Growth of Digamma function
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For $1le sigma le 2$ and $tge 2$, $s=sigma+it$ prove that $displaystyle frac{Gamma'(s)}{Gamma(s)}=O(log t)$.
From Stirling's formula we have, $displaystyle Gamma(s)approx sqrt{2pi}exp{slog s-s-frac 12 log s}$.
Then, $displaystyle frac{Gamma'(s)}{Gamma(s)}approxlog s-frac{1}{2s}$. From here I'm unable to estimate !! Any hint. ?
Where can I get rigorous proof ?
Edit: Wikipedia links below the question are NOT clear enough to me.
complex-analysis analytic-number-theory gamma-function riemann-zeta digamma-function
asked Nov 8 at 15:52
Topo
273214
This question has an open bounty worth +50
reputation from Topo ending in 2 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
Wikipedia links given in comments below the question are not clear enough to me. I want a detail proof or some reference which includes the rigorous proof.
|
show 4 more comments
up vote
3
down vote
favorite
For $1le sigma le 2$ and $tge 2$, $s=sigma+it$ prove that $displaystyle frac{Gamma'(s)}{Gamma(s)}=O(log t)$.
From Stirling's formula we have, $displaystyle Gamma(s)approx sqrt{2pi}exp{slog s-s-frac 12 log s}$.
Then, $displaystyle frac{Gamma'(s)}{Gamma(s)}approxlog s-frac{1}{2s}$. From here I'm unable to estimate !! Any hint. ?
Where can I get rigorous proof ?
Edit: Wikipedia links below the question are NOT clear enough to me.
complex-analysis analytic-number-theory gamma-function riemann-zeta digamma-function
asked Nov 8 at 15:52
Topo
273214
This question has an open bounty worth +50
reputation from Topo ending in 2 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
Wikipedia links given in comments below the question are not clear enough to me. I want a detail proof or some reference which includes the rigorous proof.
See Digamma function
– lhf
Nov 8 at 15:58
3
Stirling's approximation is literally an approximation, which you cannot differentiate to derive asymptotic formula for $Gamma'(s)/Gamma(s)$. (For instance, differentiating both sides of $sin(e^x)=mathcal{O}(1)$ leads to nothing.) In this case, the series representation of the digamma function would help.
– Sangchul Lee
Nov 8 at 15:59
@SangchulLee But in my book it is written that "from Stirling's approximation we can get...." the above expression..!!
– Topo
Nov 8 at 16:02
As pointed out above differentiating asymptotics is problematic in general (as small oscillations affect the derivative, but not as much the function value), but it usually works for cases where the function is monotonic like here (though making it rigorous is probably not straight forward)
– Winther
Nov 8 at 16:04
@Winther For your 1st comment, here $s$ is not real..
– Topo
Nov 8 at 16:18
|
show 4 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
For $1le sigma le 2$ and $tge 2$, $s=sigma+it$ prove that $displaystyle frac{Gamma'(s)}{Gamma(s)}=O(log t)$.
From Stirling's formula we have, $displaystyle Gamma(s)approx sqrt{2pi}exp{slog s-s-frac 12 log s}$.
Then, $displaystyle frac{Gamma'(s)}{Gamma(s)}approxlog s-frac{1}{2s}$. From here I'm unable to estimate !! Any hint. ?
Where can I get rigorous proof ?
Edit: Wikipedia links below the question are NOT clear enough to me.
complex-analysis analytic-number-theory gamma-function riemann-zeta digamma-function
asked Nov 8 at 15:52
Topo
273214
For $1le sigma le 2$ and $tge 2$, $s=sigma+it$ prove that $displaystyle frac{Gamma'(s)}{Gamma(s)}=O(log t)$.
From Stirling's formula we have, $displaystyle Gamma(s)approx sqrt{2pi}exp{slog s-s-frac 12 log s}$.
Then, $displaystyle frac{Gamma'(s)}{Gamma(s)}approxlog s-frac{1}{2s}$. From here I'm unable to estimate !! Any hint. ?
Where can I get rigorous proof ?
Edit: Wikipedia links below the question are NOT clear enough to me.
complex-analysis analytic-number-theory gamma-function riemann-zeta digamma-function
complex-analysis analytic-number-theory gamma-function riemann-zeta digamma-function
asked Nov 8 at 15:52
Topo
273214
asked Nov 8 at 15:52
Topo
273214
edited Nov 14 at 6:21
asked Nov 8 at 15:52
Topo
273214
asked Nov 8 at 15:52
Topo
273214
asked Nov 8 at 15:52
Topo
273214
273214
This question has an open bounty worth +50
reputation from Topo ending in 2 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
Wikipedia links given in comments below the question are not clear enough to me. I want a detail proof or some reference which includes the rigorous proof.
