Mixing
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Riemann zeta-function regularization in string theory
$begingroup$
First of all let me say that I am a physicist and therefore it is sometimes hard for me to understand some mathematical steps... Now, I've been trying to obtain the well known result for the zeta function $zeta(-1)=-frac{1}{12}$, but my string theory teacher is asking the students to show this in a very specific way. To do so, we consider the gamma function:
$$Gamma(s)=int_0^infty dt t^{s-1}e^{-t}$$
and the zeta function:
$$zeta(s)=sum_{n=1}^{infty}n^{-s}$$
with $s$ a complex number with real part greater than 1. The firststep of the demonstration is to show that using a transformation $t rightarrow nt$ we can write
$$Gamma(s)zeta(s)=int_0^infty dt frac{t^{s-1}}{e^t-1}$$.
I had no problems here, I just wrote the definition of a geometric series of argument $e^-t$ to obtain the result. Then, the second step consisted on showing that we can also write
$$Gamma(s)zeta(s)=int_0^1 dtleft(frac{1}{e^t-1}-frac{1}{t}+frac{1}{2}-frac{t}{12}right)+int_1^infty dt left(frac{t^{s-1}}{e^t-1}right)+frac{1}{s-1}-frac{1}{2s}+frac{1}{12(s+1)}.$$
I also did not have any problems here because the integrals of the parcels inside the first brackets cancel the terms depending on $s$ outside the integrals and so it was just a matter of verifying that this result was the same as before.
Now, this is where I am stuck: we started by saying that $s$ is a complex number with real part greater than 1. Now, we are supposed to explain why this last result i wrote is in fact well-behaved for any complex $s$ with real part greater than $-2$, and I don't understand how is that possible since we have terms that diverge for $s={-1,0,1}$.
Finally, the last step of the demonstration consists of using the previous result, that is well-behaved for $s$ with real part greater than $-2$, and also the fact that the gamma function as pole at $s=-1$ with residue $-1$ to conclude that $zeta(-1)=-frac{1}{12}$. I am also not able to conclude anything because I really don't know how. A friend of mine said he did something using the Bernoulli numbers, but I am not sure about what he did... Can any one help me here?
gamma-function riemann-zeta regularization analytic-continuation string-theory
asked Mar 21 '17 at 16:30
Joao LuisJoao Luis
312
$endgroup$
add a comment |
$begingroup$
First of all let me say that I am a physicist and therefore it is sometimes hard for me to understand some mathematical steps... Now, I've been trying to obtain the well known result for the zeta function $zeta(-1)=-frac{1}{12}$, but my string theory teacher is asking the students to show this in a very specific way. To do so, we consider the gamma function:
$$Gamma(s)=int_0^infty dt t^{s-1}e^{-t}$$
and the zeta function:
$$zeta(s)=sum_{n=1}^{infty}n^{-s}$$
with $s$ a complex number with real part greater than 1. The firststep of the demonstration is to show that using a transformation $t rightarrow nt$ we can write
$$Gamma(s)zeta(s)=int_0^infty dt frac{t^{s-1}}{e^t-1}$$.
I had no problems here, I just wrote the definition of a geometric series of argument $e^-t$ to obtain the result. Then, the second step consisted on showing that we can also write
$$Gamma(s)zeta(s)=int_0^1 dtleft(frac{1}{e^t-1}-frac{1}{t}+frac{1}{2}-frac{t}{12}right)+int_1^infty dt left(frac{t^{s-1}}{e^t-1}right)+frac{1}{s-1}-frac{1}{2s}+frac{1}{12(s+1)}.$$
I also did not have any problems here because the integrals of the parcels inside the first brackets cancel the terms depending on $s$ outside the integrals and so it was just a matter of verifying that this result was the same as before.
Now, this is where I am stuck: we started by saying that $s$ is a complex number with real part greater than 1. Now, we are supposed to explain why this last result i wrote is in fact well-behaved for any complex $s$ with real part greater than $-2$, and I don't understand how is that possible since we have terms that diverge for $s={-1,0,1}$.
Finally, the last step of the demonstration consists of using the previous result, that is well-behaved for $s$ with real part greater than $-2$, and also the fact that the gamma function as pole at $s=-1$ with residue $-1$ to conclude that $zeta(-1)=-frac{1}{12}$. I am also not able to conclude anything because I really don't know how. A friend of mine said he did something using the Bernoulli numbers, but I am not sure about what he did... Can any one help me here?
gamma-function riemann-zeta regularization analytic-continuation string-theory
asked Mar 21 '17 at 16:30
Joao LuisJoao Luis
312
$endgroup$
1
$begingroup$
I guess you meant $zeta(-1)=-frac{1}{12}$, instead of $zeta(-1)=0$ in your first sentence?
