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The central binomial coefficient satisfies $4^m/(2m+1)\le \left(\genfrac{}{}{0ex}{}{2m}{m}\right)\le 4^m$. Prime factors between $m$ and $2m$ give a dyadic upper bound for the Chebyshev theta function. The prime-power exponents in the factorial quotient are sums of zeros and ones, so the central coefficient gives a linear lower bound for the Second Chebyshev function. Higher prime powers contribute only $O(\sqrt{x}\log x)$; hence the prime-counting function has upper and lower bounds of order $x/\log x$.

The two Chebyshev functions measure the total natural logarithm weight of prime numbers or of prime powers greater than one below a real cutoff $x\ge 0$:Here $p$ ranges over prime numbers, $n$ over positive integers, and $k$ over positive integers. The Chebyshev theta function $\vartheta$, also written $\theta$, weights each prime number once. The Second Chebyshev function $\psi$ weights every positive power of each prime number separately. Both vanish for $0\le x<2$; the inclusive cutoff makes them right-continuous functions and nondecreasing functions, constant between their jumps. ^[ref]

At $x=10$, the prime numbers are $2,3,5,7$, while the extra prime powers greater than one are $4,8,9$. Exactly,At each prime number $p$, both Chebyshev functions jump by $\log p$. At a prime power $p^k$ with $k\ge 2$, only the Second Chebyshev function jumps, by $\log p$. At $6$ and $10$ neither jumps. Notice that the gap is already positive at $5$ and $7$: a shared jump does not remove the earlier contribution from $4$.

Grouping the prime powers by their exponent gives, for $x\ge 2$,The floor function cutoff is exact: $p\ge 2$ forces $2^k\le x$ whenever $p^k\le x$. For each exponent, $p^k\le x$ is equivalent to $p\le x^{1/k}$. An upper cutoff smaller than the lower one means an empty sum. ^[ref]

For the Second Chebyshev function, finite summation gives ${\Psi }_1(x)={\sum }_{n\le x}(x-n)\Lambda (n)$. For $c>1$, Mellin inversion of the kernel $1/(s(s+1))$ givesThe extra denominator makes this smoothing useful: the resulting sum over Nontrivial zeros of the Riemann zeta function is absolutely convergent. This integral differs from the integrated Chebyshev sum, which denotes a sum of the values $\psi (x/n)$.

Write $\theta (x)={\sum }_{p\le x}\log p$ for the Chebyshev theta function and $\psi (x)={\sum }_{n\le x}\Lambda (n)$ for the Second Chebyshev function, where $\Lambda$ is the Von Mangoldt function. We first establish the Chebyshev estimate. Every prime $n<p\le 2n$ divides the central binomial coefficient $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$, soApply this at successive powers of two and use monotonicity to obtain $\theta (x)=O(x)$. Sincethe terms $r\ge 2$ contribute $O(\sqrt{x}\log x)$ and therefore $\psi (x)=O(x)$. This Chebyshev estimate from central binomial coefficients is all the information about the distribution of primes needed here.

The Von Mangoldt divisor identity $\log n={\sum }_{d\mid n}\Lambda (d)$ gives the factorial proof of Mertens first theorem. With $N=\lfloor x\rfloor$, interchange the finite sums to obtainComparison of ${\sum }_{n\le N}\log n$ with ${\int }_1^N\log tdt$ gives $\log N!=N\log N-N+O(\log N)$. Dividing the preceding identity by $x$, using $N=x+O(1)$ and the Chebyshev estimate, givesThe contribution of higher prime powers is bounded independently of $x$, becauseThus the Mertens first theorem is

For the square-root logarithmic sum over primes, apply partial summation to $A(t)$ with weight $(\log t{)}^{-1/2}$. The lower endpoint is understood as $A(2^{-})=0$, soThe main terms are $\sqrt{\log x}$ and $\sqrt{\log x}-\sqrt{\log 2}$. The bounded error in $A(t)$ produces a bounded integral, since ${\int }_2^{\infty }dt/(t(\log t{)}^{3/2})<\infty$. ThereforeNeither argument uses the Prime number theorem.