This question has an open bounty worth +50
reputation from Topo ending in 2 days.
The question is widely applicable to a large audience. A detailed canonical answer is required to address all the concerns.
Wikipedia links given in comments below the question are not clear enough to me. I want a detail proof or some reference which includes the rigorous proof.
See Digamma function
– lhf
Nov 8 at 15:58
3
Stirling's approximation is literally an approximation, which you cannot differentiate to derive asymptotic formula for $Gamma'(s)/Gamma(s)$. (For instance, differentiating both sides of $sin(e^x)=mathcal{O}(1)$ leads to nothing.) In this case, the series representation of the digamma function would help.
– Sangchul Lee
Nov 8 at 15:59
@SangchulLee But in my book it is written that "from Stirling's approximation we can get...." the above expression..!!
– Topo
Nov 8 at 16:02
As pointed out above differentiating asymptotics is problematic in general (as small oscillations affect the derivative, but not as much the function value), but it usually works for cases where the function is monotonic like here (though making it rigorous is probably not straight forward)
– Winther
Nov 8 at 16:04
@Winther For your 1st comment, here $s$ is not real..
– Topo
Nov 8 at 16:18
|
show 4 more comments
See Digamma function
– lhf
Nov 8 at 15:58
3
Stirling's approximation is literally an approximation, which you cannot differentiate to derive asymptotic formula for $Gamma'(s)/Gamma(s)$. (For instance, differentiating both sides of $sin(e^x)=mathcal{O}(1)$ leads to nothing.) In this case, the series representation of the digamma function would help.
– Sangchul Lee
Nov 8 at 15:59
@SangchulLee But in my book it is written that "from Stirling's approximation we can get...." the above expression..!!
– Topo
Nov 8 at 16:02
As pointed out above differentiating asymptotics is problematic in general (as small oscillations affect the derivative, but not as much the function value), but it usually works for cases where the function is monotonic like here (though making it rigorous is probably not straight forward)
– Winther
Nov 8 at 16:04
@Winther For your 1st comment, here $s$ is not real..
– Topo
Nov 8 at 16:18
See Digamma function
– lhf
Nov 8 at 15:58
See Digamma function
– lhf
Nov 8 at 15:58
3
3
Stirling's approximation is literally an approximation, which you cannot differentiate to derive asymptotic formula for $Gamma'(s)/Gamma(s)$. (For instance, differentiating both sides of $sin(e^x)=mathcal{O}(1)$ leads to nothing.) In this case, the series representation of the digamma function would help.
– Sangchul Lee
Nov 8 at 15:59
Stirling's approximation is literally an approximation, which you cannot differentiate to derive asymptotic formula for $Gamma'(s)/Gamma(s)$. (For instance, differentiating both sides of $sin(e^x)=mathcal{O}(1)$ leads to nothing.) In this case, the series representation of the digamma function would help.
– Sangchul Lee
Nov 8 at 15:59
@SangchulLee But in my book it is written that "from Stirling's approximation we can get...." the above expression..!!
– Topo
Nov 8 at 16:02
@SangchulLee But in my book it is written that "from Stirling's approximation we can get...." the above expression..!!
– Topo
Nov 8 at 16:02
As pointed out above differentiating asymptotics is problematic in general (as small oscillations affect the derivative, but not as much the function value), but it usually works for cases where the function is monotonic like here (though making it rigorous is probably not straight forward)
– Winther
Nov 8 at 16:04
As pointed out above differentiating asymptotics is problematic in general (as small oscillations affect the derivative, but not as much the function value), but it usually works for cases where the function is monotonic like here (though making it rigorous is probably not straight forward)
– Winther
Nov 8 at 16:04
@Winther For your 1st comment, here $s$ is not real..
– Topo
Nov 8 at 16:18
@Winther For your 1st comment, here $s$ is not real..
– Topo
Nov 8 at 16:18
|
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
1
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If you need to differentiate a holomorphic function, try to integrate instead.
Let $g(s)=logsqrt{2pi}+slog s-s-tfrac12log s$; so $logGamma(s)=g(s)+o(1)$.