$endgroup$
– Agno
Mar 21 '17 at 17:01
$begingroup$
yes you are right, im going to edit that.
$endgroup$
– Joao Luis
Mar 21 '17 at 18:30
$begingroup$
Good thing you didn't write $1+2+3+cdots=-frac{1}{12}$ in your title.
$endgroup$
– arctic tern
Mar 21 '17 at 19:19
1
$begingroup$
One of the easiest explanations I found on how the analytic continuation of $zeta(s)$ can be done (without Bernouill-numbers) : youtube.com/watch?v=K6L4Ez4ZVZc. This yields the famous functional equation that relates $zeta(s)$ to $zeta(1-s)$ (hence also $zeta(2)$ to $zeta(-1)$).
$endgroup$
– Agno
Mar 21 '17 at 23:11
add a comment |
$begingroup$
First of all let me say that I am a physicist and therefore it is sometimes hard for me to understand some mathematical steps... Now, I've been trying to obtain the well known result for the zeta function $zeta(-1)=-frac{1}{12}$, but my string theory teacher is asking the students to show this in a very specific way. To do so, we consider the gamma function:
$$Gamma(s)=int_0^infty dt t^{s-1}e^{-t}$$
and the zeta function:
$$zeta(s)=sum_{n=1}^{infty}n^{-s}$$
with $s$ a complex number with real part greater than 1. The firststep of the demonstration is to show that using a transformation $t rightarrow nt$ we can write
$$Gamma(s)zeta(s)=int_0^infty dt frac{t^{s-1}}{e^t-1}$$.
I had no problems here, I just wrote the definition of a geometric series of argument $e^-t$ to obtain the result. Then, the second step consisted on showing that we can also write
$$Gamma(s)zeta(s)=int_0^1 dtleft(frac{1}{e^t-1}-frac{1}{t}+frac{1}{2}-frac{t}{12}right)+int_1^infty dt left(frac{t^{s-1}}{e^t-1}right)+frac{1}{s-1}-frac{1}{2s}+frac{1}{12(s+1)}.$$
I also did not have any problems here because the integrals of the parcels inside the first brackets cancel the terms depending on $s$ outside the integrals and so it was just a matter of verifying that this result was the same as before.
Now, this is where I am stuck: we started by saying that $s$ is a complex number with real part greater than 1. Now, we are supposed to explain why this last result i wrote is in fact well-behaved for any complex $s$ with real part greater than $-2$, and I don't understand how is that possible since we have terms that diverge for $s={-1,0,1}$.
Finally, the last step of the demonstration consists of using the previous result, that is well-behaved for $s$ with real part greater than $-2$, and also the fact that the gamma function as pole at $s=-1$ with residue $-1$ to conclude that $zeta(-1)=-frac{1}{12}$. I am also not able to conclude anything because I really don't know how. A friend of mine said he did something using the Bernoulli numbers, but I am not sure about what he did... Can any one help me here?
gamma-function riemann-zeta regularization analytic-continuation string-theory
asked Mar 21 '17 at 16:30
Joao LuisJoao Luis
312
$endgroup$
First of all let me say that I am a physicist and therefore it is sometimes hard for me to understand some mathematical steps... Now, I've been trying to obtain the well known result for the zeta function $zeta(-1)=-frac{1}{12}$, but my string theory teacher is asking the students to show this in a very specific way. To do so, we consider the gamma function:
$$Gamma(s)=int_0^infty dt t^{s-1}e^{-t}$$
and the zeta function:
$$zeta(s)=sum_{n=1}^{infty}n^{-s}$$
with $s$ a complex number with real part greater than 1. The firststep of the demonstration is to show that using a transformation $t rightarrow nt$ we can write
$$Gamma(s)zeta(s)=int_0^infty dt frac{t^{s-1}}{e^t-1}$$.
I had no problems here, I just wrote the definition of a geometric series of argument $e^-t$ to obtain the result. Then, the second step consisted on showing that we can also write
$$Gamma(s)zeta(s)=int_0^1 dtleft(frac{1}{e^t-1}-frac{1}{t}+frac{1}{2}-frac{t}{12}right)+int_1^infty dt left(frac{t^{s-1}}{e^t-1}right)+frac{1}{s-1}-frac{1}{2s}+frac{1}{12(s+1)}.$$
I also did not have any problems here because the integrals of the parcels inside the first brackets cancel the terms depending on $s$ outside the integrals and so it was just a matter of verifying that this result was the same as before.
