First, this preparation. We have an earlier result, that
\begin{equation*}
\sum_{n \leq x} \log n = x \log x - x + O(\log x)\text{,}
\end{equation*}
see
(4.2.10). (The details of the proof were left as an exercise, but basically this followed from comparing the sum with the integral
\(\int_1^x \log t dt \text{.}\)) We will combine this with the following lemma.
Proof of (6.3.1).
We claim that \(\sum_{n \leq x} \frac{\Lambda(n)}{n} = \log x + O(1)\text{.}\)
Start from
\begin{equation*}
\sum_{d \leq x} \Lambda(d) \left\lfloor \frac{x}{d} \right\rfloor = \sum_{n \leq x} \log n = x \log x - x + O(\log x)\text{.}
\end{equation*}
Dropping the floor operator gives
\begin{align*}
x \log x - x + O(\log x) \amp = \sum_{d \leq x} \Lambda(d) \left( \frac{x}{d} + O(1) \right) \\
\amp = x \sum_{d \leq x} \frac{\Lambda(d)}{d} + \sum_{d \leq x} \Lambda(d) O(1) \\
\amp = x \sum_{d \leq x} \frac{\Lambda(d)}{d} + O(\psi(x)) \text{.}
\end{align*}
By Chebyshev’s estimate, \(\psi(x) = O(x)\text{.}\) Therefore
\begin{equation*}
x \sum_{d \leq x} \frac{\Lambda(d)}{d} = x \log x - x + O(\log x) + O(x) = x \log x + O(x) \text{.}
\end{equation*}
Dividing by \(x\) gives the result.
Proof of (6.3.2).
We claim that \(\sum_{p \leq x} \frac{\log p}{p} = \log x + O(1) \text{.}\)
We have
\begin{align*}
\sum_{n \leq x} \frac{\Lambda(n)}{n} \amp = \sum_{p^k \leq x, k \geq 1} \frac{\log p}{p^k} \\
\amp = \sum_{p \leq x} \frac{\log p}{p} + \sum_{p^k \leq x, k \geq 2} \frac{\log p}{p^k} \text{.}
\end{align*}
Our goal is to show that the last sum is bounded. We have
\begin{align*}
\sum_{p^k \leq x, k \geq 2} \frac{\log p}{p^k}
\amp \leq \sum_{p \leq \sqrt{x}} \sum_{k=2}^{\infty} \frac{\log p}{p^k} \\
\amp = \sum_{p \leq \sqrt{x}} (\log p) \frac{1/p^2}{1-1/p} \\
\amp = \sum_{p \leq \sqrt{x}} \frac{\log p}{p(p-1)} \\
\amp \leq \sum_{n \leq \sqrt{x}} \frac{\log n}{n(n-1)} \text{.}
\end{align*}
This last sum is convergent: for large enough \(n\text{,}\) \(\log n \leq \sqrt{n}\text{,}\) and we can compare the resulting series \(\sum \frac{\sqrt{n}}{n(n-1)}\) to a \(p\)-series.
So, there is a constant \(c\) with \(\sum_{n \leq x} \frac{\log n}{n(n-1)} \leq c\text{,}\) for all \(x\text{.}\) In other words, this last sum is \(O(1)\) (as a function of \(x\)). Therefore
\begin{equation*}
\sum_{p \leq x} \frac{\log p}{p} = \sum_{n \leq x} \frac{\Lambda(n)}{n} + O(1) = \log x + O(1)
\end{equation*}
Proof of (6.3.3).
We claim that \(\sum_{p \leq x} \frac{1}{p} = \log \log x + b + O\left(\frac{1}{\log x}\right) \text{.}\)
Set \(A(x) = \sum_{p \leq x} \frac{\log p}{p} \text{.}\) We have just seen that \(A(x) = \log x + O(1)\text{.}\) We will write this as
\begin{equation*}
A(x) = \log x + R(x)\text{,}
\end{equation*}
where \(R = O(1)\text{.}\)
We can write the sum we are interested in as follows:
\begin{equation*}
\sum_{p \leq x} \frac{1}{p} = \sum_{p \leq x} \frac{\log p}{p} \frac{1}{\log p}\text{.}
\end{equation*}
We can rewrite this as an integral, specifically as a Riemann-Stieltjes integral. First, we find the differential \(dA(t)\text{.}\) For \(\Delta \gt 0\text{,}\)
\begin{equation*}
A(t+\Delta) - A(t) = \sum_{p \leq t + \Delta} \frac{\log p}{p} - \sum_{p \leq t} \frac{\log p}{p}
= \sum_{t \lt p \leq t + \Delta} \frac{\log p}{p} \text{.}
\end{equation*}
But, for any value of \(t\text{,}\) and small enough \(\Delta\text{,}\) there are no primes (in fact, not even any integers) in the interval \((t,t+\Delta]\text{.}\) On the other hand, for \(\Delta \gt 0\text{,}\)
\begin{equation*}
A(t) - A(t-\Delta) = \sum_{t-\Delta \lt p \leq t} \frac{\log p}{p} \text{.}
\end{equation*}
In the limit as \(\Delta \to 0^+\text{,}\) this is equal to \(\frac{\log p}{p}\) if \(t = p\) is prime, or equal to \(0\) if \(t\) is not equal to a prime.
