A. I. VINOGRADOV

ON THE REMAINDER IN MERTENS’ FORMULA

(Presented by Academician I. M. Vinogradov on 30 VI 1962)

In the note \((^2)\) an improvement was given of the remainder term in the classical Mertens formula for a segment of the Euler product at the point one. Here it will be shown how, using essentially the same method with a small modification, the remainder term in Mertens’ formula can be replaced by a quantity of order

\[ \exp[-a(\ln x)^{3/5}], \]

where \(a\) is an absolute constant.

Instead of the simple product \(\Pi(s,x)\) of the note \((^2)\), consider the averaged product

\[ \Pi_c(s,x)=\left[\prod_{n\le x}\prod_{p\le n}\left(1-\frac{1}{p^s}\right)^{-1}\right]^{1/x} \tag{1} \]

under the condition that \(x\) is an integer. For it the following is true.

Theorem. Let \(s=\sigma+it,\ |t|\le x,\ \sigma\ge 1-\dfrac{\lambda}{(\ln x)^{2/3}}\); then in this domain the equality

\[ \Pi_c^{-1}(s,x)\zeta(s)(s-1)=\frac{e^{-c}}{\ln x}\,e^{\omega_0(s,x)}(1+\theta(s,x)), \tag{2} \]

holds, where

\[ \omega_0(s,x)=\int_{L_0}\frac{x^{1-w}-1}{w-1}\,dw+\int_{L_1}\frac{x^{1-w}}{2-w}\,dw, \]

\(L_0\) is the straight line segment joining the points \(s\) and \((1,0)\); \(L_1\) is a straight ray issuing from the point \(s\) and not passing through the point \((2,0)\); \(\theta(s,x)\) is a function analytic in \(s\) and having in the indicated domain the estimate

\[ |\theta(s,x)|\ll \exp[-a(\ln x)^{3/5}]. \]

Proof. The product (1) is transformed into the form

\[ \Pi_c(s,x)=\prod_{p\le x}\left(1-\frac{1}{p^s}\right)^{-(1-p/x)}. \]

Let us find the logarithmic derivative of \(\Pi_c(s,x)\):

\[ \frac{\Pi_c'}{\Pi_c}(s,x)=-\sum_{p\le x}\frac{(1-p/x)}{p^s-1}. \]

The sum, with the aid of the operator

\[ \frac{1}{2\pi i}\int_{1-iT}^{1+iT}\frac{Y^w}{w(w+1)}\,dw \]

is easily brought to the form

\[ \frac{1}{2\pi i}\int_{1-iT}^{1+iT}\frac{x^w}{w(w+1)}\frac{\zeta'}{\zeta}(s+w)\,dw+\theta_1(s,x), \tag{3} \]

where \(\theta_1(s,x)\) is an analytic function having the estimate

\[ |\theta_1(s,x)|\ll x^{-\sigma+1/2}\ln x. \]

Next, applying to the integral (3) the classical contour-shifting technique, set out in detail in (³), and the technique of the note (²), we obtain the theorem.

Corollary. The relation

\[ \prod_{p \leqslant x}\left(1-\frac{1}{p}\right) = \frac{e^{-c}}{\ln x} \left(1+O\{\exp[-a(\ln x)^{3/5}]\}\right). \tag{4} \]

is valid.

Proof. Letting \(s \to 1\) in (2), we obtain

\[ \prod_{p \leqslant x}\left(1-\frac{1}{p}\right)^{1-p/x} = \frac{e^{-c}}{\ln x} \exp\left(\int_{L_1}\frac{x^{1-w}}{2-w}\,dw\right) \left(1+O\{\exp[-a(\ln x)^{3/5}]\}\right), \tag{5} \]

where \(L_1\) is a ray issuing from the unit point. The integral over \(L_1\) is transformed into the form

\[ \int_{L_1}\frac{x^{1-w}}{2-w}\,dw = \frac{1}{x}\int_{2}^{x}\frac{dy}{\ln y} + O\left(\frac{\ln x}{x}\right). \tag{6} \]

On the other hand,

\[ \prod_{p \leqslant x}\left(1-\frac{1}{p}\right)^{1-p/x} = \prod_{p \leqslant x}\left(1-\frac{1}{p}\right) \exp\left(\frac{\pi(x)}{x}+O\left(\frac{\ln x}{x}\right)\right). \tag{7} \]

Substituting (6) and (7) into (5), we obtain

\[ \prod_{p \leqslant x}\left(1-\frac{1}{p}\right) = \frac{e^{-c}}{\ln x} \exp\left( \frac{\pi(x)-\displaystyle\int_{2}^{x}\frac{dy}{\ln y}}{x} \right) \left(1+O\{\exp[-a(\ln x)^{3/5}]\}\right). \]

But, as is known from the latest theorems of I. M. Vinogradov on the boundary of the zeros for \(\zeta(s)\),

\[ \pi(x)-\int_{2}^{x}\frac{dy}{\ln y} = O\left(x\exp[-a(\ln x)^{3/5}]\right). \]

Consequently, equality (4) is indeed valid.

CITED LITERATURE

¹ I. M. Vinogradov, Izv. AN SSSR, ser. matem., 22, No. 2, 161 (1958).
² A. I. Vinogradov, DAN, 143, No. 5 (1962).
³ A. Ingham, Distribution of Prime Numbers, IL, 1936.