# Prove that $\sum\Lambda (n)\sim \frac{x}{\varphi(q)}$ where the sum is over every $n\leq x$ such that $n\equiv a\pmod q$

Prove that $\sum\Lambda (n)\sim \frac{x}{\varphi(q)}$ where the sum is over every $n\leq x$ such that $n\equiv a\pmod q$

Attempts

Let $\chi$ a character of Dirichlet modulo $q$ non trivial. I proved that $$\sum_{n=1}^\infty \frac{\Lambda (n)\chi(n)}{n}=\mathcal O(1)$$

I want now prove that $$\sum_{n\leq x}\Lambda (n)\chi(n)=o(x).$$ Using Abel summation, I get $$\frac{1}{x}\sum_{n\leq x}\Lambda (n)\chi(n)=\sum_{n\leq x}\frac{\Lambda (n)\chi(n)}{n}-\int_1^x\frac{1}{t^2}\sum_{n\leq t}\Lambda (n)\chi(n)dt$$

Q1) How can I get that $$\int_1^x \frac{1}{t^2}\sum_{n\leq t}\Lambda (n)\chi(n)dt=o(x),$$ to conclude ?

I suppose the previous result correct.

$$\sum_{\substack{n\leq x\\ n\equiv a\pmod q}}\Lambda (n)=\sum_{n\leq x}\Lambda (n)\delta_{1\equiv a\pmod q}=\frac{1}{\varphi(q)}\sum_{n\leq x}\Lambda (n)\sum_{\chi\pmod \chi}\chi(n)\overline{\chi(a)}$$ $$=\underbrace{\frac{1}{\varphi(q)}\sum_{n\leq x}\Lambda (n)}_{\sim\frac{x}{\varphi(q)}}+\frac{1}{\varphi(q)}\sum_{\substack{\chi\pmod q\\ \chi\neq \chi_0}}\overline{\chi(a)}\sum_{n\leq x}\Lambda (n)\chi(n)$$ $$=\frac{x}{\varphi(q)}+o\left(\frac{x}{\varphi(q)}\right)+\frac{1}{\varphi(q)}\sum_{\substack{\chi\pmod q\\ \chi\neq \chi_0 }}\overline{\chi(a)}o(x)$$ $$=\frac{x}{\varphi(q)}+o\left(\frac{x}{\varphi(q)}\right)+o\left(\frac{x}{\varphi(q)}\sum_{\substack{\chi\pmod q\\ \chi\neq \chi_0 }}\overline{\chi(a)}\right)$$

Q2) How can I conclude ? i.e. that $$o\left(\frac{x}{\varphi(q)}\sum_{\substack{\chi\pmod q\\ \chi\neq \chi_0 }}\overline{\chi(a)}\right)=o\left(\frac{x}{\varphi(q)}\right)\ \ ?$$

$\sum_{n < x, n \equiv a\bmod q} \Lambda(n) \sim \frac{x}{\phi(q)}$ is the prime number theorem in arithmetic progressions.

Assuming you showed $L(s,\chi)$ doesn't vanish at $s=1$,

You'll need to prove that $L(s,\chi)$ doesn't have any zeros on $\Re(s) = 1$.

For this let $$F_q(s) = \prod_{\chi \bmod q} L(s,\chi) = \prod_{p \,\nmid\, q,\ p^k \equiv 1 \bmod q} \frac{1}{(1-p^{-sk})^{\phi(q)/k}}$$ which is a Dirichlet series analytic except a simple pole at $s=1$, and whose logarithm $$\log F_q(s) = \phi(q) \sum_{n \equiv 1 \bmod q} \frac{\Lambda(n)}{\ln n} n^{-s}$$ has non-negative coefficients.

Thus we can apply the same argument as for $\zeta(s)$, to obtain a contradiction from $F_q(1+iy)=0$. Thus each $L(s,\chi)$ doesn't vanish on $\Re(s) \ge 1$,

and this is all we need. Let

$$G_{a,q}(s) = \frac{1}{\phi(q)}\sum_{\chi \bmod q} \chi(a) \frac{L'(s,\chi)}{L(s,\chi)}= \sum_{n \equiv a \bmod q} \Lambda(n) n^{-s}\\=s \int_1^\infty (\sum_{qm+a < x} \Lambda(qm+a))x^{-s-1}dx$$ By Mellin inversion $$\int_1^x (\sum_{qm+a< y} \Lambda(qm+a)-y) dy \qquad\qquad\qquad\qquad\\= \frac{1}{2i\pi} \int_{\sigma(T)-i T}^{\sigma(T)+iT} (\frac{1}{s-1}-G_{a,q}(s))\frac{x^{s+1}}{s(s+1)}ds +\mathcal{O}(x^{1+\sigma(T)} T^{-1+\epsilon})= o(x^2)$$ (we needed also some bounds $G_{a,q}(\sigma(t)+it)= \mathcal{O}((\log t)^k)$ when $\sigma(t) \to 1$ fast enough)