First alternative. The Euler product of the Riemann zeta function is absolutely convergent for $\sigma >1$. Its logarithm and the nonnegative trigonometric polynomialgiveConsequently $\zeta (\sigma {)}^3|\zeta (\sigma +it){|}^4|\zeta (\sigma +2it)|\ge 1$. Suppose $t\ne 0$ and the Riemann zeta function had a zero of order $m\ge 1$ at $1+it$. As $\sigma \downarrow 1$, its simple pole at one gives $\zeta (\sigma )=O((\sigma -1{)}^{-1})$, the alleged zero gives $\zeta (\sigma +it)=O((\sigma -1{)}^m)$, and $\zeta (\sigma +2it)$ remains bounded. The product would be $O((\sigma -1{)}^{4m-3})\to 0$, contradicting its lower bound. Thus there are no zeta zeros on $\Re s=1$; the point $s=1$ itself is a pole, not a zero.

For the first integral of the Chebyshev function, defineThe logarithmic derivative identity $-{\zeta }^{\prime }/\zeta (s)={\sum }_n\Lambda (n)n^{-s}$, valid for $\Re s>1$, leads toTo verify the kernel rather than assume an inversion formula, integrate $y^s/(s(s+1))$ on the vertical line $\Re s=c$. Closing to the left when $y>1$ gives the residues at zero and minus one, namely $1-y^{-1}$; closing to the right when $0<y<1$ gives zero. At $y=1$ the integral is zero as well. The horizontal integrals tend to zero because of the quadratic denominator, and the distant vertical integrals vanish in the indicated direction. Multiplying this kernel by $x$ with $y=x/n$ gives $(x-n{)}^{+}$. The absolute convergence of the Dirichlet series on $\Re s=c$, together with the integrable denominator, permits interchange with the integral and proves the formula.

Here is a quantitative contour route from this relation to the Prime number theorem. Question 3 establishes a zero-free region of the Riemann zeta function and the accompanying bounds for its logarithmic derivative, independently of the Prime number theorem. In a sufficiently narrow half-width region they implyon the left and horizontal sides of the rectangle with heights $\pm T$ and left edge ${\sigma }_L=1-b/(2\log (T+3))$, after making $b>0$ small enough also for bounded heights. The explanation of this bound is included at the end of Question 3. Take $c=1+1/\log x$ and $T=\exp (\sqrt{\log x})$. On the original line, the omitted tails aresince the Euler product bounds $|{\zeta }^{\prime }/\zeta (c+it)|$ by $-{\zeta }^{\prime }/\zeta (c)=O(\log x)$. Shift the truncated integral to the left edge. The only enclosed pole is at $s=1$, with residue $x^2/2$; the kernel poles at zero and minus one lie outside the rectangle. The left edge contributes $O(x^{2-b/(2\log (T+3))}{\log }^2(T+3))$, since $|s(s+1){|}^{-1}\ll (1+t^2{)}^{-1}$ there. The two horizontal edges contribute $O(x^2{\log }^2(T+3)/T^2)$. In particular, for some $a>0$,

It remains to remove the smoothing; differentiating an asymptotic formula would not justify this step. The monotonicity of the Second Chebyshev function gives, for $h=\epsilon x$ with $0<\epsilon <1$,First let $x\to \infty$ with $\epsilon$ fixed. After division by $x$, the lower and upper bounds tend to $1-\epsilon /2$ and $1+\epsilon /2$. Let $\epsilon \downarrow 0$ to obtain $\psi (x)\sim x$. Higher prime powers contribute $O(\sqrt{x}\log x)$ by the Chebyshev estimate, so $\theta (x)\sim x$. Finally partial summation givesThe integral is $O(x/{\log }^{2}x)$, by splitting at $\sqrt{x}$ and using $\theta (t)=O(t)$. This proves the Prime number theorem with $\pi (x)\sim x/\log x$.

