Asymptotics for $\sum\frac{1}{1+\log (x/p)}$ I would like an asymptotic estimate for $$\sum_{p\leqslant x}\frac{1}{1+\log (x/p)},$$where $p$ always represents a prime number.
I tried making a Stieltjes integral out of the sum and evaluate it, but it didn't really work out. Any ideas?
 A: Converting it to an integral is a good idea. Indeed, we have
$$
\sum_{p\leq x}\frac{1}{1+\log (x/p)} =\pi (x) - \int_2^x {\frac{{\pi (t)}}{{(1 + \log (x/t))^2 }}\frac{{{\rm d}t}}{t}} .
$$
We shall estimate the integral. By the prime number theorem
$$
\pi (t) = \frac{t}{{\log t}} + \mathcal{O} \bigg( {\frac{t}{{\log ^2 t}}} \bigg),
$$
and hence
\begin{align*}
\int_2^x {\frac{{\pi (t)}}{{(1 + \log (x/t))^2 }}\frac{{{\rm d}t}}{t}} &= \int_2^x {\frac{1}{{(1 + \log (x/t))^2 }}\frac{{{\rm d}t}}{{\log t}}}  + \mathcal{O} (1)\int_2^x {\frac{1}{{(1 + \log (x/t))^2 }}\frac{{{\rm d}t}}{{\log ^2 t}}} 
\\ & = \int_2^x {\frac{1}{{(1 + \log (x/t))^2 }}\frac{{{\rm d}t}}{{\log t}}}  + \mathcal{O} \bigg( {\frac{x}{{\log ^2 x}}} \bigg).
\end{align*}
The estimate for the second integral may be shown by splitting the range of integration at $t=\sqrt{x}$ and estimating the two integrals separately. To estimate the first integral, we perform a change of variables from $t$ to $s$ via $t = x/s$:
$$
\int_2^x {\frac{1}{{(1 + \log (x/t))^2 }}\frac{{{\rm d}t}}{{\log t}}}  = \frac{x}{{\log x}}\int_1^{x/2} {\frac{1}{{(1 + \log s)^2 }}\frac{1}{{s^2 }}\frac{{{\rm d}s}}{{1 - \log s/\log x}}} .
$$
By the dominated convergence theorem
$$
\mathop {\lim }\limits_{x \to  + \infty } \int_1^{x/2} {\frac{1}{{(1 + \log s)^2 }}\frac{1}{{s^2 }}\frac{{{\rm d}s}}{{1 - \log s/\log x}}}  = \int_1^{ + \infty } {\frac{1}{{(1 + \log s)^2 }}\frac{{{\rm d}s}}{{s^2 }}} =\int_0^{ + \infty } {\frac{{{\rm e}^{ - t} }}{{(1 + t)^2 }}{\rm d}t},
$$
where in the second step we made a change of variables from $s$ to $t$ via $s={\rm e}^t$. Therefore,
$$
\sum_{p\leq x}\frac{1}{1+\log (x/p)} \sim \pi (x) - \int_0^{ + \infty } {\frac{{{\rm e}^{ - t} }}{{(1 + t)^2 }}{\rm d}t}\cdot \frac{x}{{\log x}}
$$
as $x\to +\infty$. Finally, since via integration by parts,
$$
\int_0^{ + \infty } {\frac{{{\rm e}^{ - t} }}{{(1 + t)^2 }}{\rm d}t}  = 1 - \int_0^{ + \infty } {\frac{{{\rm e}^{ - t} }}{{1 + t}}{\rm d}t} 
$$
and by the prime number theorem $\pi(x)\sim x/\log x$, we can write
$$\boxed{
\sum_{p\leq x}\frac{1}{1+\log (x/p)} \sim c\, \frac{x}{{\log x}}}
$$
as $x\to+\infty$, with
$$
c=\int_0^{ + \infty } {\frac{{{\rm e}^{ - t} }}{{1 + t}}{\rm d}t} =0.596347362323194\ldots \,.
$$
This constant $c$ is sometimes called the Gompertz constant.
Addendum. With the aid of this question,
$$
\frac{1}{{\pi (x)}}\sum\limits_{p \le x} {\frac{1}{{1 - \log (p/x)}}}  \to \int_0^1 {\frac{{{\rm d}s}}{{1 - \log s}}}  = \int_0^{ + \infty } {\frac{{{\rm e}^{ - t} }}{{1 + t}}{\rm d}s} .
$$