So, our sum is equal to
\begin{equation*}
\sum_{p \leq x} dA(p) \frac{1}{\log p} = \int_{2^-}^x \frac{1}{\log t} dA(t)\text{.}
\end{equation*}
Here the lower bound of integration \(2^-\) means \(2-\epsilon\) for sufficiently small \(\epsilon\text{,}\) in fact the limit as \(\epsilon \to 0\text{.}\) (We want the interval of integration to start to the left of \(2\) because we are using left-hand differentials for \(dA\text{.}\))
Next, we evaluate the integral, starting with integration by parts:
\begin{equation*}
\int_{2^-}^x \frac{1}{\log t} dA(t) = \left. \frac{1}{\log t} A(t) \right|_{2^-}^x - \int_{2^-}^x \frac{-1}{t(\log t)^2} A(t) dt\text{.}
\end{equation*}
We have \(A(2-\epsilon) = 0\) so the first term is simply \(\frac{A(x)}{\log x}\text{.}\) From this point, we can take the integral to be simply \(\int_2^x\text{.}\) We use \(A(x) = \log x + R(x) = \log x + O(1)\text{:}\)
\begin{equation*}
\frac{A(x)}{\log x} = \frac{\log x + O(1)}{\log x} = 1 + O\left(\frac{1}{\log x}\right)
\end{equation*}
and
\begin{equation*}
\int_2^x \frac{A(t)}{t(\log t)^2} dt = \int_2^x \frac{1}{t(\log t)} + \frac{R(t)}{t(\log t)^2} dt\text{.}
\end{equation*}
The integral of the first term can be evaluated (by substitution with \(u = \log t\)):
\begin{equation*}
\int_2^x \frac{1}{t(\log t)} dt = \left. \log \log t \right|_2^x = \log \log x - \log \log 2\text{.}
\end{equation*}
Before we address the integral of the second term, let us put together what we have seen so far:
\begin{equation*}
\sum_{p \leq x} \frac{1}{p} = 1 + O\left(\frac{1}{\log x}\right) + \log \log x - \log \log 2 + \int_2^x \frac{R(t)}{t(\log t)^2} dt\text{.}
\end{equation*}
Now we consider the integral appearing as the last term on the right. By the same substitution again we see
\begin{equation*}
\int \frac{1}{t(\log t)^2} dt = - \frac{1}{\log t} + C\text{.}
\end{equation*}
Hence the following improper integral converges:
\begin{equation*}
\int_2^{\infty} \frac{1}{t(\log t)^2} dt = \lim_{t \to \infty} \frac{1}{\log 2} - \frac{1}{\log t} = \frac{1}{\log 2}\text{.}
\end{equation*}
This implies \(\int_x^{\infty} \frac{1}{t(\log t)^2} dt\) converges as well. Since \(R(t) = O(1)\text{,}\) it implies that the following improper integrals also converge:
\begin{equation*}
\int_2^{\infty} \frac{R(t)}{t(\log t)^2} dt, \int_x^{\infty} \frac{R(t)}{t(\log t)^2} dt
\end{equation*}
(but of course, the value might not be \(\frac{1}{\log 2}\) any more). Since the integrals converge, it is valid to write
\begin{equation*}
\int_2^x \frac{R(t)}{t(\log t)^2} dt = \int_2^{\infty} \frac{R(t)}{t (\log t)^2} dt - \int_x^{\infty} \frac{R(t)}{t (\log t)^2} dt\text{.}
\end{equation*}
The integral from \(2\) to \(\infty\) is constant with respect to \(x\text{.}\) For the other interval, since \(R(t) \ll 1\) (recall that this notation means \(R(t) = O(1)\)), we have
\begin{equation*}
\int_x^{\infty} \frac{R(t)}{t (\log t)^2} dt \ll \int_x^{\infty} \frac{1}{t(\log t)^2} dt = \frac{1}{\log x}
\end{equation*}
or in other words the integral on the left hand side is \(O(1/\log x)\text{.}\)
Putting together our results so far we have
\begin{equation*}
\sum_{p \leq x} \frac{1}{p} = 1 + O\left(\frac{1}{\log x}\right) + \log \log x - \log \log 2
+ \int_2^{\infty} \frac{R(t)}{t(\log t)^2} dt + O\left(\frac{1}{\log x}\right)\text{.}
\end{equation*}
To finish, we combine big-O terms, and set
\begin{equation*}
b = 1 - \log \log 2 + \int_2^{\infty} \frac{R(t)}{t (\log t)^2} dt
\end{equation*}
to get
\begin{equation*}
\sum_{p \leq x} \frac{1}{p} = \log \log x + b + O\left(\frac{1}{\log x}\right)
\end{equation*}
as claimed.
