Mixing
Estimating complex integral
$begingroup$
I am reading chapter 12 of "Lectures on the Riemann Zeta function" by H. Iwaniec.
I am stuck at understanding the computation done at the beginning of page 45.
We want to estimate the following integral
$$
I_1(s)=frac{1}{2pi i}int_{(-1/4)}Gamma(w)y^{-w}zeta(1-s-w) ,dw
$$
where $(-1/4)$ should denote the vertical line at $text{Re}(w)=-1/4$ and $s=1/2+it$. He says that to compute this we have to move the integration to the line $text{Re}(w)=-3/4$ and then expand $zeta(1-s-w)$ into its Dirichlet series, interchange the summation and integration to eventually get
$$
I_1(s)=-sum_n n^{s-1}e^{-n/y}
$$
Here are my doubts
- When we move the line of integration from $text{Re}(w)=-1/4$ to $text{Re}(w)=-3/4$ don't we pass through the pole of $zeta$ at $w=-s$? What about the residue?
- Once we expand $zeta$ into its Dirichlet series and interchange the summation and integral we are left with computing
$$
sum_n n^{s-1}frac{1}{2pi i}int_{(-3/4)}Gamma(w)left(frac{y}{n}right)^{-w},dw
$$
I know that
$$
e^{-y}=frac{1}{2pi i}int_{(c)}Gamma(w)left(yright)^{-w},dw
$$
for all c>0. So the difference between the two should be given by the residue of $Gamma(w)y^{-w}$ at $w=1$, that is 1. But what they claim is that we pick a minus sign.
I am surely overlooking something but I cannot figure out what. I really appreciate any advice. Thank you.
complex-analysis residue-calculus complex-integration
asked Jan 24 at 17:12
asdasd
30729
$endgroup$
add a comment |
$begingroup$
I am reading chapter 12 of "Lectures on the Riemann Zeta function" by H. Iwaniec.
I am stuck at understanding the computation done at the beginning of page 45.
We want to estimate the following integral
$$
I_1(s)=frac{1}{2pi i}int_{(-1/4)}Gamma(w)y^{-w}zeta(1-s-w) ,dw
$$
where $(-1/4)$ should denote the vertical line at $text{Re}(w)=-1/4$ and $s=1/2+it$. He says that to compute this we have to move the integration to the line $text{Re}(w)=-3/4$ and then expand $zeta(1-s-w)$ into its Dirichlet series, interchange the summation and integration to eventually get
$$
I_1(s)=-sum_n n^{s-1}e^{-n/y}
$$
Here are my doubts
- When we move the line of integration from $text{Re}(w)=-1/4$ to $text{Re}(w)=-3/4$ don't we pass through the pole of $zeta$ at $w=-s$? What about the residue?
- Once we expand $zeta$ into its Dirichlet series and interchange the summation and integral we are left with computing
$$
sum_n n^{s-1}frac{1}{2pi i}int_{(-3/4)}Gamma(w)left(frac{y}{n}right)^{-w},dw
$$
I know that
$$
e^{-y}=frac{1}{2pi i}int_{(c)}Gamma(w)left(yright)^{-w},dw
$$
for all c>0. So the difference between the two should be given by the residue of $Gamma(w)y^{-w}$ at $w=1$, that is 1. But what they claim is that we pick a minus sign.
I am surely overlooking something but I cannot figure out what. I really appreciate any advice. Thank you.
complex-analysis residue-calculus complex-integration
asked Jan 24 at 17:12
asdasd
30729
$endgroup$
$begingroup$
What is the estimate you are trying to get in the end?
$endgroup$
– 0x539
Jan 24 at 17:39
$begingroup$
That's the end. Then he just uses this result together with other estimates to prove a proposition. Basically I want to prove that $-sum_n n^{s-1}e^{-n/y}=frac{1}{2pi i}int_{(-1/4)}Gamma(w)y^{-w}zeta(1-s-w)dw$.
$endgroup$
– asd
Jan 24 at 17:42
$begingroup$
You mean my question?
