[proofplan] We first compare $\psi$ and $\vartheta$ by isolating the contribution of proper prime powers $p^k$ with $k \geq 2$; this contribution is $o(x)$, so the two Chebyshev functions have the same first-order asymptotic. We then prove two partial summation identities connecting $\vartheta$ and $\pi$. These identities show that $\vartheta(x) \sim x$ implies $\pi(x) \sim x/\log x$, and conversely that $\pi(x) \sim x/\log x$ implies $\vartheta(x) \sim x$, because the integral error terms are smaller than the main terms after splitting at $x^{1/2}$. [/proofplan]

[step:Bound the contribution of proper prime powers in $\psi(x) - \vartheta(x)$]For $x \geq 2$, the difference between $\psi(x)$ and $\vartheta(x)$ is \begin{align*} \psi(x) - \vartheta(x) &= \sum_{\substack{p^k \leq x\\ p \text{ prime},\ k \geq 2}} \log p. \end{align*} Indeed, the terms with $k = 1$ in the definition of $\psi(x)$ are exactly the terms in $\vartheta(x)$. Define the exponent-counting function \begin{align*} K: [2, \infty) &\to \mathbb{N} \\ x &\mapsto \left\lfloor \frac{\log x}{\log 2} \right\rfloor. \end{align*} If $p^k \leq x$ and $k \geq 2$, then $2 \leq k \leq K(x)$ and $p \leq x^{1/k}$. Therefore \begin{align*} 0 \leq \psi(x) - \vartheta(x) &= \sum_{k=2}^{K(x)} \sum_{\substack{p \leq x^{1/k}\\ p \text{ prime}}} \log p\\ &\leq \sum_{k=2}^{K(x)} \sum_{2 \leq m \leq x^{1/k}} \log x. \end{align*} For every $k \geq 2$, we have $x^{1/k} \leq x^{1/2}$. Since $K(x) \leq \log x / \log 2$ for $x \geq 2$, this gives \begin{align*} 0 \leq \psi(x) - \vartheta(x) &\leq K(x) x^{1/2} \log x\\ &\leq \frac{x^{1/2}(\log x)^2}{\log 2}. \end{align*} Thus \begin{align*} 0 \leq \frac{\psi(x) - \vartheta(x)}{x} \leq \frac{(\log x)^2}{(\log 2)x^{1/2}} \to 0 \end{align*} as $x \to \infty$. Hence \begin{align*} \psi(x) - \vartheta(x) = o(x). \end{align*}[/step]

[guided]The function $\psi$ counts prime powers, while $\vartheta$ counts only primes. Thus the only possible difference comes from the terms $p^k$ with exponent $k \geq 2$. More precisely, \begin{align*} \psi(x) - \vartheta(x) &= \sum_{\substack{p^k \leq x\\ p \text{ prime},\ k \geq 2}} \log p. \end{align*} We now show that this is much smaller than $x$. Define the exponent-counting function \begin{align*} K: [2, \infty) &\to \mathbb{N} \\ x &\mapsto \left\lfloor \frac{\log x}{\log 2} \right\rfloor. \end{align*} If $p^k \leq x$, then $2^k \leq p^k \leq x$, so $k \leq \log x/\log 2$. Thus every exponent that occurs satisfies $2 \leq k \leq K(x)$. Also, for such a fixed $k$, the primes that can occur satisfy $p \leq x^{1/k}$. Therefore \begin{align*} 0 \leq \psi(x) - \vartheta(x) &= \sum_{k=2}^{K(x)} \sum_{\substack{p \leq x^{1/k}\\ p \text{ prime}}} \log p. \end{align*} To get an upper bound, we discard the primality condition and use $\log p \leq \log x$ whenever $p \leq x$. This gives \begin{align*} 0 \leq \psi(x) - \vartheta(x) &\leq \sum_{k=2}^{K(x)} \sum_{2 \leq m \leq x^{1/k}} \log x. \end{align*} Since $k \geq 2$, we have $x^{1/k} \leq x^{1/2}$. Hence each inner sum has at most $x^{1/2}$ terms, and there are at most $\log x/\log 2$ possible exponents. Consequently \begin{align*} 0 \leq \psi(x) - \vartheta(x) &\leq \frac{x^{1/2}(\log x)^2}{\log 2}. \end{align*} After division by $x$, the right-hand side becomes \begin{align*} \frac{(\log x)^2}{(\log 2)x^{1/2}}, \end{align*} which tends to $0$ as $x \to \infty$. Therefore \begin{align*} \psi(x) - \vartheta(x) = o(x). \end{align*}[/guided]

