The Chebyshev’s $\psi$-function and Chebshev’s $\theta$-function are defined as
\[
\psi(x) = \sum_{p^k \leq x} \log p, \qquad \theta(x) = \sum_{p \leq x} \log p.
\] We can rewrite $\psi(x)$ in terms of von Mangoldt function as
\[
\psi(x) = \sum_{n \leq x} \Lambda(n).
\] The Chebyshev functions are related to each other via Möbius inversion as shown in the following result.
Theorem. For $x > 0$ we have
\[\begin{align}
\psi(x) &= \sum_{k} \theta(x^{1/k}), \tag{1} \\
\theta(x) &= \sum_{k} \mu(k) \psi(x^{1/k}) \tag{2}
\end{align}\]
Proof. Note that
\[
\psi(x) = \sum_{p^k \leq x} \log p = \sum_{k} \sum_{p \leq x^{1/k}} \log p = \sum_{k} \theta(x^{1/k}).
\] The identity (2) follows either by substituting (1) for the sum or by using divisor sum of $\mu$ as
\[\begin{align} \theta(x) &= \sum_{p \leq x} \log p = \sum_{p^k \leq x} \log p \sum_{d | k} \mu(d) = \sum_{d} \mu(d) \sum_{\scriptstyle p^k \leq x \atop \scriptstyle d | k} \log p \\ &= \sum_{d} \mu(d) \sum_{p^{d\ell} \leq x} \log p = \sum_{d} \mu(d) \sum_{p^{\ell} \leq x^{1/d}} \log p = \sum_{d} \mu(d) \psi(x^{1/d}). \end{align}\]
Theorem. For $x \geq 2$ we have
\[
\psi(x) = \theta(x) + O( \sqrt{x} \log x). \tag{3}
\]
Proof. Note that
\[
\psi(x) = \sum_{k} \theta(x^{1/k}) = \sum_{k \leq \log_2 x} \theta(x^{1/k})
\] as $x^{1/k} < 2$ for $k > \log_2 x$ in which case $\theta(x^{1/k}) = 0$. Now using the crude estimate $\theta(x) \ll x \log x$ we see that
\[
\psi(x) -\theta(x) = \sum_{2 \leq k \leq \log_2 x} \theta(x^{1/k}) \ll \sum_{2 \leq k \leq \log_2 x} x^{1/k} \log x^{1/k} \ll \sqrt{x} \log x,
\] where the last estimate follows as the first term is of order $\sqrt{x} \log x$ and the sum of the rest of the terms is $\ll x^{1/3} \log^2 x $.
The Chebyshev functions are closely related to prime counting function, $\pi(x)$. For instance, the prime number theorem
\[
\pi(x) \sim \frac{x}{\log x}
\] is equivalent to both $\psi(x) \sim x$ and $\theta(x) \sim x$. The equivalence of the latter two is obvious from the estimate (3). Note that by Abel’s summation by parts formula we have
\[
\theta(x) = \pi(x) \log x -\int_{2}^{x} \frac{\pi(u)}{u} \, du.
\] If prime number theorem holds, then we have
\[
\int_{2}^{x} \frac{\pi(u)}{u} \, du \ll \int_{2}^{x} \frac{du}{\log u} \ll \frac{x}{ \log x}
\] and so
\[
\theta(x) = \pi(x) \log x + O \left( \frac{x}{\log x} \right).
\] This now immediately leads to $\theta(x) \sim x$. As for the other direction we can write $\pi$ in terms of $\theta$ as
\[
\pi(x) = \frac{\theta(x)}{\log x} + \int_{2}^{x} \frac{\theta(u)}{u (\log u)^2} \, du.
\] If we now assume $\theta(x) \sim x$, then we have
\[
\int_{2}^{x} \frac{\theta(u)}{u (\log u)^2} \, du \ll \int_{2}^{x} \frac{du}{(\log u)^2} \ll \frac{x}{(\log x)^2}
\] and hence
\[
\pi(x) = \frac{\theta(x)}{\log x} + O\left( \frac{x}{(\log x)^2} \right).
\] The prime number theorem now follows immediately.
We now show that $\theta(x) \ll x$. The proof makes use of the properties of the middle binomial coefficient. First we note that
\[
\prod_{n < p \leq 2n} p \, | \, \binom{2n}{n}
\] and so in particular we have $\prod_{n < p \leq 2n} p \leq 4^n$. Taking the logarithm we get
\[
\theta(2n) -\theta(n) \leq 2n \log 2.