$endgroup$
– asd
Jan 24 at 18:05
$begingroup$
nevermind, seems to have been a bug on my end
$endgroup$
– 0x539
Jan 24 at 18:05
add a comment |
$begingroup$
I am reading chapter 12 of "Lectures on the Riemann Zeta function" by H. Iwaniec.
I am stuck at understanding the computation done at the beginning of page 45.
We want to estimate the following integral
$$
I_1(s)=frac{1}{2pi i}int_{(-1/4)}Gamma(w)y^{-w}zeta(1-s-w) ,dw
$$
where $(-1/4)$ should denote the vertical line at $text{Re}(w)=-1/4$ and $s=1/2+it$. He says that to compute this we have to move the integration to the line $text{Re}(w)=-3/4$ and then expand $zeta(1-s-w)$ into its Dirichlet series, interchange the summation and integration to eventually get
$$
I_1(s)=-sum_n n^{s-1}e^{-n/y}
$$
Here are my doubts
- When we move the line of integration from $text{Re}(w)=-1/4$ to $text{Re}(w)=-3/4$ don't we pass through the pole of $zeta$ at $w=-s$? What about the residue?
- Once we expand $zeta$ into its Dirichlet series and interchange the summation and integral we are left with computing
$$
sum_n n^{s-1}frac{1}{2pi i}int_{(-3/4)}Gamma(w)left(frac{y}{n}right)^{-w},dw
$$
I know that
$$
e^{-y}=frac{1}{2pi i}int_{(c)}Gamma(w)left(yright)^{-w},dw
$$
for all c>0. So the difference between the two should be given by the residue of $Gamma(w)y^{-w}$ at $w=1$, that is 1. But what they claim is that we pick a minus sign.
I am surely overlooking something but I cannot figure out what. I really appreciate any advice. Thank you.
complex-analysis residue-calculus complex-integration
asked Jan 24 at 17:12
asdasd
30729
$endgroup$
I am reading chapter 12 of "Lectures on the Riemann Zeta function" by H. Iwaniec.
I am stuck at understanding the computation done at the beginning of page 45.
We want to estimate the following integral
$$
I_1(s)=frac{1}{2pi i}int_{(-1/4)}Gamma(w)y^{-w}zeta(1-s-w) ,dw
$$
where $(-1/4)$ should denote the vertical line at $text{Re}(w)=-1/4$ and $s=1/2+it$. He says that to compute this we have to move the integration to the line $text{Re}(w)=-3/4$ and then expand $zeta(1-s-w)$ into its Dirichlet series, interchange the summation and integration to eventually get
$$
I_1(s)=-sum_n n^{s-1}e^{-n/y}
$$
Here are my doubts
- When we move the line of integration from $text{Re}(w)=-1/4$ to $text{Re}(w)=-3/4$ don't we pass through the pole of $zeta$ at $w=-s$? What about the residue?
- Once we expand $zeta$ into its Dirichlet series and interchange the summation and integral we are left with computing
$$
sum_n n^{s-1}frac{1}{2pi i}int_{(-3/4)}Gamma(w)left(frac{y}{n}right)^{-w},dw
$$
I know that
$$
e^{-y}=frac{1}{2pi i}int_{(c)}Gamma(w)left(yright)^{-w},dw
$$
for all c>0. So the difference between the two should be given by the residue of $Gamma(w)y^{-w}$ at $w=1$, that is 1. But what they claim is that we pick a minus sign.
I am surely overlooking something but I cannot figure out what. I really appreciate any advice. Thank you.
complex-analysis residue-calculus complex-integration
complex-analysis residue-calculus complex-integration
asked Jan 24 at 17:12
asdasd
30729
asked Jan 24 at 17:12
asdasd
30729
asked Jan 24 at 17:12
asdasd
30729
asked Jan 24 at 17:12
asdasd
30729
asked Jan 24 at 17:12
asdasd
30729
30729
$begingroup$
What is the estimate you are trying to get in the end?
$endgroup$
– 0x539
Jan 24 at 17:39
$begingroup$
That's the end. Then he just uses this result together with other estimates to prove a proposition. Basically I want to prove that $-sum_n n^{s-1}e^{-n/y}=frac{1}{2pi i}int_{(-1/4)}Gamma(w)y^{-w}zeta(1-s-w)dw$.
