5
$\begingroup$

Let $k\geqslant 2$ be an integer, suppose that $p_1,p_2,\dotsc,p_k$ are primes not exceeding $x$. Write $$ S_{k}(x) = \sum_{p_1 \leqslant x} \dotsb \sum_{p_k \leqslant x} \frac{1}{p_1+\dotsb +p_k}. $$ By AM-GM inequality, $p_{1}+\dotsb + p_{k} \geqslant k \sqrt[k]{p_{1}\dotsm p_{k}}$, we have $$ S_{k}(x) \leqslant \frac{1}{k} \sum_{p_{1}\leqslant x}\dotsb \sum_{p_{k} \leqslant x} \frac{1}{\sqrt[k]{p_{1}\dotsm p_{k}}} = \frac{1}{k} \left( \sum_{p \leqslant x} p^{-\frac{1}{k}} \right)^{k}. $$ By Prime Number Theorem and summation by parts we see that $$ \sum_{p \leqslant x} p^{-\frac{1}{k}} = \mathrm{Li}\big( x^{1-\frac{1}{k}} \big) + O \left( x^{1-\frac{1}{k}}\mathrm{e}^{-c\sqrt{\log x}} \right), $$ Here $\mathrm{Li}(x)$ is the logarithmic integral, and $\mathrm{Li}(x)\sim x/\log x$. Hence $$ S_{k}(x) \leqslant \left( \frac{k^{k-1}}{(k-1)^{k}} +o(1) \right) \frac{x^{k-1}}{\log^{k} x}. $$ On the other hand, $p_{1}+\dotsb +p_{k} \leqslant kx$, we have $$ S_{k}(x) \geqslant \frac{1}{kx} \sum_{p_{1} \leqslant x} \dotsb \sum_{p_{k} \leqslant x} 1 = \frac{1}{kx} \left( \sum_{p \leqslant x} 1 \right)^{k} = \frac{\pi^{k}(x)}{kx} = \frac{(1+o(1))}{k} \frac{x^{k-1}}{ \log^{k} x}. $$ My question is how to determine the coefficient of the main term of $S_{k}(x)$? Thanks!

$\endgroup$

2 Answers 2

6
$\begingroup$

Denote $\pi(x)=M\sim x/\log x$. Then $j$ varies between 1 and $M$, $p_j=j\log j+o(M\log M)$, and for $j_1,\ldots j_k$, denoting $j_i=Mt_i$ we have $$p_{j_1}+\ldots+p_{j_k}=\sum j_i\log j_i+o(M\log M)=M\log M\sum t_i+o(M\log M),$$ so your sum is the Riemann sum approximation of a certain integral: $$ (1+o(1))M^{k-1}(\log M)^{-1}\int_0^1\ldots \int_0^1 \frac{dt_1\ldots dt_k}{t_1+\ldots +t_k} $$ Thus the asymptotics of your sum is $c x^{k-1}/\log^{k} x$, where $c$ equals $$ c=\int_0^1\ldots \int_0^1 \frac{dt_1\ldots dt_k}{t_1+\ldots +t_k}= \int_0^1\ldots \int_0^1 {dt_1\ldots dt_k} \int_0^\infty e^{-(t_1+\ldots+t_k)x}dx= \int_0^\infty \left(\frac{1-e^{-x}}x\right)^kdx. $$

It may be evaluated using this method: Integral $\int_0^1 \int_0^1 \cdots \int_0^1\frac{x_{1}^2+x_{2}^2+\cdots+x_{n}^2}{x_{1}+x_{2}+\cdots+x_{n}}dx_{1}\, dx_{2}\cdots \, dx_{n}=?$

Namely, integrating by parts $k-1$ times we get $$c=\int_0^\infty \left(\frac{1-e^{-x}}x\right)^k dx=\frac1{(k-1)!}\int_0^\infty \frac{(d/dx)^{k-1}(1-e^{-x})^{k}}x dx.$$

Denote $\frac1{(k-1)!}(d/dx)^{k-1}(1-e^{-x})^{k}=\sum_{j=1}^k a_j e^{-jx}$. Then $\sum a_j=0$ (substitute $x=0$), so $(d/dx)^{k-1}(1-e^{-x})^{k}=\sum_{j=1}^k a_j (e^{-jx}-e^{-x})$ and we may integrate using the Frullani integral $\int_0^\infty \frac{e^{-jx}-e^{-x}}xdx=-\log j$. We get $$c=\sum_{j=2}^k -a_j\log j= \frac{1}{(k-1)!} \sum_{j=2}^k(-1)^{j+k}{k\choose j}j^{k-1} \log j.$$

This is probably not what you are happy with: it is not even seen from the explicit answer why $c$ is positive. For estimating $c$ for large $k$, you may use the Law of Large Numbers which ensures that $t_1+\ldots +t_k$ concentrates near $k/2$ that gives $c=2/k+o(1)$. It agrees with your bounds $1/k\leqslant c\leqslant (e+o(1))/k$.

$\endgroup$
1
  • $\begingroup$ @Petrov Thank you so much. $\endgroup$
    – YInt
    Commented Jun 27, 2020 at 14:53
2
$\begingroup$

Thank you, Mr. Petrov, but you made a little mistake.