Second alternative. We prove the functional equation of the Riemann zeta function from a theta function. Set $\Theta (u)={\sum }_{n\in \mathbb{Z}}{e}^{-\pi n^{2}u}$ for $u>0$. With Fourier transform convention $\hat{f}(\xi )={\int }_{\mathbb{R}}f(v)e^{-2\pi iv\xi }dv$, the Gaussian function $f(v)=e^{-\pi u{v}^2}$ hasIndeed, differentiation under the integral and integration by parts give ${\hat{f}}^{\prime }(\xi )=-2\pi \xi \hat{f}(\xi )/u$, while the Gaussian integral gives $\hat{f}(0)=u^{-1/2}$. Periodize $f$: the smooth period-one function ${\sum }_{n}f(v+n)$ has Fourier coefficients $\hat{f}(k)$, by integrating on $[0,1]$ and interchanging its absolutely convergent sum with the integral. Its absolutely convergent Fourier series, evaluated at zero, proves the Poisson summation formula in this case. HenceFor $\Re s>1$, termwise integration and the defining integral of the Gamma function give the Mellin representation of the completed Riemann zeta function:Split at one and substitute $u=1/v$ in the lower integral. The elementary part isand the remaining part becomes an integral over $[1,\infty )$. ThusThe integral is an entire function of $s$, because $\Theta (u)-1$ decays exponentially and uniform convergence holds on every compact subset of the complex plane. This proves meromorphic continuation of the completed Riemann zeta function, with only the displayed simple poles, and the right side is invariant under $s\mapsto 1-s$. ThereforeThe Gamma reflection formula and Gamma duplication formula give the equivalent form $\zeta (s)=2^s{\pi }^{s-1}\sin (\pi s/2)\Gamma (1-s)\zeta (1-s)$, understood by meromorphic continuation. Multiplication by $s(s-1)/2$ gives the entire Riemann xi function $\xi (s)=\xi (1-s)$.

Let $N(T)$ count positive ordinates of Nontrivial zeros of the Riemann zeta function up to height $T$, including multiplicity. The Riemann–von Mangoldt formula isMore precisely, away from ordinates of zeros, the constant term is $7/8$, the additional term is $S(T)={\pi }^{-1}\arg \zeta (1/2+iT)$, and the remaining error is $O(T^{-1})$. Here the argument is continued along $2\to 2+iT\to 1/2+iT$, and $S(T)=O(\log T)$. The coarse formula displayed above holds at all large heights with the usual right-continuous counting convention. Subtract it at $T+1$ and $T$: the main term changes by $O(\log T)$ and each error is $O(\log T)$. Hence

For infinitely many reciprocal-logarithmic gaps between zeta zeros, fix any $0<c<2\pi$, for example $c=\pi$. Suppose all sufficiently late gaps satisfied ${\gamma }_{n+1}-{\gamma }_n\le c/\log n$. Summing and usinggives ${\gamma }_n\le (c+o(1))n/\log n$. The last elementary estimate follows by integral comparison with $1/\log t$ and one integration by parts. The Riemann–von Mangoldt formula would implya contradiction. This reasoning is valid whether repeated ordinates are listed according to multiplicity or only distinct ordinates are listed, since in either case $n\le N({\gamma }_n)$. Consequently ${\gamma }_{n+1}-{\gamma }_n>c/\log n$ infinitely often, for any fixed $0<c<2\pi$.

Write $\vartheta (x)={\sum }_{p\le x}\log p$ for the Chebyshev theta function and $\psi (x)={\sum }_{p^j\le x}\log p$ for the Second Chebyshev function. Throughout, $p$ denotes a prime and all logarithms are natural. Each prime $n<p\le 2n$ divides the central binomial coefficient, soApply this at $n=1,2,4,\dots ,2^{r-1}$ and telescope: $\vartheta (2^r)\le (2^{r+1}-2)\log 2$. Bounding a general $x$ by the next power of two proves $\vartheta (x)=O(x)$. Splitting the prime-counting function at $\sqrt{x}$ now givesThis is the required Chebyshev estimate from central binomial coefficients, with no use of a prime asymptotic.

