt $$
Any help? Even just a reference is enough, if you know how this identity is called and/or where to find it.
integration definite-integrals
edited Jan 14 at 18:03
gt6989b
35k22557
asked Jan 14 at 18:01
Luca MartinoiaLuca Martinoia
33
$endgroup$
I found this identity in a statistical physics book, but I can't find a way to demonstrate it and I don't like to take thing as true without knowing why.
The identity is:
$$ |z|=frac{1}{pi} int_0^{pi}log(2cosh z - 2cos t)dt $$
Any help? Even just a reference is enough, if you know how this identity is called and/or where to find it.
integration definite-integrals
integration definite-integrals
edited Jan 14 at 18:03
gt6989b
35k22557
asked Jan 14 at 18:01
Luca MartinoiaLuca Martinoia
33
edited Jan 14 at 18:03
gt6989b
35k22557
asked Jan 14 at 18:01
Luca MartinoiaLuca Martinoia
33
edited Jan 14 at 18:03
gt6989b
35k22557
edited Jan 14 at 18:03
gt6989b
35k22557
edited Jan 14 at 18:03
gt6989b
35k22557
35k22557
asked Jan 14 at 18:01
Luca MartinoiaLuca Martinoia
33
asked Jan 14 at 18:01
Luca MartinoiaLuca Martinoia
33
asked Jan 14 at 18:01
Luca MartinoiaLuca Martinoia
33
33
closed as off-topic by Namaste, Cesareo, Xander Henderson, Leucippus, mrtaurho Jan 15 at 5:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Cesareo, Xander Henderson, Leucippus, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Namaste, Cesareo, Xander Henderson, Leucippus, mrtaurho Jan 15 at 5:46
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Namaste, Cesareo, Xander Henderson, Leucippus, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
1
$begingroup$
I think this appeared here on MSE before, but it's quite hard to find it. Anyway something similar on AoPS: artofproblemsolving.com/community/c7h17769p120601
$endgroup$
– Zacky
Jan 14 at 18:09
$begingroup$
I tried to look for it, but I'm kinda new to the side and it was hard to find an integral like this. Sadly AoPS seems down now, but thanks for the help.
$endgroup$
– Luca Martinoia
Jan 14 at 18:15
add a comment |
1
$begingroup$
I think this appeared here on MSE before, but it's quite hard to find it. Anyway something similar on AoPS: artofproblemsolving.com/community/c7h17769p120601
$endgroup$
– Zacky
Jan 14 at 18:09
$begingroup$
I tried to look for it, but I'm kinda new to the side and it was hard to find an integral like this. Sadly AoPS seems down now, but thanks for the help.
$endgroup$
– Luca Martinoia
Jan 14 at 18:15
1
1
$begingroup$
I think this appeared here on MSE before, but it's quite hard to find it. Anyway something similar on AoPS: artofproblemsolving.com/community/c7h17769p120601
$endgroup$
– Zacky
Jan 14 at 18:09
$begingroup$
I think this appeared here on MSE before, but it's quite hard to find it. Anyway something similar on AoPS: artofproblemsolving.com/community/c7h17769p120601
$endgroup$
– Zacky
Jan 14 at 18:09
$begingroup$
I tried to look for it, but I'm kinda new to the side and it was hard to find an integral like this. Sadly AoPS seems down now, but thanks for the help.
$endgroup$
– Luca Martinoia
Jan 14 at 18:15
$begingroup$
I tried to look for it, but I'm kinda new to the side and it was hard to find an integral like this. Sadly AoPS seems down now, but thanks for the help.
