Im looking for closed forms for $\lim_{x \rightarrow \infty} \ln(x) \prod_{x>(p-a)>0}(1-(p-a)^{-1})$ where $x$ is a positive real, $a$ is a given real, $p$ is the set of primes such that the inequation is valid. Lets call this Limit $L(a)$. Mertens gave a closed form for $L(0)$. Are there others possible ?
Also the inverse function $L^{-1}(b)=a$ intrests me. What is the value of $L^{-1}(0)$ ??
I always take the log when I run into a product, so, pressing on regardless, let's look at $\begin{align} f(x, a) &=\sum_{x>(p-a)>0}\ln(1-(p-a)^{-1})\\ &=\sum_{a < p < x+a}\ln(1-(p-a)^{-1})\\ &=\sum_{a < p < x+a}-(\frac1{p-a}+\frac1{2(p-a)^2}+...)\\ &=-\sum_{a < p < x+a}\frac1{p-a}+C\\ &=C-\sum_{a < p < x+a}\frac1{p(1-a/p)}\\ &=C-\sum_{a < p < x+a}\frac1{p}(1+a/p+(a/p)^2+...)\\ &=C-\sum_{a < p < x+a}\frac1{p}+\sum_{a < p < x+a}\frac{a}{p^2} +\sum_{a < p < x+a}\frac{a^2}{p^3} +...\\ &=C_1-\sum_{a < p < x+a}\frac1{p} \end{align} $
where I have blithely absorbed the various convergent sums ($ \sum_{a < p < x+a}1/(p-a)^k$ and $\sum_{a < p < x+a}1/p^k$ for $k \ge 2$) into constants $C$ and $C_1$ which will depend on $a$.
Using the known estimate $\sum_{p < x} \frac1{p} \approx \ln \ln x$, $f(x, a) \approx C_1-\ln \ln(x+a)+\ln\ln a \approx C_2-\ln \ln(x+a) $.
Since $\ln(x+a) = \ln x + \ln(1+a/x) \approx \ln x+a/x $,
$\begin{align} \ln\ln(x+a) &\approx \ln(\ln x+a/x)\\ &= \ln(\ln x(1+a/(x \ln x))\\ &= \ln\ln x+ \ln(1+a/(x \ln x))\\ &\approx \ln\ln x+ a/(x \ln x)\\ \end{align} $
so $f(x, a) \approx C_2-\ln \ln(x+a) \approx C_2-\ln\ln x- a/(x \ln x) $.
So, $\ln x e^{f(x, a)} \approx e^{\ln \ln x + C_2-\ln\ln x} \approx C_3 $.
All this shows is that the limit approaches a constant, without giving much help in evaluating the constant.