[step:Conclude that $\psi(x) \sim x$ is equivalent to $\vartheta(x) \sim x$] From the previous step, \begin{align*} \frac{\psi(x)}{x} - \frac{\vartheta(x)}{x} = \frac{\psi(x) - \vartheta(x)}{x} \to 0. \end{align*} Therefore $\psi(x)/x \to 1$ if and only if $\vartheta(x)/x \to 1$. Equivalently, \begin{align*} \psi(x) \sim x \quad \Longleftrightarrow \quad \vartheta(x) \sim x. \end{align*} [/step]

[step:Derive the partial summation identities relating $\vartheta$ and $\pi$]For $x \geq 2$, the functions $\pi$ and $\vartheta$ satisfy the following identities. Throughout the rest of the proof, $\mathcal{L}^1$ denotes one-dimensional Lebesgue measure on $\mathbb{R}$. \begin{align*} \vartheta(x) &= \pi(x)\log x - \int_2^x \frac{\pi(t)}{t}\,d\mathcal{L}^1(t), \end{align*} and \begin{align*} \pi(x) &= \frac{\vartheta(x)}{\log x} + \int_2^x \frac{\vartheta(t)}{t(\log t)^2}\,d\mathcal{L}^1(t). \end{align*} We verify the first identity. Let $p_1 < p_2 < \cdots < p_N$ be the primes not exceeding $x$, where $N = \pi(x)$. On each interval $[p_j,p_{j+1})$ the function $\pi$ is constant and equal to $j$, and on $[p_N,x]$ it is constant and equal to $N$. Telescoping gives \begin{align*} \int_2^x \frac{\pi(t)}{t}\,d\mathcal{L}^1(t) &= \sum_{j=1}^{N-1} j \int_{p_j}^{p_{j+1}} \frac{1}{t}\,d\mathcal{L}^1(t) + N \int_{p_N}^{x} \frac{1}{t}\,d\mathcal{L}^1(t)\\ &= \sum_{j=1}^{N-1} j(\log p_{j+1} - \log p_j) + N(\log x - \log p_N)\\ &= N\log x - \sum_{j=1}^{N} \log p_j\\ &= \pi(x)\log x - \vartheta(x). \end{align*} Rearranging gives the first identity. We verify the second identity directly. Define \begin{align*} w: [2, \infty) &\to \mathbb{R} \\ t &\mapsto \frac{1}{\log t}. \end{align*} For $1 \leq j \leq N$, set $A_j := \sum_{i=1}^{j} \log p_i$, and set $A_0 := 0$. Thus $A_j = \vartheta(p_j)$ for each $j$. By finite summation by parts, \begin{align*} \pi(x) &= \sum_{j=1}^{N} \log p_j\, w(p_j)\\ &= A_N w(x) - \sum_{j=1}^{N-1} A_j\bigl(w(p_{j+1}) - w(p_j)\bigr) - A_N\bigl(w(x) - w(p_N)\bigr). \end{align*} The function $\vartheta$ is constant and equal to $A_j$ on $[p_j, p_{j+1})$ for $1 \leq j \leq N-1$, and constant and equal to $A_N$ on $[p_N,x]$. Since $w$ is continuously differentiable on $[2,x]$ and \begin{align*} w'(t) = -\frac{1}{t(\log t)^2}, \end{align*} the [fundamental theorem of calculus](/theorems/632) on each interval gives \begin{align*} - \sum_{j=1}^{N-1} A_j\bigl(w(p_{j+1}) - w(p_j)\bigr) - A_N\bigl(w(x) - w(p_N)\bigr) &= \int_2^x \frac{\vartheta(t)}{t(\log t)^2}\,d\mathcal{L}^1(t). \end{align*} Because $A_N = \vartheta(x)$, we obtain \begin{align*} \pi(x) &= \frac{\vartheta(x)}{\log x} + \int_2^x \frac{\vartheta(t)}{t(\log t)^2}\,d\mathcal{L}^1(t). \end{align*}[/step]