Now, this is where I am stuck: we started by saying that $s$ is a complex number with real part greater than 1. Now, we are supposed to explain why this last result i wrote is in fact well-behaved for any complex $s$ with real part greater than $-2$, and I don't understand how is that possible since we have terms that diverge for $s={-1,0,1}$.
Finally, the last step of the demonstration consists of using the previous result, that is well-behaved for $s$ with real part greater than $-2$, and also the fact that the gamma function as pole at $s=-1$ with residue $-1$ to conclude that $zeta(-1)=-frac{1}{12}$. I am also not able to conclude anything because I really don't know how. A friend of mine said he did something using the Bernoulli numbers, but I am not sure about what he did... Can any one help me here?
gamma-function riemann-zeta regularization analytic-continuation string-theory
gamma-function riemann-zeta regularization analytic-continuation string-theory
asked Mar 21 '17 at 16:30
Joao LuisJoao Luis
312
asked Mar 21 '17 at 16:30
Joao LuisJoao Luis
312
edited Mar 21 '17 at 18:30
Joao Luis
asked Mar 21 '17 at 16:30
Joao LuisJoao Luis
312
asked Mar 21 '17 at 16:30
Joao LuisJoao Luis
312
asked Mar 21 '17 at 16:30
Joao LuisJoao Luis
312
312
1
$begingroup$
I guess you meant $zeta(-1)=-frac{1}{12}$, instead of $zeta(-1)=0$ in your first sentence?
$endgroup$
– Agno
Mar 21 '17 at 17:01
$begingroup$
yes you are right, im going to edit that.
$endgroup$
– Joao Luis
Mar 21 '17 at 18:30
$begingroup$
Good thing you didn't write $1+2+3+cdots=-frac{1}{12}$ in your title.
$endgroup$
– arctic tern
Mar 21 '17 at 19:19
1
$begingroup$
One of the easiest explanations I found on how the analytic continuation of $zeta(s)$ can be done (without Bernouill-numbers) : youtube.com/watch?v=K6L4Ez4ZVZc. This yields the famous functional equation that relates $zeta(s)$ to $zeta(1-s)$ (hence also $zeta(2)$ to $zeta(-1)$).
$endgroup$
– Agno
Mar 21 '17 at 23:11
add a comment |
1
$begingroup$
I guess you meant $zeta(-1)=-frac{1}{12}$, instead of $zeta(-1)=0$ in your first sentence?
$endgroup$
– Agno
Mar 21 '17 at 17:01
$begingroup$
yes you are right, im going to edit that.
$endgroup$
– Joao Luis
Mar 21 '17 at 18:30
$begingroup$
Good thing you didn't write $1+2+3+cdots=-frac{1}{12}$ in your title.
$endgroup$
– arctic tern
Mar 21 '17 at 19:19
1
$begingroup$
One of the easiest explanations I found on how the analytic continuation of $zeta(s)$ can be done (without Bernouill-numbers) : youtube.com/watch?v=K6L4Ez4ZVZc. This yields the famous functional equation that relates $zeta(s)$ to $zeta(1-s)$ (hence also $zeta(2)$ to $zeta(-1)$).
$endgroup$
– Agno
Mar 21 '17 at 23:11
1
1
$begingroup$
I guess you meant $zeta(-1)=-frac{1}{12}$, instead of $zeta(-1)=0$ in your first sentence?
$endgroup$
– Agno
Mar 21 '17 at 17:01
$begingroup$
I guess you meant $zeta(-1)=-frac{1}{12}$, instead of $zeta(-1)=0$ in your first sentence?
$endgroup$
– Agno
Mar 21 '17 at 17:01
$begingroup$
yes you are right, im going to edit that.
$endgroup$
– Joao Luis
Mar 21 '17 at 18:30
$begingroup$
yes you are right, im going to edit that.
$endgroup$
– Joao Luis
Mar 21 '17 at 18:30
$begingroup$
Good thing you didn't write $1+2+3+cdots=-frac{1}{12}$ in your title.
$endgroup$
– arctic tern
Mar 21 '17 at 19:19
$begingroup$
Good thing you didn't write $1+2+3+cdots=-frac{1}{12}$ in your title.
$endgroup$
– arctic tern
Mar 21 '17 at 19:19
1
1
$begingroup$
One of the easiest explanations I found on how the analytic continuation of $zeta(s)$ can be done (without Bernouill-numbers) : youtube.com/watch?v=K6L4Ez4ZVZc. This yields the famous functional equation that relates $zeta(s)$ to $zeta(1-s)$ (hence also $zeta(2)$ to $zeta(-1)$).