$endgroup$
– Luca Martinoia
Jan 14 at 18:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It can't be true for complex $z$, because the right side is analytic while the left side is not. So let's assume $z > 0$ (by symmetry, the result for $z$ and $-z$ will be the same). We have, after careful attention to branches,
$$ eqalign{frac{1}{pi} int_0^pi log left(e^z + e^{-z} - e^{it} - e^{-it}right); dt
&= frac{1}{pi} int_0^pi logleft(e^z left(1 - e^{it-z}right) left(1 - e^{-it-z}right)right); dt cr
&= frac{1}{pi} int_0^pi left(z + log left(1 - e^{it-z}right) + log left(1 - e^{-it - z}right)right) ; dtcr
&= z + frac{2}{pi} text{Re} int_0^pi log(1 - e^{it-z}); dtcr
&= z + frac{2}{pi} text{Im} (text{dilog}(1-e^{-z}) - text{dilog}(1+e^{-z}))cr & = z}$$
since the dilog function is real on $(0,2)$.
answered Jan 14 at 19:49
Robert IsraelRobert Israel
329k23217470
$endgroup$
$begingroup$
Could you comment on the explicit expression for the integral in the case of complex $z$ and on its dependence on the choice of the $log $ branch?
$endgroup$
– user
Jan 14 at 20:46
$begingroup$
Yes, sorry, forgot to say that $z$ is a real number. Thanks for the help.
$endgroup$
– Luca Martinoia
Jan 14 at 20:55
$begingroup$
Given a particular branch of $log$ whose branch cut stays away from the positive reals, the integrand $log(2cosh(z)-2cos(t))$ is analytic for $z$ in some neighbourhood $U$ of the positive reals, and thus the integral will be analytic there. Since the result is $z$ for positive real $z$, it will still be $z$ for $z in U$. But of course this can't be true in $-U$: by symmetry the result is $-z$ there. $U$ and $-U$ are separated by some set in which $2cosh(z)-2cos(t)$ intersects the branch cut.
$endgroup$
– Robert Israel
Jan 14 at 21:52
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It can't be true for complex $z$, because the right side is analytic while the left side is not. So let's assume $z > 0$ (by symmetry, the result for $z$ and $-z$ will be the same). We have, after careful attention to branches,
$$ eqalign{frac{1}{pi} int_0^pi log left(e^z + e^{-z} - e^{it} - e^{-it}right); dt
&= frac{1}{pi} int_0^pi logleft(e^z left(1 - e^{it-z}right) left(1 - e^{-it-z}right)right); dt cr
&= frac{1}{pi} int_0^pi left(z + log left(1 - e^{it-z}right) + log left(1 - e^{-it - z}right)right) ; dtcr
&= z + frac{2}{pi} text{Re} int_0^pi log(1 - e^{it-z}); dtcr
&= z + frac{2}{pi} text{Im} (text{dilog}(1-e^{-z}) - text{dilog}(1+e^{-z}))cr & = z}$$
since the dilog function is real on $(0,2)$.
answered Jan 14 at 19:49
Robert IsraelRobert Israel
329k23217470
$endgroup$
$begingroup$
Could you comment on the explicit expression for the integral in the case of complex $z$ and on its dependence on the choice of the $log $ branch?
$endgroup$
– user
Jan 14 at 20:46
$begingroup$
Yes, sorry, forgot to say that $z$ is a real number. Thanks for the help.
$endgroup$
– Luca Martinoia
Jan 14 at 20:55
$begingroup$
Given a particular branch of $log$ whose branch cut stays away from the positive reals, the integrand $log(2cosh(z)-2cos(t))$ is analytic for $z$ in some neighbourhood $U$ of the positive reals, and thus the integral will be analytic there. Since the result is $z$ for positive real $z$, it will still be $z$ for $z in U$. But of course this can't be true in $-U$: by symmetry the result is $-z$ there. $U$ and $-U$ are separated by some set in which $2cosh(z)-2cos(t)$ intersects the branch cut.