The higher prime powers also satisfyFor an integer $N$, the prime factorization of a factorial givesThe Stirling formula, or integral comparison for $\log t$, says $\log N!=N\log N-N+O(\log N)$. After division by $N$ we obtain ${\sum }_{p^j\le N}(\log p)/p^j=\log N+O(1)$. MoreoverSubtract this bounded contribution and replace $N$ by $\lfloor x\rfloor$. The factorial proof of Mertens first theorem therefore yields

For fixed $\lambda >1$, apply partial summation to $A(t)$ and $f(t)=(\log t{)}^{\lambda -1}$:The main contribution isThe bounded error in $A(t)$ contributes $O((\log x{)}^{\lambda -1})$ at the endpoint and in the integral, since ${\int }_2^x(\log t{)}^{\lambda -2}dt/t=O_{\lambda }((\log x{)}^{\lambda -1})$. Thus the positive logarithmic moments of reciprocal primes satisfy

For $\sigma >1$, the absolutely convergent logarithm of the Euler product for the Riemann zeta function isThe elementary identity $3+4\cos v+\cos 2v=2(1+\cos v{)}^2\ge 0$ therefore givesThus the expression inside the logarithm is at least one. The Meromorphic continuation of the Riemann zeta function to the right half-plane has only a simple pole at one, of residue one. It follows, for example, fromwhose integral is holomorphic there.

Suppose $t\ne 0$ and $\zeta$ has a zero of order $m\ge 1$ at $1+it$. As $\sigma \downarrow 1$, the real factor is $O((\sigma -1{)}^{-3})$, the middle factor is $O((\sigma -1{)}^{4m})$, and the factor at $1+2it$ is bounded. Their product would tend to zero, contradicting its lower bound one. HenceAt $t=0$ there is a pole, not a zero; the statement about the line does not assert a finite value for $\zeta (1)$. This proves the three-four-one product proof of zeta boundary nonvanishing.

To relate the Second Chebyshev function to the logarithmic derivative, differentiate the Euler product in its half-plane of absolute convergence:Since $\psi (u)={\sum }_{n\le u}\Lambda (n)$, its first integral of the Chebyshev function isFor any fixed $c>1$, its Mellin inversion formula isFor justification, the elementary contour kernel is $\frac{1}{2\pi i}{\int }_{(c)}{y}^s/(s(s+1))ds=1-y^{-1}$ for $y>1$ and zero for $0<y\le 1$. Close left or right and take the residues at zero and minus one; at $y=1$ the continuous value is zero. Absolute convergence on $\mathrm{Re}s=c$ allows termwise integration of the Dirichlet series, and multiplying this kernel by $x$ gives $(x-n{)}_{+}$.

Here is how the relation yields the asymptotic. Move the contour to the left, using the functional equation of the Riemann zeta function to control the left-hand side. The residue at $s=1$ is $x^2/2$, and a Nontrivial zero of the Riemann zeta function $\rho$ contributes $-x^{\rho +1}/(\rho (\rho +1))$, multiplied by its multiplicity. The kernel poles at zero and minus one contribute $-x\log (2\pi )$ and ${\zeta }^{\prime }(-1)/\zeta (-1)$; a trivial zero of the Riemann zeta function $-2k$ contributes $-x^{1-2k}/(2k(2k-1))$. Thus the smoothed explicit formula for the Chebyshev function isFor the contour argument, use heights avoiding the zero ordinates and then let those heights increase. The local zero bound from the Riemann–von Mangoldt formula, together with the Local partial-fraction expansion of the Riemann zeta logarithmic derivative, permits heights with logarithmic-derivative bound $O({\log }^{2}T)$ in each fixed vertical strip. The horizontal integrals then vanish because the denominator is of size $T^2$. After that, send the left edge through negative odd integers to minus infinity; the functional equation of the Riemann zeta function bounds the logarithmic derivative there by a logarithm, and $x^{s+1}$ makes the left integral vanish for fixed $x>1$. This explains why this smoothing allows a convergent contour calculation without a quantitative zero-free region.