$endgroup$
– Agno
Mar 21 '17 at 23:11
$begingroup$
One of the easiest explanations I found on how the analytic continuation of $zeta(s)$ can be done (without Bernouill-numbers) : youtube.com/watch?v=K6L4Ez4ZVZc. This yields the famous functional equation that relates $zeta(s)$ to $zeta(1-s)$ (hence also $zeta(2)$ to $zeta(-1)$).
$endgroup$
– Agno
Mar 21 '17 at 23:11
add a comment |
1 Answer
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oldest
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$begingroup$
You write
"Now, we are supposed to explain why this last result I wrote is in fact well-behaved for any complex $s$ with real part greater than $−2$, and I don't understand how is that possible since we have terms that diverge for $sin{−1,0,1}$."
The exceptional values for $s$ are as you say; so the explanatory task you have been set is, if you have stated it faithfully, wrongly posed to say the least. It looks like mathematical monkey business to me.
That said, it is a perfectly good result that $zeta(-1)=-frac1{12}$.
answered Jan 10 at 14:21
John BentinJohn Bentin
11.3k22554
$endgroup$
add a comment |
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$begingroup$
You write
"Now, we are supposed to explain why this last result I wrote is in fact well-behaved for any complex $s$ with real part greater than $−2$, and I don't understand how is that possible since we have terms that diverge for $sin{−1,0,1}$."
The exceptional values for $s$ are as you say; so the explanatory task you have been set is, if you have stated it faithfully, wrongly posed to say the least. It looks like mathematical monkey business to me.
That said, it is a perfectly good result that $zeta(-1)=-frac1{12}$.
answered Jan 10 at 14:21
John BentinJohn Bentin
11.3k22554
$endgroup$
add a comment |
$begingroup$
You write
"Now, we are supposed to explain why this last result I wrote is in fact well-behaved for any complex $s$ with real part greater than $−2$, and I don't understand how is that possible since we have terms that diverge for $sin{−1,0,1}$."
The exceptional values for $s$ are as you say; so the explanatory task you have been set is, if you have stated it faithfully, wrongly posed to say the least. It looks like mathematical monkey business to me.
That said, it is a perfectly good result that $zeta(-1)=-frac1{12}$.
answered Jan 10 at 14:21
John BentinJohn Bentin
11.3k22554
$endgroup$
add a comment |
$begingroup$
You write
"Now, we are supposed to explain why this last result I wrote is in fact well-behaved for any complex $s$ with real part greater than $−2$, and I don't understand how is that possible since we have terms that diverge for $sin{−1,0,1}$."
The exceptional values for $s$ are as you say; so the explanatory task you have been set is, if you have stated it faithfully, wrongly posed to say the least. It looks like mathematical monkey business to me.
That said, it is a perfectly good result that $zeta(-1)=-frac1{12}$.
answered Jan 10 at 14:21
John BentinJohn Bentin
11.3k22554
$endgroup$
You write
"Now, we are supposed to explain why this last result I wrote is in fact well-behaved for any complex $s$ with real part greater than $−2$, and I don't understand how is that possible since we have terms that diverge for $sin{−1,0,1}$."
The exceptional values for $s$ are as you say; so the explanatory task you have been set is, if you have stated it faithfully, wrongly posed to say the least. It looks like mathematical monkey business to me.
That said, it is a perfectly good result that $zeta(-1)=-frac1{12}$.
answered Jan 10 at 14:21
John BentinJohn Bentin
11.3k22554
answered Jan 10 at 14:21
John BentinJohn Bentin
11.3k22554
answered Jan 10 at 14:21
John BentinJohn Bentin
11.3k22554
answered Jan 10 at 14:21
John BentinJohn Bentin
11.3k22554
11.3k22554
add a comment |
add a comment |
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$begingroup$
I guess you meant $zeta(-1)=-frac{1}{12}$, instead of $zeta(-1)=0$ in your first sentence?
$endgroup$
– Agno
Mar 21 '17 at 17:01
$begingroup$
yes you are right, im going to edit that.
$endgroup$
– Joao Luis
Mar 21 '17 at 18:30
$begingroup$
Good thing you didn't write $1+2+3+cdots=-frac{1}{12}$ in your title.
$endgroup$
– arctic tern
Mar 21 '17 at 19:19
1
$begingroup$
One of the easiest explanations I found on how the analytic continuation of $zeta(s)$ can be done (without Bernouill-numbers) : youtube.com/watch?v=K6L4Ez4ZVZc. This yields the famous functional equation that relates $zeta(s)$ to $zeta(1-s)$ (hence also $zeta(2)$ to $zeta(-1)$).
$endgroup$
– Agno
Mar 21 '17 at 23:11