$endgroup$
– Robert Israel
Jan 14 at 21:52
add a comment |
$begingroup$
It can't be true for complex $z$, because the right side is analytic while the left side is not. So let's assume $z > 0$ (by symmetry, the result for $z$ and $-z$ will be the same). We have, after careful attention to branches,
$$ eqalign{frac{1}{pi} int_0^pi log left(e^z + e^{-z} - e^{it} - e^{-it}right); dt
&= frac{1}{pi} int_0^pi logleft(e^z left(1 - e^{it-z}right) left(1 - e^{-it-z}right)right); dt cr
&= frac{1}{pi} int_0^pi left(z + log left(1 - e^{it-z}right) + log left(1 - e^{-it - z}right)right) ; dtcr
&= z + frac{2}{pi} text{Re} int_0^pi log(1 - e^{it-z}); dtcr
&= z + frac{2}{pi} text{Im} (text{dilog}(1-e^{-z}) - text{dilog}(1+e^{-z}))cr & = z}$$
since the dilog function is real on $(0,2)$.
answered Jan 14 at 19:49
Robert IsraelRobert Israel
329k23217470
$endgroup$
$begingroup$
Could you comment on the explicit expression for the integral in the case of complex $z$ and on its dependence on the choice of the $log $ branch?
$endgroup$
– user
Jan 14 at 20:46
$begingroup$
Yes, sorry, forgot to say that $z$ is a real number. Thanks for the help.
$endgroup$
– Luca Martinoia
Jan 14 at 20:55
$begingroup$
Given a particular branch of $log$ whose branch cut stays away from the positive reals, the integrand $log(2cosh(z)-2cos(t))$ is analytic for $z$ in some neighbourhood $U$ of the positive reals, and thus the integral will be analytic there. Since the result is $z$ for positive real $z$, it will still be $z$ for $z in U$. But of course this can't be true in $-U$: by symmetry the result is $-z$ there. $U$ and $-U$ are separated by some set in which $2cosh(z)-2cos(t)$ intersects the branch cut.
$endgroup$
– Robert Israel
Jan 14 at 21:52
add a comment |
$begingroup$
It can't be true for complex $z$, because the right side is analytic while the left side is not. So let's assume $z > 0$ (by symmetry, the result for $z$ and $-z$ will be the same). We have, after careful attention to branches,
$$ eqalign{frac{1}{pi} int_0^pi log left(e^z + e^{-z} - e^{it} - e^{-it}right); dt
&= frac{1}{pi} int_0^pi logleft(e^z left(1 - e^{it-z}right) left(1 - e^{-it-z}right)right); dt cr
&= frac{1}{pi} int_0^pi left(z + log left(1 - e^{it-z}right) + log left(1 - e^{-it - z}right)right) ; dtcr
&= z + frac{2}{pi} text{Re} int_0^pi log(1 - e^{it-z}); dtcr
&= z + frac{2}{pi} text{Im} (text{dilog}(1-e^{-z}) - text{dilog}(1+e^{-z}))cr & = z}$$
since the dilog function is real on $(0,2)$.
answered Jan 14 at 19:49
Robert IsraelRobert Israel
329k23217470
$endgroup$
It can't be true for complex $z$, because the right side is analytic while the left side is not. So let's assume $z > 0$ (by symmetry, the result for $z$ and $-z$ will be the same). We have, after careful attention to branches,
$$ eqalign{frac{1}{pi} int_0^pi log left(e^z + e^{-z} - e^{it} - e^{-it}right); dt
&= frac{1}{pi} int_0^pi logleft(e^z left(1 - e^{it-z}right) left(1 - e^{-it-z}right)right); dt cr
&= frac{1}{pi} int_0^pi left(z + log left(1 - e^{it-z}right) + log left(1 - e^{-it - z}right)right) ; dtcr
&= z + frac{2}{pi} text{Re} int_0^pi log(1 - e^{it-z}); dtcr
&= z + frac{2}{pi} text{Im} (text{dilog}(1-e^{-z}) - text{dilog}(1+e^{-z}))cr & = z}$$
since the dilog function is real on $(0,2)$.
answered Jan 14 at 19:49
Robert IsraelRobert Israel
329k23217470
answered Jan 14 at 19:49
Robert IsraelRobert Israel
329k23217470
answered Jan 14 at 19:49
Robert IsraelRobert Israel
329k23217470
answered Jan 14 at 19:49
Robert IsraelRobert Israel
329k23217470
329k23217470
$begingroup$
Could you comment on the explicit expression for the integral in the case of complex $z$ and on its dependence on the choice of the $log $ branch?