$endgroup$
– asd
Jan 24 at 17:42
$begingroup$
You mean my question?
$endgroup$
– asd
Jan 24 at 18:05
$begingroup$
nevermind, seems to have been a bug on my end
$endgroup$
– 0x539
Jan 24 at 18:05
add a comment |
$begingroup$
What is the estimate you are trying to get in the end?
$endgroup$
– 0x539
Jan 24 at 17:39
$begingroup$
That's the end. Then he just uses this result together with other estimates to prove a proposition. Basically I want to prove that $-sum_n n^{s-1}e^{-n/y}=frac{1}{2pi i}int_{(-1/4)}Gamma(w)y^{-w}zeta(1-s-w)dw$.
$endgroup$
– asd
Jan 24 at 17:42
$begingroup$
You mean my question?
$endgroup$
– asd
Jan 24 at 18:05
$begingroup$
nevermind, seems to have been a bug on my end
$endgroup$
– 0x539
Jan 24 at 18:05
$begingroup$
What is the estimate you are trying to get in the end?
$endgroup$
– 0x539
Jan 24 at 17:39
$begingroup$
What is the estimate you are trying to get in the end?
$endgroup$
– 0x539
Jan 24 at 17:39
$begingroup$
That's the end. Then he just uses this result together with other estimates to prove a proposition. Basically I want to prove that $-sum_n n^{s-1}e^{-n/y}=frac{1}{2pi i}int_{(-1/4)}Gamma(w)y^{-w}zeta(1-s-w)dw$.
$endgroup$
– asd
Jan 24 at 17:42
$begingroup$
That's the end. Then he just uses this result together with other estimates to prove a proposition. Basically I want to prove that $-sum_n n^{s-1}e^{-n/y}=frac{1}{2pi i}int_{(-1/4)}Gamma(w)y^{-w}zeta(1-s-w)dw$.
$endgroup$
– asd
Jan 24 at 17:42
$begingroup$
You mean my question?
$endgroup$
– asd
Jan 24 at 18:05
$begingroup$
You mean my question?
$endgroup$
– asd
Jan 24 at 18:05
$begingroup$
nevermind, seems to have been a bug on my end
$endgroup$
– 0x539
Jan 24 at 18:05
$begingroup$
nevermind, seems to have been a bug on my end
$endgroup$
– 0x539
Jan 24 at 18:05
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For $c > 0,y > 0$ $$e^{-y}=frac{1}{2pi i}int_{(c)}Gamma(w)y^{-w},dw$$ you can see it from the residue theorem (summing over the poles at negative integers and showing $lim_{a to -infty}frac{1}{2pi i}int_{(a)}Gamma(w)left(yright)^{-w},dw = 0$)
or from the inverse Fourier/Laplace/Mellin transform of $Gamma(w) = int_0^infty y^{w-1} e^{-y}dy=int_{-infty}^infty e^{-uw} e^{-e^{-u}}du$ given $e^{-uc} e^{-e^{-u}}$ is Schwartz thus so is $Gamma(c+it)$.