A detailed calculation of $c$ is as follows:

Write $g(x)=(1-\mathrm{e}^{-x})^k= \sum\limits_{j=0}^{k} \binom{k}{j} (-1)^{j} \mathrm{e}^{-jx}$, integrating by parts we get \begin{align} \int_{0}^{\infty} g(x) x^{-k} \,\mathrm{d} x & = \int_{0}^{\infty} g(x) \,\mathrm{d} \left( \frac{x^{-k+1}}{-k+1} \right) \nonumber \\ & = \left. \frac{g(x)}{(-k+1)x^{k-1}} \right|_{0}^{\infty} + \frac{1}{k-1} \int_{0}^{\infty} \frac{g'(x)}{x^{k-1}} \mathrm{d} x, \end{align} since $\lim\limits_{x\to 0} \dfrac{g(x)}{x^{k-1}} = \lim\limits_{x\to +\infty} \dfrac{g(x)}{x^{k-1}} = 0$, so that \begin{align*} \frac{1}{k-1} \int_{0}^{\infty} \frac{g'(x)}{x^{k-1}} \mathrm{d} x & = \frac{1}{k-1} \int_{0}^{\infty} g'(x) \, \mathrm{d} \left( \frac{x^{-k+2}}{-k+2} \right) \\ & = - \left. \frac{g'(x)}{(k-1)(k-2)x^{k-2}} \right|_{0}^{\infty} + \frac{1}{(k-1)(k-2)} \int_{0}^{\infty} \frac{g''(x)}{x^{k-2}} \mathrm{d} x, \end{align*} where $g'(x)=k(1-\mathrm{e}^{-x})^{k-1}\cdot \mathrm{e}^{-x}$ and $\lim\limits_{x\to 0} \dfrac{-g'(x)}{(k-1)(k-2)x^{k-2}}= \lim\limits_{x\to +\infty} \dfrac{-g'(x)}{(k-1)(k-2)x^{k-2}}=0$. Hence, integrating by parts $k-1$ times gives \begin{align} \int_{0}^{\infty} \frac{\sum\limits_{j=0}^{k} \binom{k}{j} (-1)^{j}\mathrm{e}^{-jx}}{x^k} \, \mathrm{d} x & =\frac{1}{(k-1)!}\int_{0}^{\infty} \frac{\sum\limits_{j=0}^{k} \binom{k}{j} (-1)^j(-j)^{k-1} \mathrm{e}^{-jx}}{x} \,\mathrm{d} x \nonumber \\ & =\frac{1}{(k-1)!}\int_{0}^{\infty} \sum\limits_{j=1}^{k} \binom{k}{j} (-1)^{k+j-1}j^{k-1} \frac{\mathrm{e}^{-jx}}{x} \, \mathrm{d} x. \quad (\ast) \end{align} Notice that $(-1)^{k+j-1}=(-1)^{k+j+1}=-(-1)^{k-j}$, and consider the Stirling number of the second kind, we get \begin{align} \frac{1}{(k-1)!} \sum_{j=1}^{k} (-1)^{k+j-1} \binom{k}{j} j^{k-1} & = -k \cdot \frac{1}{k!} \sum_{j=1}^{k} (-1)^{k-j} \binom{k}{j} j^{k-1} \\ & = -k\cdot S(k-1,k)=0. \end{align} Set $\displaystyle a_{j} = \frac{(-1)^{k+j-1}j^{k-1}}{(k-1)!} \binom{k}{j}$, then $\sum\limits_{j=1}^{k} a_{j}=0$.

Using the Frullani's integral formula $\int_{0}^{\infty} \frac{\mathrm{e}^{-jx}- \mathrm{e}^{-Ax}}{x} \mathrm{d} x = \log A - \log j$ with $0<j<A$.

Write $(\ast)$ as \begin{align*} \int_{0}^{\infty} \sum_{j=1}^{k} a_{j} \frac{\mathrm{e}^{-jx}}{x} \mathrm{d} x & = \lim_{A\to + \infty} \int_{0}^{\infty} \sum_{j=1}^{k} a_{j} \frac{\mathrm{e}^{-jx}- \mathrm{e}^{-Ax}}{x} \mathrm{d} x \\ & = \lim_{A\to +\infty} \sum_{j=1}^{k} a_{j} (\log A - \log j) = - \sum_{j=1}^{k} a_{j} \log j, \end{align*} where $\lim\limits_{A\to +\infty} \sum\limits_{j=1}^{k} a_{j} \log A =0$. We obtain $$ \int_{0}^{\infty} \left(\frac{1-\mathrm{e}^{-x}}{x}\right)^k \,\mathrm{d}x = c = \frac{1}{(k-1)!} \sum_{j=2}^{k} (-1)^{k+j} j^{k-1} \binom{k}{j} \log j. $$

$\endgroup$
1
  • $\begingroup$ You are correct, I was sick in differentiating the exponent. Now fixed in my post too. $\endgroup$ Commented Jun 27, 2020 at 18:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.