• You are probably correct, but I don't now any notion you are talking about, so it should be an other way. Jun 11, 2017 at 15:56
• @user380364 There are no "other ways". Continue reading your books, you'll see what I meant. Jun 11, 2017 at 15:57
• There is a much easier way ! See my answer.
– Surb
Jun 14, 2017 at 10:11

1) You can use Abel with $a_n=\frac{\Lambda (n)\chi(n)}{n}$ and $f(t)=t$ what will gives you $$\sum_{n\leq x}\Lambda(n)\chi(n)=\sum_{n\leq x}\frac{\Lambda (n)\chi(n)n}{n}=x\sum_{n\leq x}\frac{\chi(n)\Lambda (n)}{n}-\int_1^x\sum_{n\leq t}\frac{\Lambda (n)\chi(n)}{n}\mathrm dt.$$ Since $$\lim_{x\to \infty }\sum_{n\leq x}\frac{\Lambda (n)\chi(n)}{n}$$ exist, there is $\ell\geq 0$ s.t. $$\sum_{n\leq x}\frac{\Lambda (n)\chi(n)}{n}=\ell+o(1),$$ and thus $$\sum_{n\leq x}\Lambda (n)\chi(n)= \ell x+o(x)-\int_1^x(\ell+o(1))\mathrm dt=o(x).$$

2) You missed the condition $(n,q)=1$. \begin{alignat*}{1} \sum_{\substack{n\leqslant x \\ n\equiv a \mod (q)}}\Lambda(n) = & \ \frac{1}{\varphi(q)}\sum_{\chi \mod (q)}\overline{\chi(a)}\sum_{n\leqslant x}\Lambda(n)\chi(n) \\ = & \frac{1}{\varphi(q)}\sum_{\substack{n\leqslant x \\ (n,q)=1}}\Lambda(n)+\frac{1}{\phi(q)}\sum_{\chi\neq 1}\sum_{n\leqslant x}\Lambda(n)\chi(n), \end{alignat*} the RHS second term of the las equality gives you a $o\left(\frac{x}{\phi(q)}\right)$ by the previous question, and the first term is $$\sum_{\substack{n\leq x\\ (n,q)= 1}}\Lambda (n)=\sum_{n\leq x}\Lambda (n)-\sum_{\substack{n\leq x\\ (n,q)>1}}\Lambda (n)$$ and $$\sum_{\substack{n\leq x\\ (n,q)>1}}\Lambda (n)=\sum_{p\mid q}\log(p)\sum_{p^k\leq x} 1\ll \log(x)\log(q).$$ The claim follow.

EDIT

The existence of $\lim_{x\to \infty }\sum_{n\leq x}\frac{\Lambda (n)\chi(n)}{n}$ comme from the fact that $$-\frac{L'(\chi,s)}{L(chi,s)}=\sum_{n=1}^\infty \frac{\Lambda (n)\chi(n)}{n^s}$$ for $\Re(s)>1$ and since $\frac{L'(\chi,s)}{L(\chi,s)}$ is analytic on $\mathbb Re(s)>0$ and that $L(\chi,1)\neq 0$, using Neuwman theorem allow you to conclude on the existence of $\sum_{n=1}^\infty \frac{\Lambda (n)\chi(n)}{n}.$

• $\sum_{n\leq x}\frac{\Lambda (n)\chi(n)}{n}=\mathcal O(1)$ is not the same as $\sum_{n\leq x}\frac{\Lambda (n)\chi(n)}{n}=l + o(1)$. Jun 14, 2017 at 10:21
• @user1952009: Yes it is ! In this case it's equivalent !
– Surb
Jun 14, 2017 at 10:23
• No. $\sum_{n \le x} \frac{\Lambda(n)-1}{n} = \mathcal{O}(1)$ is the Mertens theorem. Similary we obtain $\sum_{n \le x} \frac{\Lambda(n)\chi(n)-\chi(n)}{n} = \mathcal{O}(1)$ and since $\sum_n \frac{\chi(n)}{n}$ converges we obtain $\sum_{n \le x} \frac{\Lambda(n)\chi(n)}{n} = \mathcal{O}(1)$. But this doesn't imply the PNT $\sum_{n \le x} \Lambda(n)\chi(n) = o(x)$ Jun 14, 2017 at 10:29
• Proof : $\sum_{n \le x} \frac{\Lambda(n)\chi(n)+n^{it}}{n} = \mathcal{O}(1)$ (but not $l+o(1)$) and $\sum_{n \le x} (\Lambda(n)\chi(n)+n^{it})$ is not $o(x)$ Jun 14, 2017 at 10:30
• $\sum_{n \le x} \frac{\Lambda(n)-1}{n} = \mathcal{O}(1)$ and $\sum_{n \le x} \frac{\Lambda(n)\chi(n)}{n} = \mathcal{O}(1)$ are equivalent to $\frac{-\zeta'(s)}{\zeta(s)}-\frac{1}{s-1}$ and $\frac{-L'(s,\chi)}{L(s,\chi)}$ being analytic at $s=1$ Jun 14, 2017 at 10:38