Thus for $Re(s) < 0,-Re(s) > c > 0$ as everything converges absolutely $$sum_{n=1}^infty n^{s-1} e^{-y/n}=frac{1}{2pi i}int_{(c)}Gamma(w)sum_{n=1}^infty n^{s-1} left(y/nright)^{-w},dw=frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw$$
When $Re(s) < 0$ and shifting the contour to the left, $zeta(1-s-w)$ stays bounded and $Gamma(w)$ stays fast decreasing (because $Gamma(w+1) =w Gamma(w)$ and $Gamma(w)$ is Schwartz-fast decreasing on vertical lines for $Re(w)>0$)
so there is no fear to say for $Re(s) < 0,-Re(s) > c > 0$
$$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw\= frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw-text{Res}_{w=0}(Gamma(w)zeta(1-s-w)y^{-w})\=sum_{n=1}^infty n^{s-1} e^{-y/n}-zeta(1-s)$$
- What happens for $Re(s) < 0, c > -Re(s)$ ? Even if $zeta(1-s-w)$ isn't bounded $Gamma(w)zeta(1-s-w)$ stays fast decreasing. So there is just to add the residue at $w=-s$ giving
$$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw\=frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw-text{Res}_{w=0}(Gamma(w)zeta(1-s-w)y^{-w})-text{Res}_{w=-s}(Gamma(w)zeta(1-s-w)y^{-w})$$
- What happens for $Re(s) > 0$ ? Something similar holds except that you'll need to regularize $F_y(s)=sum_{n=1}^infty n^{s-1} e^{-y/n}$, for example $sum_{n=1}^infty n^{s-1} (e^{-y/n}-1) = F_y(s)-zeta(1-s)$ for $Re(s)in (0,1),y > 0$ and hence $$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw=sum_{n=1}^infty n^{s-1} (e^{-y/n}-1)$$
answered Jan 25 at 2:41
reunsreuns
21.4k21352
$endgroup$
$begingroup$
so, basically what I want to prove is wrong?
$endgroup$
– asd
Jan 25 at 17:35
$begingroup$
@asd What exactly and where are you stuck at checking if it's wrong or not ?
$endgroup$
– reuns
Jan 26 at 9:12
$begingroup$
I don't know how I can prove that $sum_{ninmathbb{N}}n^{s-1}(e^{-y/n}-1)=-sum_{ninmathbb{N}}n^{s-1}e^{-n/y}$ for $Re(s)=1/2$.
$endgroup$
– asd
Jan 26 at 9:37
$begingroup$
Actually the $y$ in the Proposition I have to prove is given by $y=frac{2pi}{sqrt{s(1-s)}}$. Does it make any difference?
$endgroup$
– asd
Jan 26 at 9:42
$begingroup$
@asd $y$ non-real makes obviously a difference as then $y^{-w}$ isn't bounded on vertical lines. So you need to look at the Stirling approximation to see if $Gamma(w)y^{-w}$ is fast decreasing or not. If it is, everything works the same way.
$endgroup$
– reuns
Jan 26 at 9:48
|
show 3 more comments
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1 Answer
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1 Answer
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$begingroup$
For $c > 0,y > 0$ $$e^{-y}=frac{1}{2pi i}int_{(c)}Gamma(w)y^{-w},dw$$ you can see it from the residue theorem (summing over the poles at negative integers and showing $lim_{a to -infty}frac{1}{2pi i}int_{(a)}Gamma(w)left(yright)^{-w},dw = 0$)
or from the inverse Fourier/Laplace/Mellin transform of $Gamma(w) = int_0^infty y^{w-1} e^{-y}dy=int_{-infty}^infty e^{-uw} e^{-e^{-u}}du$ given $e^{-uc} e^{-e^{-u}}$ is Schwartz thus so is $Gamma(c+it)$.
Thus for $Re(s) < 0,-Re(s) > c > 0$ as everything converges absolutely $$sum_{n=1}^infty n^{s-1} e^{-y/n}=frac{1}{2pi i}int_{(c)}Gamma(w)sum_{n=1}^infty n^{s-1} left(y/nright)^{-w},dw=frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw$$
When $Re(s) < 0$ and shifting the contour to the left, $zeta(1-s-w)$ stays bounded and $Gamma(w)$ stays fast decreasing (because $Gamma(w+1) =w Gamma(w)$ and $Gamma(w)$ is Schwartz-fast decreasing on vertical lines for $Re(w)>0$)
so there is no fear to say for $Re(s) < 0,-Re(s) > c > 0$
$$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw\= frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw-text{Res}_{w=0}(Gamma(w)zeta(1-s-w)y^{-w})\=sum_{n=1}^infty n^{s-1} e^{-y/n}-zeta(1-s)$$
- What happens for $Re(s) < 0, c > -Re(s)$ ? Even if $zeta(1-s-w)$ isn't bounded $Gamma(w)zeta(1-s-w)$ stays fast decreasing. So there is just to add the residue at $w=-s$ giving
$$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw\=frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw-text{Res}_{w=0}(Gamma(w)zeta(1-s-w)y^{-w})-text{Res}_{w=-s}(Gamma(w)zeta(1-s-w)y^{-w})$$
- What happens for $Re(s) > 0$ ? Something similar holds except that you'll need to regularize $F_y(s)=sum_{n=1}^infty n^{s-1} e^{-y/n}$, for example $sum_{n=1}^infty n^{s-1} (e^{-y/n}-1) = F_y(s)-zeta(1-s)$ for $Re(s)in (0,1),y > 0$ and hence $$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw=sum_{n=1}^infty n^{s-1} (e^{-y/n}-1)$$
answered Jan 25 at 2:41
reunsreuns
21.4k21352
$endgroup$
$begingroup$
so, basically what I want to prove is wrong?
