Looking an asymptotic for $\sum_{k\leq x}\Lambda(k)e^k$, where $\Lambda (n)$ is the von Mangoldt function

Using Abel's identity (see Theorem 4.2 in page 77 of [1]) and Prime Number Theorem (Theorem 4.4 in page 75) I compute

$$\frac{1}{x}\sum_{k\leq x}\Lambda(k)e^k\sim 1\cdot e^x-\frac{1}{x}\int_1^x \psi(t)e^t dt,$$

for the Mangoldt function $\Lambda (n)$ which is equal to zero unless that $n=p^a$ for some prime number $p$ and an integer $a\geq 1$, in this case its value is $\log p$ (see Wikipedia), and $\psi(n)$ is the second Chebyshev function defined for $x>0$ as $\sum_{n\leq x}\Lambda(n)$. My goal is to refresh and learn more mathematics, this is my

Question. What is a good asymptotic in the infinity (to continue previous computations with the equivalence $\sim$) for the term $(1/x)\cdot\int_1^x \psi(t)e^t dt$? Thanks in advance.

My attempt at this moment is tedious and non-effective: split $\int_1^x=\int_1^A+\int_A^x$, then another time by PNT, $\forall \epsilon>0$ $\exists A>0$ (the previous quantity) such that $\forall t>A$ we have $|\frac{\psi (t)}{t}-1|<\epsilon$, thus $|\psi(t)-t|<\epsilon t$. Now $\psi(t)e^t=(\psi(t)-t+t)e^t$ and I can write after several computations when I use inequalities involving Chebyshev $\vartheta-$function (Theorem 4.1, page 76, and using $\vartheta(n)\leq 4n\log 2$ in page 83 as $\vartheta(x)\leq 4x\log 2$)

$$\int_1^A(\psi(t)-t)e^t dt\leq \int_1^A t\log t e^t dt+ \int_1^A\frac{\sqrt{t}(\log t)^2}{2\log 2}e^t dt-\int_1^A t e^t dt.$$

Still I don't use Prime Number Theorem, perhaps in another way I can use. Remains compute the asyptotic behaviour of previous integrals and of course the terms $\int_A^x$. Thus I believe that perhaps isn't a good way.

Appendix: From previous identity we obtain by substitution $x=\psi(n)$ for a fixed integer $n>1$, thus $e^{\psi(n)}=l.c.m(1,2,\cdots,n)$ is an integer,

$$\psi(\psi(n))\cdot l.c.m(1,2,\cdots,n)=\sum_{k\leq \psi(n)}\Lambda(k)e^k+\int_1^{\psi(n)} \psi(t)e^t dt,$$

or by substitution $x=e^{\psi(n)}$ (you too can write even $l.c.m(1,2,\cdots,n)$)

$$\psi(e^{\psi(n)})e^{l.c.m(1,2,\cdots,n)}=\sum_{k\leq l.c.m(1,2,\cdots,n)}\Lambda(k)e^k+\int_1^{e^{\psi(n)}} \psi(t)e^t dt.$$

Question (Optional). How obtain bounds for $\sum_{k\leq y} \Lambda(k)e^k$, for a real $y=x$ and for a positive integer $y=N$?

References:

[1] Apostol, Introduction to Analytic Number Theory, Springer.

[2] Wikipedia.

There are a lot of questions in your post, but I think I can answer a couple.

The behavior of this sum appears to be highly irregular, so it does not seem like there will be a simple asymptotic for it. Here's a plot of

$$\frac{1}{\log(x)e^x} \sum_{k \leq x} \Lambda(k) e^k$$

for $10 \leq x \leq 2000$:

As far as an upper bound goes, we know that

$$\Lambda(2k+1) \leq \log(2k+1),$$

and

$$\Lambda(2k) \leq \log(2),$$

so

\begin{align} \sum_{k \leq x} \Lambda(k)e^k &= \sum_{\substack{k \leq x \\ k \text{ odd}}} \Lambda(k)e^k + \sum_{\substack{k \leq x \\ k \text{ even}}} \Lambda(k)e^k \\ &\leq \sum_{\substack{k \leq x \\ k \text{ odd}}} \log(k)e^k + \log(2)\sum_{\substack{k \leq x \\ k \text{ even}}} e^k. \tag{1} \end{align}

Now

\begin{align} \sum_{\substack{k \leq x \\ k \text{ even}}} e^k &= \sum_{k \leq x/2} e^{2k} \\ &= \frac{e^2}{e^2-1} \left( e^{2\lfloor x/2\rfloor} - 1 \right) \\ &< \frac{e^2}{e^2-1} e^x \tag{2} \end{align}

and similarly

\begin{align} \sum_{\substack{k \leq x \\ k \text{ odd}}} \log(k)e^k &= \sum_{k \leq (x-1)/2} \log(2k+1)e^{2k+1} \\ &< \log(x)\sum_{k \leq (x-1)/2} e^{2k+1} \\ &< \frac{e^2}{e^2-1} \log(x) e^x. \tag{3} \end{align}

In fact it can be shown that

$$\sum_{\substack{k \leq x \\ k \text{ odd}}} \log(k)e^k \sim \frac{e^2}{e^2-1} \log(x)e^{2\lfloor (x-1)/2 \rfloor + 1} \quad \text{as } x \to \infty,$$

so this bound is sharp.

By substituting $(2)$ and $(3)$ in $(1)$ we get the upper bound

$$\sum_{k \leq x} \Lambda(k)e^k < \frac{e^2}{e^2-1} \log(2x)e^x. \tag{4}$$

It follows that

$$\limsup_{x \to \infty} \frac{1}{\log(x)e^x} \sum_{k \leq x} \Lambda(k)e^k \leq \frac{e^2}{e^2-1}. \tag{5}$$

To illustrate the sharpness of this bound, here is a plot of

$$\frac{1}{\log(x)e^x} \sum_{k \leq x} \Lambda(k) e^k$$

in $\color{blue}{\text{blue}}$ versus

$$\frac{e^2}{e^2-1}$$

in $\color{red}{\text{red}}$:

Notes.

I think it should be possible to show that $(5)$ is true without the $\limsup$. Specifically, that

$$\frac{1}{\log(x)e^x} \sum_{k \leq x} \Lambda(k)e^k < \frac{e^2}{e^2-1} \tag{6}$$

for all $x > 1$.

• I understand virtually all calculations and it is very nice and useful the method that you show to compute sums involving von Mangoldt's function. I feel strongly surprised by your ability and time that you taken in computations and reasoning. These are your maths and are wonderful (I'd taken notes from your method), very thanks much you @AntonioVargas. – user243301 Sep 29 '15 at 5:54
• Thanks Juan, I really appreciate the kind words. I'm glad to help. I also added that part to the end of my post that I mentioned last night (from equation $(7)$ to equation $(9)$). – Antonio Vargas Sep 29 '15 at 15:40
• I will study this in detail, very thanks much. – user243301 Sep 29 '15 at 18:30
• There is an error in the last part, I'm working on a correction. – Antonio Vargas Sep 29 '15 at 22:01
• @AntonioVargas That won't be the case either. It's going to be based on the admissible set constructed inductively starting at 2 by adding the smallest element so that the set is still admissible, and then taking a summing over this set in a limit as its size goes to infinity. – Eric Naslund Sep 30 '15 at 15:16