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)$? 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
$$
A=\sum_{i=1}^{\lambda}\varphi(p_i-1)
$$
 A: 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
$$\begin{align}
\sum_{p\leq x}\phi(p-1)&=(x-1)\sum_{d\leq x}\frac{\mu(d)}d\pi(x,1,d)-\int_2^x \left(li(t) \sum_{d=1}^{\infty} \frac{\mu(d)}{d\phi(d)} + O\left(\frac t{\log^C t}\right) \right)dt\\
&=(x-1)li(x)\sum_{d=1}^{\infty}\frac{\mu(d)}{d\phi(d)}-\int_2^x li(t)\sum_{d=1}^{\infty} \frac{\mu(d)}{d\phi(d)}dt + O\left(\frac x{\log^C x}\right)\\
&=\mathcal{A}\cdot\left((x-1)li(x)-\int_2^x li(t) dt\right)+O\left(\frac x{\log^C x}\right)\\
&=\mathcal{A}\cdot\int_2^x \frac{t-1}{\log t} dt + O\left(\frac x{\log^C x}\right).
\end{align}$$
