# Vonmangoldt sums

The dirichlet series for the Vonmangoldt function, $\Lambda(n)$, which is equal to zero when $n$ is not a prime a power, and $\ln(p)$ when it is a prime power say, $n=p^j$, is

$$-\frac{\zeta'(s)}{\zeta(s)}=\sum_{k=1}^{\infty}\frac{\Lambda(k)}{k^s}$$

Where $\zeta(s)$ is the zeta function, and $\zeta'(s)$ is the derivative of the zeta function with respect to $s$,

This can be re-written as $$-\frac{\zeta'(s)}{\zeta(s)}=\sum_{k=1}^{\infty}\frac{\Lambda(k)}{k^s}=\sum_{p}\frac{\ln(p)}{p^s-1},$$ with the last sum ranging over all primes p,

Can someone help me brake this sum, up into prime congruence sums similarly $$\sum_{k=0}^{\infty}\frac{\Lambda(5k+1)}{(5k+1)^s}$$, ie re-write it as prime sums, where the primes are congruent to some b modulo $5$. I have done it before modulo $4$, and $3$ so I know it can be done, I am just having trouble restricting the powers appropietly to account for cases when $p^a\equiv1$ mod $5$, has no solutions.

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See Greg Martin's answer on this thread: Von mangoldt function dirichlet series

In general,

$$\sum_{n=0}^\infty \frac{\Lambda(qn+b)}{(qn+b)^s}=\frac{1}{\phi(q)}\sum_{\chi \pmod{q}}\overline{\chi}(b)\sum_{p}\frac{\chi(p)\log p}{p^{s}-\chi(p)}.$$ We may also write this as $$\frac{-1}{\phi(q)}\sum_{\chi\pmod{q}}\overline{\chi}(b)\frac{L^{'}}{L}(s,\chi).$$ Looking at a specific example, this yields $$\sum_{n\equiv1\ (3)}^{\infty}\frac{\Lambda(n)}{n^{s}}=\sum_{p\equiv1\ (3)}\frac{\log p}{p^{s}-1}+\sum_{p\equiv2\ (3)}\frac{\log p}{p^{2s}-1} .$$

To see this, notice that $$\sum_{n=0}^\infty \frac{\Lambda(qn+b)}{(qn+b)^s}=\sum_{n\equiv b\pmod{q}} \frac{\Lambda(n)}{n^s}.$$ Using the orthogonality relations of the Dirichlet Characters, that is the fact that $$\frac{1}{\phi(q)}\sum_{\chi \pmod{q}} \chi(a)=\left\{ \begin{array}{c} 1\ \text{when } a\equiv 1 \pmod{q} \\ 0\ \text{otherwise} \end{array}\right\},$$ it follows that

$$\sum_{n=0}^\infty \frac{\Lambda(qn+b)}{(qn+b)^s}=\frac{1}{\phi(q)}\sum_{\chi\pmod{q}}\overline{\chi}(b)\sum_{n=1}^{\infty}\frac{\Lambda(n)\chi(n)}{n^{s}}$$ Recalling the function $L(s,\chi)$, this last sum then equals $$\frac{-1}{\phi(q)}\sum_{\chi\pmod{q}}\overline{\chi}(b)\frac{L^{'}}{L}(s,\chi).$$

From here, we obtain the first equation by expanding we obtain the first equation by using the fact that $$L(s,\chi)=\prod_{p}\left(1-\frac{\chi(p)}{p^{s}}\right)^{-1}.$$

Key Idea: Use Dirichlet characters to isolate an arithmetic progression. This is how most major results on primes in arithmetic progressions are proven.

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I appreciate the work you spent, but Im not trying to evaluate in terms of L functions, I want to re-write it as a series of primes. – Ethan Dec 17 '12 at 23:02
@Ethan: $L$-functions allow you to do exactly that. I'll expand a bit more. – Eric Naslund Dec 17 '12 at 23:17
@Ethan: Using the formula with Dirichlet characters I wrote above, if you look at the case $p=3k+1$, for example, we get that $$\sum_{n\equiv1\ (3)}^{\infty}\frac{\Lambda(n)}{n^{s}}=\sum_{p\equiv1\ (3)}\frac{\log p}{p^{s}-1}+\sum_{p\equiv2\ (3)}\frac{\log p}{p^{2s}-1} .$$ There will be a similar formula for all other moduli if you want to write it over sums of primes based on congruence class. – Eric Naslund Dec 22 '12 at 23:01
I do, but I want a general way of doing so, its a pain to check which individula congruences are satisfied between 1 and a, and then using that to re-write the sum – Ethan Dec 23 '12 at 1:57
@Ethan: For a general modulus $b$, we have that $$\sum_{a\in\left(\mathbb{Z}/b\mathbb{Z}\right)^{\times}}\sum_{p\equiv a\ (b)}\frac{\log p}{p^{s\cdot \text{ord}_{b}(a)}-1}$$ where $\text{ord}_{b}(a)$ is the order of $a$ in the multiplicative group $\left(\mathbb{Z}/b\mathbb{Z}\right)^{\times},$ that is the smallest positive integer $r$ such that $a^r\equiv 1\text{ mod }b$. I doubt there will be any further simplification since $\text{ord}_b(a)$ is a rather complicated function. – Eric Naslund Dec 23 '12 at 4:57