$endgroup$
– user
Jan 14 at 20:46
$begingroup$
Yes, sorry, forgot to say that $z$ is a real number. Thanks for the help.
$endgroup$
– Luca Martinoia
Jan 14 at 20:55
$begingroup$
Given a particular branch of $log$ whose branch cut stays away from the positive reals, the integrand $log(2cosh(z)-2cos(t))$ is analytic for $z$ in some neighbourhood $U$ of the positive reals, and thus the integral will be analytic there. Since the result is $z$ for positive real $z$, it will still be $z$ for $z in U$. But of course this can't be true in $-U$: by symmetry the result is $-z$ there. $U$ and $-U$ are separated by some set in which $2cosh(z)-2cos(t)$ intersects the branch cut.
$endgroup$
– Robert Israel
Jan 14 at 21:52
add a comment |
$begingroup$
Could you comment on the explicit expression for the integral in the case of complex $z$ and on its dependence on the choice of the $log $ branch?
$endgroup$
– user
Jan 14 at 20:46
$begingroup$
Yes, sorry, forgot to say that $z$ is a real number. Thanks for the help.
$endgroup$
– Luca Martinoia
Jan 14 at 20:55
$begingroup$
Given a particular branch of $log$ whose branch cut stays away from the positive reals, the integrand $log(2cosh(z)-2cos(t))$ is analytic for $z$ in some neighbourhood $U$ of the positive reals, and thus the integral will be analytic there. Since the result is $z$ for positive real $z$, it will still be $z$ for $z in U$. But of course this can't be true in $-U$: by symmetry the result is $-z$ there. $U$ and $-U$ are separated by some set in which $2cosh(z)-2cos(t)$ intersects the branch cut.
$endgroup$
– Robert Israel
Jan 14 at 21:52
$begingroup$
Could you comment on the explicit expression for the integral in the case of complex $z$ and on its dependence on the choice of the $log $ branch?
$endgroup$
– user
Jan 14 at 20:46
$begingroup$
Could you comment on the explicit expression for the integral in the case of complex $z$ and on its dependence on the choice of the $log $ branch?
$endgroup$
– user
Jan 14 at 20:46
$begingroup$
Yes, sorry, forgot to say that $z$ is a real number. Thanks for the help.
$endgroup$
– Luca Martinoia
Jan 14 at 20:55
$begingroup$
Yes, sorry, forgot to say that $z$ is a real number. Thanks for the help.
$endgroup$
– Luca Martinoia
Jan 14 at 20:55
$begingroup$
Given a particular branch of $log$ whose branch cut stays away from the positive reals, the integrand $log(2cosh(z)-2cos(t))$ is analytic for $z$ in some neighbourhood $U$ of the positive reals, and thus the integral will be analytic there. Since the result is $z$ for positive real $z$, it will still be $z$ for $z in U$. But of course this can't be true in $-U$: by symmetry the result is $-z$ there. $U$ and $-U$ are separated by some set in which $2cosh(z)-2cos(t)$ intersects the branch cut.
$endgroup$
– Robert Israel
Jan 14 at 21:52
$begingroup$
Given a particular branch of $log$ whose branch cut stays away from the positive reals, the integrand $log(2cosh(z)-2cos(t))$ is analytic for $z$ in some neighbourhood $U$ of the positive reals, and thus the integral will be analytic there. Since the result is $z$ for positive real $z$, it will still be $z$ for $z in U$. But of course this can't be true in $-U$: by symmetry the result is $-z$ there. $U$ and $-U$ are separated by some set in which $2cosh(z)-2cos(t)$ intersects the branch cut.
$endgroup$
– Robert Israel
Jan 14 at 21:52
add a comment |
1
$begingroup$
I think this appeared here on MSE before, but it's quite hard to find it. Anyway something similar on AoPS: artofproblemsolving.com/community/c7h17769p120601
$endgroup$
– Zacky
Jan 14 at 18:09
$begingroup$
I tried to look for it, but I'm kinda new to the side and it was hard to find an integral like this. Sadly AoPS seems down now, but thanks for the help.
$endgroup$
– Luca Martinoia
Jan 14 at 18:15