Take a circle around $s$ with radius $r=frac12$. By Cauchy's formulas,
$$
frac{Gamma'}{Gamma}(s) = g'(s) + big(logGamma(s)-g(s)big)' =
log s-frac1{2s} + frac1{2pi i}oint_{|w-s|=r} frac{logGamma(w)-g(w)}{(w-s)^2} mathrm dw;
$$
$$
bigg|frac{Gamma'}{Gamma}(s) - log s+frac1{2s} bigg| le
frac1{2pi}oint_{|w-s|=r} frac{big|logGamma(w)-g(w)big|}{r^2} |mathrm dw|
le O(1)
$$
so,
$$
frac{Gamma'}{Gamma}(s) = log s + O(1).
$$
answered Nov 12 at 16:40
user141614
12k925
How, $log Gamma (s)=g(s)+O(1)$ ? It will be $g(s)+O(1/s)$
– Topo
Nov 12 at 17:17
1
$|s|gesigma>1$
– user141614
Nov 12 at 17:35
How $log s+O(1)=O(log t)$ ?
– Topo
Nov 13 at 16:54
You should clarify your own question. Is $sigma$ fixed or it varies? What is the range of $t$?
– user141614
Nov 13 at 18:27
$1le sigma le 2$ and $tge 2$.
– Topo
Nov 14 at 6:14
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If you need to differentiate a holomorphic function, try to integrate instead.
Let $g(s)=logsqrt{2pi}+slog s-s-tfrac12log s$; so $logGamma(s)=g(s)+o(1)$.
Take a circle around $s$ with radius $r=frac12$. By Cauchy's formulas,
$$
frac{Gamma'}{Gamma}(s) = g'(s) + big(logGamma(s)-g(s)big)' =
log s-frac1{2s} + frac1{2pi i}oint_{|w-s|=r} frac{logGamma(w)-g(w)}{(w-s)^2} mathrm dw;
$$
$$
bigg|frac{Gamma'}{Gamma}(s) - log s+frac1{2s} bigg| le
frac1{2pi}oint_{|w-s|=r} frac{big|logGamma(w)-g(w)big|}{r^2} |mathrm dw|
le O(1)
$$
so,
$$
frac{Gamma'}{Gamma}(s) = log s + O(1).
$$
answered Nov 12 at 16:40
user141614
12k925
How, $log Gamma (s)=g(s)+O(1)$ ? It will be $g(s)+O(1/s)$
– Topo
Nov 12 at 17:17
1
$|s|gesigma>1$
– user141614
Nov 12 at 17:35
How $log s+O(1)=O(log t)$ ?
– Topo
Nov 13 at 16:54
You should clarify your own question. Is $sigma$ fixed or it varies? What is the range of $t$?
– user141614
Nov 13 at 18:27
$1le sigma le 2$ and $tge 2$.
– Topo
Nov 14 at 6:14
|
show 1 more comment
up vote
1
down vote
If you need to differentiate a holomorphic function, try to integrate instead.
Let $g(s)=logsqrt{2pi}+slog s-s-tfrac12log s$; so $logGamma(s)=g(s)+o(1)$.
Take a circle around $s$ with radius $r=frac12$. By Cauchy's formulas,
$$
frac{Gamma'}{Gamma}(s) = g'(s) + big(logGamma(s)-g(s)big)' =
log s-frac1{2s} + frac1{2pi i}oint_{|w-s|=r} frac{logGamma(w)-g(w)}{(w-s)^2} mathrm dw;
$$
$$
bigg|frac{Gamma'}{Gamma}(s) - log s+frac1{2s} bigg| le
frac1{2pi}oint_{|w-s|=r} frac{big|logGamma(w)-g(w)big|}{r^2} |mathrm dw|
le O(1)
$$
so,
$$
frac{Gamma'}{Gamma}(s) = log s + O(1).
$$
answered Nov 12 at 16:40
user141614
12k925
How, $log Gamma (s)=g(s)+O(1)$ ? It will be $g(s)+O(1/s)$
– Topo
Nov 12 at 17:17
1
$|s|gesigma>1$
– user141614
Nov 12 at 17:35
How $log s+O(1)=O(log t)$ ?
– Topo
Nov 13 at 16:54
You should clarify your own question. Is $sigma$ fixed or it varies? What is the range of $t$?
– user141614
Nov 13 at 18:27
$1le sigma le 2$ and $tge 2$.
– Topo
Nov 14 at 6:14
|
show 1 more comment
up vote
1
down vote
up vote
1
down vote
If you need to differentiate a holomorphic function, try to integrate instead.
Let $g(s)=logsqrt{2pi}+slog s-s-tfrac12log s$; so $logGamma(s)=g(s)+o(1)$.