[guided]The bridge between $\vartheta$ and $\pi$ is partial summation. Because $\pi$ and $\vartheta$ are step functions, we can prove the needed formulas directly by writing the integral over intervals between consecutive primes. Fix $x \geq 2$. Let $p_1 < p_2 < \cdots < p_N$ be the primes not exceeding $x$, where \begin{align*} N := \pi(x). \end{align*} The function $\pi: [2,\infty) \to \mathbb{N}$ is constant on each interval between consecutive primes: on $[p_j,p_{j+1})$ it equals $j$, and on $[p_N,x]$ it equals $N$. Therefore \begin{align*} \int_2^x \frac{\pi(t)}{t}\,d\mathcal{L}^1(t) &= \sum_{j=1}^{N-1} j \int_{p_j}^{p_{j+1}} \frac{1}{t}\,d\mathcal{L}^1(t) + N \int_{p_N}^{x} \frac{1}{t}\,d\mathcal{L}^1(t)\\ &= \sum_{j=1}^{N-1} j(\log p_{j+1} - \log p_j) + N(\log x - \log p_N). \end{align*} The finite sum telescopes. Expanding the coefficients of each $\log p_j$, every interior term cancels except for one negative copy of each $\log p_j$, while $N\log x$ remains. Hence \begin{align*} \int_2^x \frac{\pi(t)}{t}\,d\mathcal{L}^1(t) &= N\log x - \sum_{j=1}^{N} \log p_j\\ &= \pi(x)\log x - \vartheta(x). \end{align*} Rearranging gives \begin{align*} \vartheta(x) &= \pi(x)\log x - \int_2^x \frac{\pi(t)}{t}\,d\mathcal{L}^1(t). \end{align*} For the inverse formula, we use the same idea but weight the jumps of $\vartheta$ by $1/\log t$. The function $\vartheta$ has a jump of size $\log p$ at each prime $p$. Therefore summing the jump size multiplied by $1/\log p$ counts each prime once: \begin{align*} \sum_{\substack{p \leq x\\ p \text{ prime}}} \frac{\log p}{\log p} = \pi(x). \end{align*} Applying the same telescoping calculation to the step function $\vartheta$ and the differentiable weight $t \mapsto 1/\log t$, whose derivative is \begin{align*} -\frac{1}{t(\log t)^2}, \end{align*} gives \begin{align*} \pi(x) &= \frac{\vartheta(x)}{\log x} - \int_2^x \vartheta(t)\left(-\frac{1}{t(\log t)^2}\right)\,d\mathcal{L}^1(t)\\ &= \frac{\vartheta(x)}{\log x} + \int_2^x \frac{\vartheta(t)}{t(\log t)^2}\,d\mathcal{L}^1(t). \end{align*} This is the second partial summation identity.[/guided]

