The Prime Number Theorem and Dirichlet's Theorem

Step 1 — Logarithmic derivative. From the Euler product $\zeta(s)=\prod_p(1-p^{-s})^{-1}$, take $-\log$ and differentiate:

Derivation hint: $\log\zeta(s)=\sum_p\sum_k k^{-1}p^{-ks}$; differentiate term-by-term to recover $\Lambda(n)$ as Dirichlet coefficients.

Step 2 — Perron / contour inversion. Express the smoothed count as a contour integral:

Derivation hint: $\frac{1}{2\pi i}\int x^s/s\,ds$ is the discontinuous Heaviside integral, equal to $1$ for $x>1$ and $0$ for $0<x<1$; integrating term-by-term over $\Lambda(n)x^s n^{-s}$ recovers $\sum_{n\le x}\Lambda(n)$.

Step 3 — Shift the contour to $\operatorname{Re}(s)=1$. Move the line of integration leftward. The pole of $\zeta(s)$ at $s=1$ (simple, residue $1$) gives $-\zeta'/\zeta$ a simple pole with residue $1$, contributing the main term:

Derivation hint: near $s=1$, $\zeta(s)=\frac{1}{s-1}+\gamma+\cdots$, so $-\zeta'/\zeta \sim 1/(s-1)$; multiply by $x^s/s$ and read off the residue at $s=1$, giving $x^1/1=x$.

Step 4 — Non-vanishing on the line $\operatorname{Re}(s)=1$. The crux: $\zeta(1+it)\neq 0$ for all real $t\neq 0$. This uses the inequality

Derivation hint: apply this with $\phi = t\ln p$ to $\log|\zeta(\sigma)^3\zeta(\sigma+it)^4\zeta(\sigma+2it)|$; the nonnegativity forces $|\zeta(\sigma+it)|$ to not vanish faster than $(\sigma-1)$ allows, ruling out a zero at $\sigma=1$.

Step 5 — Tauberian conclusion. With no zeros on $\operatorname{Re}(s)=1$, the boundary integral over $\operatorname{Re}(s)=1$ contributes $o(x)$. The Wiener–Ikehara Tauberian theorem (or Newman's contour argument) then converts the analytic behavior into $\psi(x)\sim x$. Partial summation gives $\pi(x)\sim x/\ln x$. $\blacksquare$

Step 1 — Orthogonality of characters. For $\gcd(a,q)=1$,

Derivation hint: the characters form an orthonormal basis of functions on $(\mathbb{Z}/q\mathbb{Z})^\times$; this is the column-orthogonality relation of the character table of a finite abelian group.

Step 2 — Sum over primes. Multiply by $\Lambda(n)n^{-s}$ and sum:

Derivation hint: $-L'/L(s,\chi)=\sum_n \chi(n)\Lambda(n)n^{-s}$ by the same logarithmic-derivative computation as for $\zeta$; substitute and use orthogonality from [11].

Step 3 — Behavior at $s\to 1^+$. The principal character term contributes a pole (from $L(s,\chi_0)\sim 1/(s-1)$ up to local factors), giving divergence $\sim \frac{1}{\varphi(q)}\log\frac{1}{s-1}$:

Derivation hint: only $\chi_0$ gives a pole; the non-principal terms stay bounded — provided $L(1,\chi)\neq 0$ — so they cannot cancel the divergence.

Step 4 — The hard lemma: $L(1,\chi)\neq 0$ for $\chi\neq\chi_0$. Complex characters: a vanishing $L(1,\chi)=0$ would force $L(1,\bar\chi)=0$, producing a double cancellation contradicting $\prod_\chi L(s,\chi)\ge 1$ (a Dedekind-zeta-type product with nonnegative Dirichlet coefficients). Real characters (the delicate case): $L(1,\chi)>0$ follows from Dirichlet's class number formula $L(1,\chi)= \frac{2h\log\varepsilon}{\sqrt{D}}$ (real) or $\frac{2\pi h}{w\sqrt{|D|}}$ (imaginary), all manifestly positive. Derivation hint: the real-character case is the unique one resisting purely analytic argument; the class number formula identifies $L(1,\chi)$ with a positive geometric/arithmetic invariant of a quadratic field.

Step 5 — Conclusion. Since [13] diverges as $s\to1^+$, the sum $\sum_{p\equiv a\,(q)} p^{-s}$ has infinitely many terms — there are infinitely many such primes, and the leading constant $1/\varphi(q)$ yields equidistribution. $\blacksquare$