Take a circle around $s$ with radius $r=frac12$. By Cauchy's formulas,
$$
frac{Gamma'}{Gamma}(s) = g'(s) + big(logGamma(s)-g(s)big)' =
log s-frac1{2s} + frac1{2pi i}oint_{|w-s|=r} frac{logGamma(w)-g(w)}{(w-s)^2} mathrm dw;
$$
$$
bigg|frac{Gamma'}{Gamma}(s) - log s+frac1{2s} bigg| le
frac1{2pi}oint_{|w-s|=r} frac{big|logGamma(w)-g(w)big|}{r^2} |mathrm dw|
le O(1)
$$
so,
$$
frac{Gamma'}{Gamma}(s) = log s + O(1).
$$
answered Nov 12 at 16:40
user141614
12k925
If you need to differentiate a holomorphic function, try to integrate instead.
Let $g(s)=logsqrt{2pi}+slog s-s-tfrac12log s$; so $logGamma(s)=g(s)+o(1)$.
Take a circle around $s$ with radius $r=frac12$. By Cauchy's formulas,
$$
frac{Gamma'}{Gamma}(s) = g'(s) + big(logGamma(s)-g(s)big)' =
log s-frac1{2s} + frac1{2pi i}oint_{|w-s|=r} frac{logGamma(w)-g(w)}{(w-s)^2} mathrm dw;
$$
$$
bigg|frac{Gamma'}{Gamma}(s) - log s+frac1{2s} bigg| le
frac1{2pi}oint_{|w-s|=r} frac{big|logGamma(w)-g(w)big|}{r^2} |mathrm dw|
le O(1)
$$
so,
$$
frac{Gamma'}{Gamma}(s) = log s + O(1).
$$
answered Nov 12 at 16:40
user141614
12k925
answered Nov 12 at 16:40
user141614
12k925
answered Nov 12 at 16:40
user141614
12k925
answered Nov 12 at 16:40
user141614
12k925
12k925
How, $log Gamma (s)=g(s)+O(1)$ ? It will be $g(s)+O(1/s)$
– Topo
Nov 12 at 17:17
1
$|s|gesigma>1$
– user141614
Nov 12 at 17:35
How $log s+O(1)=O(log t)$ ?
– Topo
Nov 13 at 16:54
You should clarify your own question. Is $sigma$ fixed or it varies? What is the range of $t$?
– user141614
Nov 13 at 18:27
$1le sigma le 2$ and $tge 2$.
– Topo
Nov 14 at 6:14
|
show 1 more comment
How, $log Gamma (s)=g(s)+O(1)$ ? It will be $g(s)+O(1/s)$
– Topo
Nov 12 at 17:17
1
$|s|gesigma>1$
– user141614
Nov 12 at 17:35
How $log s+O(1)=O(log t)$ ?
– Topo
Nov 13 at 16:54
You should clarify your own question. Is $sigma$ fixed or it varies? What is the range of $t$?
– user141614
Nov 13 at 18:27
$1le sigma le 2$ and $tge 2$.
– Topo
Nov 14 at 6:14
How, $log Gamma (s)=g(s)+O(1)$ ? It will be $g(s)+O(1/s)$
– Topo
Nov 12 at 17:17
How, $log Gamma (s)=g(s)+O(1)$ ? It will be $g(s)+O(1/s)$
– Topo
Nov 12 at 17:17
1
1
$|s|gesigma>1$
– user141614
Nov 12 at 17:35
$|s|gesigma>1$
– user141614
Nov 12 at 17:35
How $log s+O(1)=O(log t)$ ?
– Topo
Nov 13 at 16:54
How $log s+O(1)=O(log t)$ ?
– Topo
Nov 13 at 16:54
You should clarify your own question. Is $sigma$ fixed or it varies? What is the range of $t$?
– user141614
Nov 13 at 18:27
You should clarify your own question. Is $sigma$ fixed or it varies? What is the range of $t$?
– user141614
Nov 13 at 18:27
$1le sigma le 2$ and $tge 2$.
– Topo
Nov 14 at 6:14
$1le sigma le 2$ and $tge 2$.
– Topo
Nov 14 at 6:14
|
show 1 more comment
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See Digamma function
– lhf
Nov 8 at 15:58
3
Stirling's approximation is literally an approximation, which you cannot differentiate to derive asymptotic formula for $Gamma'(s)/Gamma(s)$. (For instance, differentiating both sides of $sin(e^x)=mathcal{O}(1)$ leads to nothing.) In this case, the series representation of the digamma function would help.
– Sangchul Lee
Nov 8 at 15:59
@SangchulLee But in my book it is written that "from Stirling's approximation we can get...." the above expression..!!
– Topo
Nov 8 at 16:02
As pointed out above differentiating asymptotics is problematic in general (as small oscillations affect the derivative, but not as much the function value), but it usually works for cases where the function is monotonic like here (though making it rigorous is probably not straight forward)
– Winther
Nov 8 at 16:04
@Winther For your 1st comment, here $s$ is not real..
– Topo
Nov 8 at 16:18