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!


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 $1/k\leqslant c\leqslant (e+o(1))/k$.

  • $\begingroup$ @Petrov Thank you so much. $\endgroup$ – YInt Jun 27 '20 at 14:53

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 $$ \int_{0}^{\infty} \left(\frac{1-\mathrm{e}^{-x}}{x}\right)^k \,\mathrm{d}x = c = \frac{1}{(k-1)!} \sum_{j=2}^{k} (-1)^{k+j} j^{k-1} \binom{k}{j} \log j. $$

  • $\begingroup$ You are correct, I was sick in differentiating the exponent. Now fixed in my post too. $\endgroup$ – Fedor Petrov Jun 27 '20 at 18:03

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