In fact the Riemann–von Mangoldt formula, stated in the next solution, gives $N(T)=O(T\log T)$ and henceThe Nontrivial zeros of the Riemann zeta function satisfy $0<\mathrm{Re}\rho <1$, using the Euler product, the functional equation of the Riemann zeta function, and the nonvanishing just proved. For each such zero, $x^{\rho -1}\to 0$ as $x\to \infty$, whereas $|x^{\rho -1}|\le 1$ for $x\ge 1$. The dominated convergence theorem therefore makes the absolutely convergent zero sum, divided by $x^2$, tend to zero. All the other displayed terms are $o(x^2)$. We obtain

It remains to unsmooth; differentiating an asymptotic without justification would not suffice. Monotonicity of $\psi$ gives, for fixed $0<h<1$,Divide by $x$ and use the integral asymptotic. The lower and upper limits lie between $1-h/2$ and $1+h/2$. Letting $h\downarrow 0$ proves $\psi (x)\sim x$. Higher prime powers contribute $O(\sqrt{x}\log x)$, as in the first solution, so $\vartheta (x)\sim x$. Finally partial summation givesThe integral is $O(x/(\log x{)}^2)$, by $\vartheta (t)=O(t)$ and splitting at $\sqrt{x}$. Hencewhich is the Prime number theorem.

Write $F(s)=-{\zeta }^{\prime }(s)/\zeta (s)$ and ${\psi }_{\Gamma }={\Gamma }^{\prime }/\Gamma$. The global partial-fraction expansion of the zeta logarithmic derivative isHere $\gamma$ is the Euler--Mascheroni constant, and the sum includes multiplicities of the Nontrivial zeros of the Riemann zeta function. The paired terms are locally absolutely convergent away from their poles, by the Riemann–von Mangoldt formula. This identity follows by differentiating the genus-one Hadamard factorization of the Riemann xi function and using the Gamma function recurrence.

We first prove the classical zero-free region of the Riemann zeta function. The Euler-product nonvanishing of the Riemann zeta function and its functional equation place every nontrivial zero in the closed strip $0\le \beta \le 1$; no assertion about zeros on its boundary is needed yet. The real logarithmic derivative of Riemann xi and the digamma function asymptotic imply, for $1<\sigma \le 2$ and $|t|\ge 1$,Every summand being subtracted is nonnegative. If $\beta +it$ is a zero, retaining its summand gives $\Re F(\sigma +it)\le C\log (|t|+2)-1/(\sigma -\beta )$; at height $2t$ the upper bound $C\log (|t|+2)$ suffices. At real $\sigma$ the simple pole of $\zeta$ gives $F(\sigma )=1/(\sigma -1)+O(1)$.

For $\sigma >1$, the Dirichlet series $F(s)={\sum }_n\Lambda (n)n^{-s}$ has nonnegative coefficients. The identity $3+4\cos u+\cos 2u=2(1+\cos u{)}^2\ge 0$ therefore gives the three-four-one zero-free-region argumentWith $L=\log (|t|+2)$, a zero at $\beta +it$ would consequently satisfyChoose $\sigma =1+a/L$, with an absolute $a>0$ sufficiently small, and put $c=a/8$. If $\beta \ge 1-c/L$, this inequality would imply $4/(a+c)\le 3/a+C$. But $4/(a+a/8)-3/a=5/(9a)>C$ for small enough $a$, a contradiction.