Proof of (6.3.4).
We claim that \(\prod_{p \leq x} \left( 1 - \frac{1}{p} \right)^{-1} = e^c \log x + O(1) \text{.}\)
First recall the following Taylor series for logarithm:
\begin{equation}
-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \dotsb
= \sum_{n=1}^{\infty} \frac{x^n}{n}\text{.}\tag{6.3.8}
\end{equation}
Consider the logarithm of the product on the left:
\begin{equation}
\begin{split}
\log \prod_{p \leq x} \left( 1 - \frac{1}{p} \right)^{-1}
\amp = \sum_{p \leq x} - \log \left( 1 - \frac{1}{p} \right) \\
\amp = \sum_{p \leq x} \sum_{n=1}^{\infty} \frac{1}{n p^n} \\
\amp = \sum_{p \leq x} \frac{1}{p}
+ \sum_{p \leq x} \sum_{n=2}^{\infty} \frac{1}{n p^n}
\end{split}\text{.}\tag{6.3.9}
\end{equation}
Use
(6.3.3) for the first term. For the second term, observe that
\(\sum_p \sum_{n = 2}^{\infty} \frac{1}{n p^n}\) converges. To see this, start with
\begin{equation}
\sum_{n=2}^{\infty} \frac{1}{n p^n} \leq \sum_{n=2}^{\infty} \frac{1}{p^n}
= \frac{1}{p(p-1)} = \frac{1}{p-1} - \frac{1}{p}\tag{6.3.10}
\end{equation}
and then observe that the sum over all positive integers \(p\) converges by telescoping; this implies that the sum over just the primes also converges. So we can replace the sum over \(p \leq x\) with the difference between the sum over all \(p\text{,}\) and the sum over \(p \gt x\text{.}\) This gives:
\begin{equation}
\log \prod_{p \leq x} \left( 1 - \frac{1}{p} \right)^{-1}
= \log \log x + b + \sum_p \sum_{n=2}^{\infty} \frac{1}{n p^n}
- \sum_{p \gt x} \sum_{n=2}^{\infty} \frac{1}{n p^n}
+ O\left( \frac{1}{\log x} \right)\text{.}\tag{6.3.11}
\end{equation}
In fact,
\begin{equation}
\sum_{p \gt x} \sum_{n=2}^{\infty} \frac{1}{n p^n}
\leq \sum_{p \gt x} \frac{1}{p(p-1)}
\leq \sum_{n \gt x} \frac{1}{n(n-1)}
\ll \frac{1}{x} \ll \frac{1}{\log x}\text{.}\tag{6.3.12}
\end{equation}
\begin{equation}
\log \prod_{p \leq x} \left( 1 - \frac{1}{p} \right)^{-1}
= \log \log x + c + O\left( \frac{1}{\log x} \right)\tag{6.3.13}
\end{equation}
where \(c = b + \sum_p \sum_{n=2}^{\infty} \frac{1}{n p^n}\text{.}\)
Taking the exponential of each side gives
\begin{equation}
\prod_{p \leq x} \left( 1 - \frac{1}{p} \right)^{-1}
= \exp(\log \log x + c + f(x))
= e^c \log x \cdot \exp(f(x))\tag{6.3.14}
\end{equation}
where \(f(x) = O(1/\log x)\text{.}\) We have
\begin{equation}
\exp(f(x)) = 1 + f(x) + \frac{1}{2} f(x)^2 + \dotsb = 1 + O\left(\frac{1}{\log x}\right)\text{.}\tag{6.3.15}
\end{equation}
Thus
\begin{equation}
\prod_{p \leq x} \left( 1 - \frac{1}{p} \right)^{-1}
= e^c \log x \left(1 + O\left(\frac{1}{\log x}\right)\right)
= e^c \log x + O(1)\text{,}\tag{6.3.16}
\end{equation}
as claimed.
