Lets that $p_1,p_2, ...,p_\lambda>2$ be a set of prime numbers. Is there estimation for the summation of $ A=\sum_{i=1}^{\lambda}\varphi(p_i-1)$?

Question

Lets that $p_1,p_2, ...,p_\lambda>2$ be a set of primes number greater than $2$. Is there any exact formula or estimation for the summation $$ A=\sum_{i=1}^{\lambda}\varphi(p_i-1) $$

Answer 1

For simplicity of notation, let $x=p_{\lambda}$, and assuming that $p_i$'s are in an increasing order.

The idea comes from Mathematika 16 (1969) p178-188, Lemma 1, "An Average Results on Artin's Conjecture" by P. J. Stephens.

In the following, $B$ and $C$ are large positive constants, $\mathcal{A}$ is Artin's constant. That is, $$ \mathcal{A}=\sum_{d=1}^{\infty}\frac{\mu(d)}{d\phi(d)}=\prod_p \left(1-\frac1{p(p-1)} \right). $$ Denote by $\pi(x,1,d)$ the number of primes up to $x$ which are $1$ modulo $d$, and $li(x)=\int_2^x \frac1{\log t} dt$.

We begin with $$ \begin{align} \sum_{p\leq x} \phi(p-1)&=\sum_{p\leq x} (p-1)\sum_{d|p-1} \frac{\mu(d)}d\\ &=\sum_{d\leq x} \frac{\mu(d)}d\sum_{p\leq x, p\equiv 1 (d)} (p-1), \ \ (1) \end{align} $$ We treat the inner sum with Stieltjes integral $$\begin{align} \sum_{p\leq x, p\equiv 1(d) }(p-1)&=\int_{2-}^x (t-1)d\pi(t,1,d)\\ &=(t-1)\pi(t,1,d)|_{2-}^x - \int_2^x\pi(t,1,d)dt\\ &=(x-1)\pi(x,1,d)-\int_2^x\pi(t,1,d)dt, \ \ (2)\end{align} $$ Now, consider the sum over $d$ of the integral in (2). $$\begin{align} \sum_{d\leq x}\frac{\mu(d)}d \int_2^x\pi(t,1,d) dt&=\int_2^x \sum_{d\leq x}\frac{\mu(d)}d \pi(t,1,d) dt \\ &=\int_2^x\sum_{d\leq t}\frac{\mu(d)}d \pi(t,1,d)dt\end{align}, \ \ (3) $$ Splitting the sum in (3) into $d\leq \log^B t$ and $\log^B t<d\leq t$, then using Siegel Walfisz theorem, we obtain $$ \sum_{d\leq t}\frac{\mu(d)}d \pi(t,1,d)=li(t) \sum_{d=1}^{\infty} \frac{\mu(d)}{d\phi(d)}+O\left(\frac t{\log^C t}\right), \ \ (4). $$ Inserting this into (3), then (2), and then (1), we have $$\begin{align} \sum_{p\leq x}\phi(p-1)&=(x-1)\sum_{d\leq x}\frac{\mu(d)}d\pi(x,1,d)-\int_2^x \left(li(t) \sum_{d=1}^{\infty} \frac{\mu(d)}{d\phi(d)} + O\left(\frac t{\log^C t}\right) \right)dt\\ &=(x-1)li(x)\sum_{d=1}^{\infty}\frac{\mu(d)}{d\phi(d)}-\int_2^x li(t)\sum_{d=1}^{\infty} \frac{\mu(d)}{d\phi(d)}dt + O\left(\frac x{\log^C x}\right)\\ &=\mathcal{A}\cdot\left((x-1)li(x)-\int_2^x li(t) dt\right)+O\left(\frac x{\log^C x}\right)\\ &=\mathcal{A}\cdot\int_2^x \frac{t-1}{\log t} dt + O\left(\frac x{\log^C x}\right). \end{align}$$