[step:Show that $\vartheta(x) \sim x$ implies $\pi(x) \sim x/\log x$] Assume $\vartheta(x) \sim x$ as $x \to \infty$. From the inverse partial summation identity, \begin{align*} \pi(x) &= \frac{\vartheta(x)}{\log x} + \int_2^x \frac{\vartheta(t)}{t(\log t)^2}\,d\mathcal{L}^1(t). \end{align*} The first term satisfies \begin{align*} \frac{\vartheta(x)}{\log x} \sim \frac{x}{\log x}. \end{align*} It remains to show that the integral is $o(x/\log x)$. Define \begin{align*} I: [2, \infty) &\to [0, \infty) \\ x &\mapsto \int_2^x \frac{\vartheta(t)}{t(\log t)^2}\,d\mathcal{L}^1(t), \end{align*} and define $y := x^{1/2}$. For $2 \leq t \leq y$, we use the elementary bound $\vartheta(t) \leq t\log t$, since there are at most $t$ positive integers not exceeding $t$ and each prime $p \leq t$ satisfies $\log p \leq \log t$. Hence \begin{align*} \int_2^y \frac{\vartheta(t)}{t(\log t)^2}\,d\mathcal{L}^1(t) &\leq \int_2^y \frac{1}{\log t}\,d\mathcal{L}^1(t)\\ &\leq \frac{y}{\log 2} = \frac{x^{1/2}}{\log 2} = o\!\left(\frac{x}{\log x}\right). \end{align*} Let $\varepsilon > 0$. Since $\vartheta(t) \sim t$, there exists $T_\varepsilon \geq 2$ such that $\vartheta(t) \leq (1+\varepsilon)t$ for all $t \geq T_\varepsilon$. For $x$ sufficiently large, $y = x^{1/2} \geq T_\varepsilon$, and then \begin{align*} \int_y^x \frac{\vartheta(t)}{t(\log t)^2}\,d\mathcal{L}^1(t) &\leq (1+\varepsilon)\int_y^x \frac{1}{(\log t)^2}\,d\mathcal{L}^1(t)\\ &\leq (1+\varepsilon)\frac{x}{(\log y)^2}\\ &= 4(1+\varepsilon)\frac{x}{(\log x)^2}\\ &= o\!\left(\frac{x}{\log x}\right). \end{align*} Thus $I(x) = o(x/\log x)$, and therefore \begin{align*} \pi(x) \sim \frac{x}{\log x}. \end{align*} [/step]

[step:Show that $\pi(x) \sim x/\log x$ implies $\vartheta(x) \sim x$] Assume $\pi(x) \sim x/\log x$ as $x \to \infty$. From the partial summation identity, \begin{align*} \vartheta(x) &= \pi(x)\log x - \int_2^x \frac{\pi(t)}{t}\,d\mathcal{L}^1(t). \end{align*} The first term satisfies \begin{align*} \pi(x)\log x \sim x. \end{align*} It remains to show that the integral is $o(x)$. Define \begin{align*} J: [2, \infty) &\to [0, \infty) \\ x &\mapsto \int_2^x \frac{\pi(t)}{t}\,d\mathcal{L}^1(t), \end{align*} and define $y := x^{1/2}$. For $2 \leq t \leq y$, the elementary bound $\pi(t) \leq t$ gives \begin{align*} \int_2^y \frac{\pi(t)}{t}\,d\mathcal{L}^1(t) &\leq \int_2^y 1\,d\mathcal{L}^1(t)\\ &\leq y = x^{1/2} = o(x). \end{align*} Let $\varepsilon > 0$. Since $\pi(t) \sim t/\log t$, there exists $T_\varepsilon \geq 2$ such that \begin{align*} \pi(t) \leq (1+\varepsilon)\frac{t}{\log t} \end{align*} for all $t \geq T_\varepsilon$. For $x$ sufficiently large, $y = x^{1/2} \geq T_\varepsilon$, and therefore \begin{align*} \int_y^x \frac{\pi(t)}{t}\,d\mathcal{L}^1(t) &\leq (1+\varepsilon)\int_y^x \frac{1}{\log t}\,d\mathcal{L}^1(t)\\ &\leq (1+\varepsilon)\frac{x}{\log y}\\ &= 2(1+\varepsilon)\frac{x}{\log x}\\ &= o(x). \end{align*} Hence $J(x) = o(x)$. Therefore \begin{align*} \vartheta(x) = \pi(x)\log x - J(x) \sim x. \end{align*} [/step]

[step:Combine the implications] We have shown \begin{align*} \psi(x) \sim x \quad \Longleftrightarrow \quad \vartheta(x) \sim x \end{align*} and \begin{align*} \vartheta(x) \sim x \quad \Longleftrightarrow \quad \pi(x) \sim \frac{x}{\log x}. \end{align*} Combining these two equivalences proves that the three asserted forms of the [Prime Number Theorem](/theorems/1742) are equivalent. [/step]