$endgroup$
– asd
Jan 25 at 17:35
$begingroup$
@asd What exactly and where are you stuck at checking if it's wrong or not ?
$endgroup$
– reuns
Jan 26 at 9:12
$begingroup$
I don't know how I can prove that $sum_{ninmathbb{N}}n^{s-1}(e^{-y/n}-1)=-sum_{ninmathbb{N}}n^{s-1}e^{-n/y}$ for $Re(s)=1/2$.
$endgroup$
– asd
Jan 26 at 9:37
$begingroup$
Actually the $y$ in the Proposition I have to prove is given by $y=frac{2pi}{sqrt{s(1-s)}}$. Does it make any difference?
$endgroup$
– asd
Jan 26 at 9:42
$begingroup$
@asd $y$ non-real makes obviously a difference as then $y^{-w}$ isn't bounded on vertical lines. So you need to look at the Stirling approximation to see if $Gamma(w)y^{-w}$ is fast decreasing or not. If it is, everything works the same way.
$endgroup$
– reuns
Jan 26 at 9:48
|
show 3 more comments
$begingroup$
For $c > 0,y > 0$ $$e^{-y}=frac{1}{2pi i}int_{(c)}Gamma(w)y^{-w},dw$$ you can see it from the residue theorem (summing over the poles at negative integers and showing $lim_{a to -infty}frac{1}{2pi i}int_{(a)}Gamma(w)left(yright)^{-w},dw = 0$)
or from the inverse Fourier/Laplace/Mellin transform of $Gamma(w) = int_0^infty y^{w-1} e^{-y}dy=int_{-infty}^infty e^{-uw} e^{-e^{-u}}du$ given $e^{-uc} e^{-e^{-u}}$ is Schwartz thus so is $Gamma(c+it)$.
Thus for $Re(s) < 0,-Re(s) > c > 0$ as everything converges absolutely $$sum_{n=1}^infty n^{s-1} e^{-y/n}=frac{1}{2pi i}int_{(c)}Gamma(w)sum_{n=1}^infty n^{s-1} left(y/nright)^{-w},dw=frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw$$
When $Re(s) < 0$ and shifting the contour to the left, $zeta(1-s-w)$ stays bounded and $Gamma(w)$ stays fast decreasing (because $Gamma(w+1) =w Gamma(w)$ and $Gamma(w)$ is Schwartz-fast decreasing on vertical lines for $Re(w)>0$)
so there is no fear to say for $Re(s) < 0,-Re(s) > c > 0$
$$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw\= frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw-text{Res}_{w=0}(Gamma(w)zeta(1-s-w)y^{-w})\=sum_{n=1}^infty n^{s-1} e^{-y/n}-zeta(1-s)$$
- What happens for $Re(s) < 0, c > -Re(s)$ ? Even if $zeta(1-s-w)$ isn't bounded $Gamma(w)zeta(1-s-w)$ stays fast decreasing. So there is just to add the residue at $w=-s$ giving
$$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw\=frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw-text{Res}_{w=0}(Gamma(w)zeta(1-s-w)y^{-w})-text{Res}_{w=-s}(Gamma(w)zeta(1-s-w)y^{-w})$$
- What happens for $Re(s) > 0$ ? Something similar holds except that you'll need to regularize $F_y(s)=sum_{n=1}^infty n^{s-1} e^{-y/n}$, for example $sum_{n=1}^infty n^{s-1} (e^{-y/n}-1) = F_y(s)-zeta(1-s)$ for $Re(s)in (0,1),y > 0$ and hence $$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw=sum_{n=1}^infty n^{s-1} (e^{-y/n}-1)$$
answered Jan 25 at 2:41
reunsreuns
21.4k21352
$endgroup$
$begingroup$
so, basically what I want to prove is wrong?
