02/19/2023, 08:59 AM
Hmm, this might be non-sense, but I'll write it anyway.
The goal will be to construct some functions made up of lambda that allow us to compute \( \lambda(n) \) while only evaulating it on the positive integers. One very useful identity we can use is that
\[ g(n) = \sum_{n=1}^\infty \frac{\lambda(n)}{n^s} = \frac{\zeta(2s)}{\zeta(s)}\]
Now, let's consider the two functions
\[ \sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}} \qquad \sum_{k=1}^\infty \frac{\lambda(k)}{z-k}\]
These are linked in the following way. We have that
\[\star_1 \quad \sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}} = \sum_{n=0}^\infty \frac{1}{z^{n+1}} \sum_{k=1}^\infty \lambda(k)k^n = \]
\[\sum_{k=1}^\infty \frac{\lambda(k)}{z} \sum_{n=0}^\infty (\frac{k}{z})^n= \]
\[\star_2 \quad \sum_{k=1}^\infty \lambda(k) \frac{1}{z-k} \]
The stars indicate lines where I replaced a divergent series with an analytical continuation. In these lines, we have to either add or subtract the residues that were introduced. The usefulness of this approach is that we introduce a residue at \( k=z \), and its value is some multiple of \(\lambda(z)\). Therefore, we should be able to compute the value of \(\lambda(z)\).
So, going back, let's go back and carefully keep track of the residues introduced. At \( \star_1 \) we replace an analytical continuation with a divergent series. Therefore, we create some residues, so we need to add them back in to get equality. Thus, we should we should have that
\[ \sum \frac{g(-n)}{z^{n+1}}+\sum \text{Res}\left(\frac{\zeta(-2s)}{\zeta(-s)}\frac{1}{z^{s+1}(e^{2 \pi i s}-1)}, s = -\rho \right) = \sum \frac{1}{z^{n+1}} \sum \lambda(k) k^n \]
In other words, we have to sum over the residues from the non-trivial zeroes of the zeta function (notice that the trivial zeroes get canceled out since we are looking at \( \frac{\zeta(2s)}{\zeta(s)} \).)
Next, we look at \( \star_2 \). In this case, we are replacing a divergent series by an analytical continuation, so we need to subtract the residue it generates. However, that residue it generates is equal to
\[ 2 \pi i \frac{\lambda(z)}{e^{2 \pi i z} - 1}\]
But this involves \( \lambda(z) \)! Therefore, we should be able to compute \( \lambda(z) \) by evaluating these series. In particular, we now have
\[ \sum \frac{\lambda(k)}{z-k} + 2 \pi i \frac{\lambda(z)}{e^{2 \pi i z} - 1} = \sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}} +\sum \text{Res}\left(\frac{\zeta(-2s)}{\zeta(-s)}\frac{1}{z^{s+1}(e^{2 \pi i s}-1)}, s = -\rho \right)\]
Which implies that
\[ \lambda(z) = \frac{e^{2 \pi i z} -1}{2 \pi i}\left(\sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}}- \sum_{k=1}^\infty \frac{\lambda(k)}{z-k} +\sum \text{Res}\left(\frac{\zeta(-2s)}{\zeta(-s)}\frac{1}{z^{s+1}(e^{2 \pi i s}-1)}, s = -\rho \right)\right)\]
There's a much easier way to pick up all those zeta-zero residues however, we just do a contour integral at some point that picks them up. In particular, it just has to behind \(z = -\frac{1}{2}\) assuming the RH, but behind \(z=-1 \) will work no matter what.
The goal will be to construct some functions made up of lambda that allow us to compute \( \lambda(n) \) while only evaulating it on the positive integers. One very useful identity we can use is that
\[ g(n) = \sum_{n=1}^\infty \frac{\lambda(n)}{n^s} = \frac{\zeta(2s)}{\zeta(s)}\]
Now, let's consider the two functions
\[ \sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}} \qquad \sum_{k=1}^\infty \frac{\lambda(k)}{z-k}\]
These are linked in the following way. We have that
\[\star_1 \quad \sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}} = \sum_{n=0}^\infty \frac{1}{z^{n+1}} \sum_{k=1}^\infty \lambda(k)k^n = \]
\[\sum_{k=1}^\infty \frac{\lambda(k)}{z} \sum_{n=0}^\infty (\frac{k}{z})^n= \]
\[\star_2 \quad \sum_{k=1}^\infty \lambda(k) \frac{1}{z-k} \]
The stars indicate lines where I replaced a divergent series with an analytical continuation. In these lines, we have to either add or subtract the residues that were introduced. The usefulness of this approach is that we introduce a residue at \( k=z \), and its value is some multiple of \(\lambda(z)\). Therefore, we should be able to compute the value of \(\lambda(z)\).
So, going back, let's go back and carefully keep track of the residues introduced. At \( \star_1 \) we replace an analytical continuation with a divergent series. Therefore, we create some residues, so we need to add them back in to get equality. Thus, we should we should have that
\[ \sum \frac{g(-n)}{z^{n+1}}+\sum \text{Res}\left(\frac{\zeta(-2s)}{\zeta(-s)}\frac{1}{z^{s+1}(e^{2 \pi i s}-1)}, s = -\rho \right) = \sum \frac{1}{z^{n+1}} \sum \lambda(k) k^n \]
In other words, we have to sum over the residues from the non-trivial zeroes of the zeta function (notice that the trivial zeroes get canceled out since we are looking at \( \frac{\zeta(2s)}{\zeta(s)} \).)
Next, we look at \( \star_2 \). In this case, we are replacing a divergent series by an analytical continuation, so we need to subtract the residue it generates. However, that residue it generates is equal to
\[ 2 \pi i \frac{\lambda(z)}{e^{2 \pi i z} - 1}\]
But this involves \( \lambda(z) \)! Therefore, we should be able to compute \( \lambda(z) \) by evaluating these series. In particular, we now have
\[ \sum \frac{\lambda(k)}{z-k} + 2 \pi i \frac{\lambda(z)}{e^{2 \pi i z} - 1} = \sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}} +\sum \text{Res}\left(\frac{\zeta(-2s)}{\zeta(-s)}\frac{1}{z^{s+1}(e^{2 \pi i s}-1)}, s = -\rho \right)\]
Which implies that
\[ \lambda(z) = \frac{e^{2 \pi i z} -1}{2 \pi i}\left(\sum_{n=0}^\infty \frac{g(-n)}{z^{n+1}}- \sum_{k=1}^\infty \frac{\lambda(k)}{z-k} +\sum \text{Res}\left(\frac{\zeta(-2s)}{\zeta(-s)}\frac{1}{z^{s+1}(e^{2 \pi i s}-1)}, s = -\rho \right)\right)\]
There's a much easier way to pick up all those zeta-zero residues however, we just do a contour integral at some point that picks them up. In particular, it just has to behind \(z = -\frac{1}{2}\) assuming the RH, but behind \(z=-1 \) will work no matter what.