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.
 A: 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 $x > 1$.
