Asymptotic with sharp error term of $\sum_{p} \frac{\log p}{p^2} e^{-x/p^2}$ as $x\to\infty$
问题内容
Define $$S(x)=\sum_{p} \frac{\log p}{p^2} e^{-x/p^2}$$ where p denotes prime. I need asymptotic expansion of $S(x)$ with sharp error term as $x\to\infty$.
Define $v(t)=\sum_{p\leq t} \log p$, then by Euler summation we have
$$S(x)=\lim_{T\to\infty}\left(v(T)\frac{e^{-x/T^2}}{T^2}+2\int_2^T v(t)e^{-x/t^2}\left(\frac{1}{t^3}-\frac{x}{t^5}\right)dt\right)$$
Since by PNT $v(t)\sim t$ as $t\to\infty$, we have $\lim_{T\to\infty}v(T)\frac{e^{-x/T^2}}{T^2}=0$ and hence we obtain
$$S(x)=2\int_2^\infty v(t)e^{-x/t^2}\left(\frac{1}{t^3}-\frac{x}{t^5}\right)dt$$
Split the above integral as $$S(x)=S_1(x)+S_2(x)$$ where $$S_1(x)=2\int_2^\infty te^{-x/t^2}\left(\frac{1}{t^3}-\frac{x}{t^5}\right)dt$$ and $$S_2(x)=2\int_2^\infty (v(t)-t)e^{-x/t^2}\left(\frac{1}{t^3}-\frac{x}{t^5}\right)dt$$
In $S_1(x)$, substitute $t=\sqrt{x}/u$ so that $$S_1(x)=\frac{2}{\sqrt{x}}\int_0^{\sqrt{x}/2}(1-u^2)e^{-u^2}du$$
Now write $S_1(x)$ as
$$S_1(x)=\frac{2}{\sqrt{x}}\int_0^{\infty}(1-u^2)e^{-u^2}du-\frac{2}{\sqrt{x}}\int_{\sqrt{x}/2}^\infty(1-u^2)e^{-u^2}du$$
We have on integrating the first integral,
$$S_1(x)=\frac{\sqrt{\pi}}{2\sqrt{x}}-\frac{2}{\sqrt{x}}\int_{\sqrt{x}/2}^\infty(1-u^2)e^{-u^2}du$$
Write $$S_1(x)=\frac{\sqrt{\pi}}{2\sqrt{x}}+T(x)
$$
Now $$|T(x)|\leq \frac{2}{\sqrt{x}}\int_{\sqrt{x}/2}^\infty e^{-u}du=\frac{2 e^{-\sqrt{x}/2}}{\sqrt{x}}$$
So we have $$T(x)=O\left(\frac{e^{-\sqrt{x}/2}}{\sqrt{x}}\right)$$ as $x\to\infty$. We have $$S(x)=\frac{\sqrt{\pi}}{2\sqrt{x}}+S_2(x)+O\left(\frac{e^{-\sqrt{x}/2}}{\sqrt{x}}\right)$$ as $x\to\infty$.
Any help would be highly appreciated. Thanks
回答 (2)
Define$$\mathcal{P}(s) = \sum_{p} \frac{\log p}{p^s}$$ By Mellin Inversion, we have $$S(x) = \frac{1}{2\pi i} \int_{(c)} \Gamma(s) x^{-s} \mathcal{P}(2 - 2s) \, ds, \quad 0 < c < \frac{1}{2}$$ Also since $-\frac{\zeta'}{\zeta}(s) = \sum_{m \ge 1} \mathcal{P}(ms)$ $$\mathcal{P}(s) = -\frac{\zeta'}{\zeta}(s) + \frac{\zeta'}{\zeta}(2s) + H(s)$$ Where $H$ is holomorphic for $\Re(s) > 1/3$. Thus, assuming RH, we cross the poles $$s=\frac12,\,\frac34,\,1-\frac \rho2$$ If we shift the contour to $\Re(s) = 4/5$. The $s=1/2$ pole contributes $$\frac{\sqrt{\pi}}{2} x^{-1/2}$$ All other crossed singularities lie on $\Re(s) = 3/4$. Their total contribution is $O(x^{-3/4})$ since the factor $\Gamma\left(1 - \frac{\rho}{2}\right)$ decays exponentially with $\vert{}\Im(\rho)\vert{}$. The integral on the new line is $O(x^{-4/5})$, therefore $$S(x) = \frac{\sqrt{\pi}}{2}x^{-1/2} + O\left(x^{-3/4}\right)$$
One intgration by parts gives $$I=\int(1-u^2)\,e^{-u^2}\,du=\frac{1}{2} u\,e^{-u^2} +\frac{ \sqrt{\pi } }{4}\,\text{erf}(u)$$
$$S_1(x)=\frac{2}{\sqrt{x}}\int_0^{\sqrt{x}/2}(1-u^2)\,e^{-u^2}\,du=\frac 12\Bigg(e^{-x/4}+\sqrt{\frac{\pi}{x}}\,\text{erf}\left(\frac{\sqrt{x}}{2}\right) \Bigg)$$ and if $x$ is large $$S_1(x)=\frac 12\sqrt{\frac{\pi}{x}}+ \frac{e^{-x/4} } {2} \Big( 1-\frac{2}{x}+\frac{4}{x^2}+O\left(\frac{1}{x^3}\right) \Big)$$ seems to be tighter.