For any fixed $t\ne 0$, the same positivity inequality as $\sigma \downarrow 1$ excludes a zero at $1+it$: its negative contribution has coefficient at least four, whereas the real pole contributes only three. Any zero at $1+2it$ only strengthens the contradiction. The pole at $s=1$ has a zero-free punctured neighbourhood. By compactness, the remaining bounded-height portion of the line has a zero-free neighbourhood as well. Shrink $c$ to combine these facts with the large-height bound. Hencewith the understanding that $s=1$ is a pole.

The Riemann–von Mangoldt explicit formula for the Second Chebyshev function, with half weight at a prime-power endpoint, isThe zero sum uses symmetric height truncation. To estimate it, take a half-integer $x$ and use the truncated explicit formula for the second Chebyshev functionThe zero-free region gives $|x^{\rho }|\le x\exp (-c\log x/\log (T+2))$ for these zeros. The Riemann–von Mangoldt formula and partial summation give ${\sum }_{|\Im \rho |\le T}|\rho {|}^{-1}\ll {\log }^2(T+2)$. Choose $T=\exp (\sqrt{\log x})$. Both the zero sum and the truncation error are then $O(x{e}^{-c^{\prime }\sqrt{\log x}})$ after reducing the positive constant to absorb logarithmic factors. Replacing an arbitrary $X$ by $\lfloor X\rfloor +1/2$ preserves the Von Mangoldt function sum and changes the main term by at most one. This proves the requested form of the Prime number theorem:

Write $1(n)=1$ and let $\epsilon$ be the Identity for Dirichlet convolution. The Möbius function uses the empty-product convention $\mu (1)=1$; the Von Mangoldt function has $\Lambda (1)=0$. We first establish all the arithmetic function identities needed below. Every integer divisor of $n$ contributes one to $\tau (n)$, so $\tau =1\ast 1$. If $n>1$ has $r$ distinct prime factors, the squarefree divisors correspond to their subsets andAt $n=1$ the sum is one. Thus $\mu \ast 1=\epsilon$. Writing $n={\prod }_p{p}^{v_p(n)}$, the only divisors with nonzero Von Mangoldt function are $p,p^2,\dots ,p^{v_p(n)}$, and henceThe Dirichlet convolution is commutative by replacing $d$ with $n/d$, and associative because either bracketing sums $f(a)g(b)h(c)$ over $abc=n$. ConsequentlyThese are consequences of the definitions, rather than unproved invocations of Möbius inversion.

To identify the constant, integrate on each interval $[j,j+1)$:The tail is $O(1/J)$, since the integrand is at most $2/x^2$ for $x\ge 2$. Summation therefore gives $\gamma ={\lim }_{J\to \infty }(H_J-\log (J+1))$, where $H_J$ is the harmonic number, and $H_J=\log J+\gamma +O(1/J)$. Thus this is Euler's constant.

Put $r=\lfloor \sqrt{N}\rfloor$. Counting pairs of positive integers whose product is at most $N$, and subtracting the square counted twice, gives the Dirichlet hyperbola method identitySince $r=\sqrt{N}+O(1)$, the harmonic number estimate yieldsThe increasing natural logarithm also satisfies ${\sum }_{n\le N}\log n=N\log N-N+O(\log (2N))$ by comparison with its integral. Subtracting proves

For the second implication, denote the Mertens function by $M(x)$ and the Second Chebyshev function by $\psi (x)$. Apply the proved Dirichlet convolution identities to $b(n)=\tau (n)-\log n-2\gamma 1(n)$:Summing this identity over integers up to $N$ gives the exact relationFix an integer $K\ge 2$. The contribution to $E(N)$ from $d\le N/K$ has absolute value at mostFor $d>N/K$, group by $j=\lfloor N/d\rfloor$. This gives exactlyunder the assumed cancellation $M(x)=o(x)$; extending that assumption from integers to real $x$ uses $M(x)=M(\lfloor x\rfloor )$. Therefore ${\mathrm{limsup}}_{N\to \infty }|E(N)|/N\ll K^{-1/2}$. Letting $K\to \infty$ proves the prime number asymptotic from Mertens cancellation:

Let $\Delta (N)={\sum }_{n\le N}(\tau (n)-\log n-2\gamma )$. The Möbius divisor-sum identity and the Von Mangoldt divisor identity give $\mu \ast \tau =1$ and $\mu \ast \log =\Lambda$, soHere $\ast$ is Dirichlet convolution. The divisor summatory asymptotic bounds the terms with $d\le N/K$ in absolute value by $O(N/\sqrt{K})$. For $d>N/K$, group by $j=\lfloor N/d\rfloor <K$: the remaining sum is ${\sum }_{j<K}\Delta (j)(M(N/j)-M(N/(j+1)))=o_K(N)$. First let $N$ tend to infinity, then $K$ tend to infinity. This proves the Second Chebyshev function asymptotic from cancellation of the Mertens function.

The Second Chebyshev function is the partial sum of the Von Mangoldt function up to a real cutoff $x\ge 0$:For example, $\psi (3)=\log 6$, $\psi (4)=\log 12$, and $\psi (4.5)=\log 12$. At $4=2^2$ the jump has size $\log 2$; between $4$ and $5$ no new prime power greater than one enters the sum. In contrast, the Chebyshev theta function stays at $\vartheta (4)=\log 6$.

There is also an exact connection with the least common multiple. For $N=\lfloor x\rfloor \ge 1$, each prime number $p\le N$ occurs to the highest exponent $k$ for which $p^k\le N$, both in that least common multiple and in the accumulated Von Mangoldt function weights. ThusThis explains the $\log 2520$ value at $x=10$: the least common multiple is $2^3\cdot 3^2\cdot 5\cdot 7$. The Chebyshev theta function instead gives the natural logarithm of the primorial $2\cdot 3\cdot 5\cdot 7=210$. The additional prime powers retain the highest required multiplicities.

Why is this useful for the distribution of prime numbers? The Prime number theorem has the equivalent formswhere $\pi$ is the prime-counting function and $\sim$ denotes asymptotic equivalence. Thus an arithmetic function defined using prime factorization has average weight tending to one over long initial intervals of the positive integers. These are substantive asymptotic equivalences, not consequences of the definitions or the Von Mangoldt divisor identity alone. ^[ref]

The elementary reason the two Chebyshev functions have the same leading behavior is that their difference is small compared with $x$. Every additional prime power has base $p\le \sqrt{x}$, and for each such prime number its total extra weight is at most $\log x$. There are at most $\lfloor \sqrt{x}\rfloor$ such bases, soThis transfers $\psi (x)\sim x$ to $\vartheta (x)\sim x$, or conversely, once either is established. Partial summation then relates the Chebyshev theta function to the prime-counting function. The estimate controls the extra prime powers; proving the leading value itself is the harder Prime number theorem problem.

The Von Mangoldt function assigns a natural logarithm weight to each positive power of a prime number. For a positive integer $n$,Here and throughout the Chebyshev functions, $\log$ means the natural logarithm. In particular, $\Lambda (1)=0$: the exponent must be positive, even though the convention in prime power allows exponent zero. The support of a function of $\Lambda$ therefore consists precisely of the prime powers greater than one. ^[ref]

For example, $\Lambda (2)=\Lambda (4)=\Lambda (8)=\log 2$ and $\Lambda (3)=\Lambda (9)=\log 3$, whereas $\Lambda (6)=\Lambda (12)=0$. A composite number can have a nonzero weight: $8$ has just one distinct prime factor. But $12=2^2\cdot 3$ has two distinct prime factors, so its weight is zero. At $p^k$ the weight is $\log p$, not $\log (p^k)=k\log p$. This choice lets successive powers record successive copies of the same prime factor in the Von Mangoldt divisor identity.

The Von Mangoldt function is useful because it connects two views of the prime numbers: the prime factorization of an individual integer, through a divisor sum, and the accumulated weights below a cutoff, through the Second Chebyshev function. The latter turns questions about the distribution of prime numbers into questions about a partial sum of an arithmetic function.