$endgroup$
– asd
Jan 25 at 17:35
$begingroup$
@asd What exactly and where are you stuck at checking if it's wrong or not ?
$endgroup$
– reuns
Jan 26 at 9:12
$begingroup$
I don't know how I can prove that $sum_{ninmathbb{N}}n^{s-1}(e^{-y/n}-1)=-sum_{ninmathbb{N}}n^{s-1}e^{-n/y}$ for $Re(s)=1/2$.
$endgroup$
– asd
Jan 26 at 9:37
$begingroup$
Actually the $y$ in the Proposition I have to prove is given by $y=frac{2pi}{sqrt{s(1-s)}}$. Does it make any difference?
$endgroup$
– asd
Jan 26 at 9:42
$begingroup$
@asd $y$ non-real makes obviously a difference as then $y^{-w}$ isn't bounded on vertical lines. So you need to look at the Stirling approximation to see if $Gamma(w)y^{-w}$ is fast decreasing or not. If it is, everything works the same way.
$endgroup$
– reuns
Jan 26 at 9:48
|
show 3 more comments
$begingroup$
For $c > 0,y > 0$ $$e^{-y}=frac{1}{2pi i}int_{(c)}Gamma(w)y^{-w},dw$$ you can see it from the residue theorem (summing over the poles at negative integers and showing $lim_{a to -infty}frac{1}{2pi i}int_{(a)}Gamma(w)left(yright)^{-w},dw = 0$)
or from the inverse Fourier/Laplace/Mellin transform of $Gamma(w) = int_0^infty y^{w-1} e^{-y}dy=int_{-infty}^infty e^{-uw} e^{-e^{-u}}du$ given $e^{-uc} e^{-e^{-u}}$ is Schwartz thus so is $Gamma(c+it)$.
Thus for $Re(s) < 0,-Re(s) > c > 0$ as everything converges absolutely $$sum_{n=1}^infty n^{s-1} e^{-y/n}=frac{1}{2pi i}int_{(c)}Gamma(w)sum_{n=1}^infty n^{s-1} left(y/nright)^{-w},dw=frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw$$
When $Re(s) < 0$ and shifting the contour to the left, $zeta(1-s-w)$ stays bounded and $Gamma(w)$ stays fast decreasing (because $Gamma(w+1) =w Gamma(w)$ and $Gamma(w)$ is Schwartz-fast decreasing on vertical lines for $Re(w)>0$)
so there is no fear to say for $Re(s) < 0,-Re(s) > c > 0$
$$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw\= frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw-text{Res}_{w=0}(Gamma(w)zeta(1-s-w)y^{-w})\=sum_{n=1}^infty n^{s-1} e^{-y/n}-zeta(1-s)$$
- What happens for $Re(s) < 0, c > -Re(s)$ ? Even if $zeta(1-s-w)$ isn't bounded $Gamma(w)zeta(1-s-w)$ stays fast decreasing. So there is just to add the residue at $w=-s$ giving
$$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw\=frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw-text{Res}_{w=0}(Gamma(w)zeta(1-s-w)y^{-w})-text{Res}_{w=-s}(Gamma(w)zeta(1-s-w)y^{-w})$$
- What happens for $Re(s) > 0$ ? Something similar holds except that you'll need to regularize $F_y(s)=sum_{n=1}^infty n^{s-1} e^{-y/n}$, for example $sum_{n=1}^infty n^{s-1} (e^{-y/n}-1) = F_y(s)-zeta(1-s)$ for $Re(s)in (0,1),y > 0$ and hence $$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw=sum_{n=1}^infty n^{s-1} (e^{-y/n}-1)$$
answered Jan 25 at 2:41
reunsreuns
21.4k21352
$endgroup$
For $c > 0,y > 0$ $$e^{-y}=frac{1}{2pi i}int_{(c)}Gamma(w)y^{-w},dw$$ you can see it from the residue theorem (summing over the poles at negative integers and showing $lim_{a to -infty}frac{1}{2pi i}int_{(a)}Gamma(w)left(yright)^{-w},dw = 0$)
or from the inverse Fourier/Laplace/Mellin transform of $Gamma(w) = int_0^infty y^{w-1} e^{-y}dy=int_{-infty}^infty e^{-uw} e^{-e^{-u}}du$ given $e^{-uc} e^{-e^{-u}}$ is Schwartz thus so is $Gamma(c+it)$.