\] Now suppose that $2^{m} \leq n < 2^{m + 1}$. Then we have
\[\begin{align} \theta(n) \leq \theta(2^{m + 1}) &= \sum_{k = 0}^{m} (\theta(2^{k + 1}) -\theta(2^k)) \leq \sum_{k = 0}^{m} 2^{k + 1} \log 2 = (2^{m + 2} -2) \log 2. \end{align}
\] Finally note that $2^{m + 2} = 4 \cdot 2^m \leq 4n$. Thus we have $\theta(n) \leq 4n \log 2$ for every $n \in \mathbb{N}$. If $x \geq 1$, then $\theta(x) = \theta(\lfloor x \rfloor) \leq 4 \lfloor x \rfloor \log 2 \leq 4x \log 2$. Hence, we obtain the estimate $\theta(x) \ll x$. This coupled with the estimate (3) yields $\psi(x) \ll x$.
Using the bound $\psi(x) \ll x$ we now obtain some estimate involves von Mangoldt function and primes. First observe that by our all time favorite Abel’s summation formula we have
\[
\sum_{n \leq x} \log n = \lfloor x \rfloor \log x -\int_{1}^{x} \frac{\lfloor u \rfloor}{u} \, du = x \log x -x + O(\log x). \tag{4}
\] But note that we can write $\log$ in terms of $\Lambda$ to get
\[
\sum_{n \leq x} \log n = \sum_{n \leq x} \sum_{d | n} \Lambda(d) = \sum_{n \leq x} \Lambda(n) \left\lfloor \frac{x}{n} \right\rfloor = x \sum_{n \leq x} \frac{\Lambda(n)}{n } + O (x),
\] where we use the estimate $\psi(x) \ll x$ to get the error term. Using the estimate (4) we find that
\[
\sum_{n \leq x} \frac{\Lambda(n)}{n} = \log x + O(1).
\] Since higher prime powers do not contribute much we should have
\[
\sum_{p \leq x} \frac{\log p}{p} = \log x + O(1). \tag{5}
\] We can justify this by noting that
\[
\sum_{\substack{p^k \leq x \ k \geq 2}} \frac{\log p}{p^k} \leq \sum_{p} \log p \sum_{k \geq 2} \frac{1}{p^k} = \sum_{p} \frac{\log p}{p(p -1)},
\] where the series to the right converges.
Theorem. If the limit
\[
\lim_{x \to \infty} \frac{\pi(x) \log x}{x}
\] exists, then it must be $1$.
Proof. Using Abel’s summation by parts formula we obtain
\[
\sum_{p \leq x} \frac{\log p}{p} = \frac{\pi(x) \log x}{x} -\int_{2}^{x} \pi(u) \left( \frac{1- \log u}{u^2} \right) \, du.
\] Employing the estimate (5) we see that
\[
\log x + O(1) = \frac{\pi (x) \log x}{x} + \int_{2}^{x} \frac{\pi(u) \log u}{u} \left(\frac{1}{u}- \frac{1}{u \log u} \right) \, du.
\] Hence we obtain
\[
\int_{2}^{x} \frac{\pi(u) \log u}{u} \left(\frac{1}{u} -\frac{1}{u \log u} \right) \, du = \log x + O(1) \tag{6}
\] as $\pi(x) (\log x)/x \ll 1$ by assumption. Now let
\[
L = \lim_{x \to \infty} \frac{\pi(x)\log x }{x}
\] and let $\varepsilon > 0$ be fixed. Let $M > 1$ be such that $\pi(x) (\log x)/x < L + \varepsilon$ for $x > M$. Note that
\[\begin{align} \int_{2}^{M} \frac{\pi(u) \log u}{u} \left(\frac{1}{u} -\frac{1}{u \log u} \right) \, du \ll 1
\end{align}\] as $M$ is fixed and
\[
\int_{M}^{x} \frac{\pi(u) \log u}{u} \left(\frac{1}{u} -\frac{1}{u \log u} \right) \, du \leq (L + \varepsilon) \log x.
\] Hence, we have
\[
\int_{2}^{x} \frac{\pi(u) \log u}{u} \left(\frac{1}{u} -\frac{1}{u \log u} \right) \, du \leq (L + \varepsilon) \log x + O(1),
\] where the implicit constant depends on $\varepsilon$. Combining this with (6) we immediately deduce that $L \geq 1$. Similarly we have
\[
\int_{2}^{x} \frac{\pi(u) \log u}{u} \left(\frac{1}{u} -\frac{1}{u \log u} \right) \, du \geq (L- \varepsilon) \log x + O(1).
\] Again combining this with (6) it follows that $L \leq 1$. Thus we conclude that $L = 1$.