Thus for $Re(s) < 0,-Re(s) > c > 0$ as everything converges absolutely $$sum_{n=1}^infty n^{s-1} e^{-y/n}=frac{1}{2pi i}int_{(c)}Gamma(w)sum_{n=1}^infty n^{s-1} left(y/nright)^{-w},dw=frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw$$
When $Re(s) < 0$ and shifting the contour to the left, $zeta(1-s-w)$ stays bounded and $Gamma(w)$ stays fast decreasing (because $Gamma(w+1) =w Gamma(w)$ and $Gamma(w)$ is Schwartz-fast decreasing on vertical lines for $Re(w)>0$)
so there is no fear to say for $Re(s) < 0,-Re(s) > c > 0$
$$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw\= frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw-text{Res}_{w=0}(Gamma(w)zeta(1-s-w)y^{-w})\=sum_{n=1}^infty n^{s-1} e^{-y/n}-zeta(1-s)$$
- What happens for $Re(s) < 0, c > -Re(s)$ ? Even if $zeta(1-s-w)$ isn't bounded $Gamma(w)zeta(1-s-w)$ stays fast decreasing. So there is just to add the residue at $w=-s$ giving
$$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw\=frac{1}{2pi i}int_{(c)}Gamma(w)zeta(1-s-w)y^{-w},dw-text{Res}_{w=0}(Gamma(w)zeta(1-s-w)y^{-w})-text{Res}_{w=-s}(Gamma(w)zeta(1-s-w)y^{-w})$$
- What happens for $Re(s) > 0$ ? Something similar holds except that you'll need to regularize $F_y(s)=sum_{n=1}^infty n^{s-1} e^{-y/n}$, for example $sum_{n=1}^infty n^{s-1} (e^{-y/n}-1) = F_y(s)-zeta(1-s)$ for $Re(s)in (0,1),y > 0$ and hence $$frac{1}{2pi i}int_{(-1/4)}Gamma(w)zeta(1-s-w)y^{-w},dw=sum_{n=1}^infty n^{s-1} (e^{-y/n}-1)$$
answered Jan 25 at 2:41
reunsreuns
21.4k21352
edited Jan 25 at 3:10
answered Jan 25 at 2:41
reunsreuns
21.4k21352
answered Jan 25 at 2:41
reunsreuns
21.4k21352
answered Jan 25 at 2:41
reunsreuns
21.4k21352
21.4k21352
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so, basically what I want to prove is wrong?
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– asd
Jan 25 at 17:35
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@asd What exactly and where are you stuck at checking if it's wrong or not ?
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– reuns
Jan 26 at 9:12
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I don't know how I can prove that $sum_{ninmathbb{N}}n^{s-1}(e^{-y/n}-1)=-sum_{ninmathbb{N}}n^{s-1}e^{-n/y}$ for $Re(s)=1/2$.
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– asd
Jan 26 at 9:37
$begingroup$
Actually the $y$ in the Proposition I have to prove is given by $y=frac{2pi}{sqrt{s(1-s)}}$. Does it make any difference?
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– asd
Jan 26 at 9:42
$begingroup$
@asd $y$ non-real makes obviously a difference as then $y^{-w}$ isn't bounded on vertical lines. So you need to look at the Stirling approximation to see if $Gamma(w)y^{-w}$ is fast decreasing or not. If it is, everything works the same way.
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– reuns
Jan 26 at 9:48
|
show 3 more comments
$begingroup$
so, basically what I want to prove is wrong?
$endgroup$
– asd
Jan 25 at 17:35
$begingroup$
@asd What exactly and where are you stuck at checking if it's wrong or not ?
$endgroup$
– reuns
Jan 26 at 9:12
$begingroup$
I don't know how I can prove that $sum_{ninmathbb{N}}n^{s-1}(e^{-y/n}-1)=-sum_{ninmathbb{N}}n^{s-1}e^{-n/y}$ for $Re(s)=1/2$.
$endgroup$
– asd
Jan 26 at 9:37
$begingroup$
Actually the $y$ in the Proposition I have to prove is given by $y=frac{2pi}{sqrt{s(1-s)}}$. Does it make any difference?
$endgroup$
– asd
Jan 26 at 9:42
$begingroup$
@asd $y$ non-real makes obviously a difference as then $y^{-w}$ isn't bounded on vertical lines. So you need to look at the Stirling approximation to see if $Gamma(w)y^{-w}$ is fast decreasing or not. If it is, everything works the same way.
$endgroup$
– reuns
Jan 26 at 9:48
$begingroup$
so, basically what I want to prove is wrong?
$endgroup$
– asd
Jan 25 at 17:35
$begingroup$
so, basically what I want to prove is wrong?
$endgroup$
– asd
Jan 25 at 17:35
$begingroup$
@asd What exactly and where are you stuck at checking if it's wrong or not ?
$endgroup$
– reuns
Jan 26 at 9:12
$begingroup$
@asd What exactly and where are you stuck at checking if it's wrong or not ?
$endgroup$
– reuns
Jan 26 at 9:12
$begingroup$
I don't know how I can prove that $sum_{ninmathbb{N}}n^{s-1}(e^{-y/n}-1)=-sum_{ninmathbb{N}}n^{s-1}e^{-n/y}$ for $Re(s)=1/2$.
$endgroup$
– asd
Jan 26 at 9:37
$begingroup$
I don't know how I can prove that $sum_{ninmathbb{N}}n^{s-1}(e^{-y/n}-1)=-sum_{ninmathbb{N}}n^{s-1}e^{-n/y}$ for $Re(s)=1/2$.
$endgroup$
– asd
Jan 26 at 9:37
$begingroup$
Actually the $y$ in the Proposition I have to prove is given by $y=frac{2pi}{sqrt{s(1-s)}}$. Does it make any difference?
$endgroup$
– asd
Jan 26 at 9:42
$begingroup$
Actually the $y$ in the Proposition I have to prove is given by $y=frac{2pi}{sqrt{s(1-s)}}$. Does it make any difference?
$endgroup$
– asd
Jan 26 at 9:42
$begingroup$
@asd $y$ non-real makes obviously a difference as then $y^{-w}$ isn't bounded on vertical lines. So you need to look at the Stirling approximation to see if $Gamma(w)y^{-w}$ is fast decreasing or not. If it is, everything works the same way.
$endgroup$
– reuns
Jan 26 at 9:48
$begingroup$
@asd $y$ non-real makes obviously a difference as then $y^{-w}$ isn't bounded on vertical lines. So you need to look at the Stirling approximation to see if $Gamma(w)y^{-w}$ is fast decreasing or not. If it is, everything works the same way.
$endgroup$
– reuns
Jan 26 at 9:48
|
show 3 more comments
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$begingroup$
What is the estimate you are trying to get in the end?
$endgroup$
– 0x539
Jan 24 at 17:39
$begingroup$
That's the end. Then he just uses this result together with other estimates to prove a proposition. Basically I want to prove that $-sum_n n^{s-1}e^{-n/y}=frac{1}{2pi i}int_{(-1/4)}Gamma(w)y^{-w}zeta(1-s-w)dw$.
$endgroup$
– asd
Jan 24 at 17:42
$begingroup$
You mean my question?
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– asd
Jan 24 at 18:05
$begingroup$
nevermind, seems to have been a bug on my end
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– 0x539
Jan 24 at 18:05