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Fix a positive integer $q$ and an integer $a$ coprime to $q$. We use Dirichlet characters modulo $q$, extended by zero on integers not coprime to $q$, and write ${\chi }_0$ for the principal Dirichlet character. For a nonprincipal Dirichlet character, the sum over one complete period is zero. Its partial sums $A_{\chi }(u)={\sum }_{n\le u}\chi (n)$ are therefore $O(q)$. Partial summation givesThe integral converges locally uniformly there, so these Dirichlet L-functions are holomorphic at one. On the other hand,has a simple pole at one, with residue $\phi (q)/q$.

The assumed nonvanishing of a nonprincipal Dirichlet L-function at one for real Dirichlet characters also excludes vanishing for nonreal ones. To prove this, form the product over all Dirichlet charactersUse the logarithm supplied by the absolutely convergent Euler product. The Orthogonality of Dirichlet characters yieldsThus $\mathcal{F}(\sigma )\ge 1$. If a nonreal Dirichlet character $\chi$ had $L(1,\chi )=0$, its distinct conjugate $\overline{\chi }$ would also have a zero there, since $L(1,\overline{\chi })=\overline{L(1,\chi )}$. These two zeros would outweigh the single pole of the principal factor; all other factors are bounded near one. It would follow that $\mathcal{F}(\sigma )\to 0$ as $\sigma \downarrow 1$, contradicting the lower bound. Together with the assumption for real Dirichlet characters, this proves $L(1,\chi )\ne 0$ for every nonprincipal Dirichlet character.

The first-degree part of the Euler product logarithm givesThe omitted terms are bounded uniformly, since ${\sum }_p{\sum }_{m\ge 2}(m{p}^{m\sigma }{)}^{-1}\ll {\sum }_p{p}^{-2}<\infty$. For each nonprincipal factor, its logarithm remains bounded: choose a holomorphic logarithm near the now established nonzero value $L(1,\chi )$; the Euler-product branch differs from it by a fixed integral multiple of $2\pi i$ on a sufficiently small real interval. For the principal factor,Applying Orthogonality of Dirichlet characters again givesThe right side diverges as $\sigma \downarrow 1$, so the arithmetic progression contains infinitely many primes. This proves Dirichlet's theorem on arithmetic progressions: every reduced residue class modulo $q$ contains infinitely many primes. The argument also covers $q=1$ and $q=2$, when there need not be a nonreal character.

For the sharper statement about reciprocal primes in a fixed arithmetic progression, use the given prime-counting function asymptotic with $q$ fixed:The second equivalence follows by integration by parts in the logarithmic integral function. Partial summation yieldsFor any $\epsilon >0$, choose a fixed $T$ so large that the relative error in $\pi (t;q,a)$ is at most $\epsilon$ for $t\ge T$. The initial integral is a constant, while the remaining integral lies betweenThe boundary term is $O_q(1/\log x)$. Divide by $\log \log x$, let $x\to \infty$, and then let $\epsilon \downarrow 0$. We obtainOnly a relative asymptotic is deduced from the given information; an additive $O(1)$ error, or uniformity for a growing modulus $q$, would require additional estimates.

Fix integers $q\ge 1$ and $a$ with $(a,q)=1$. We prove the Dirichlet theorem on primes in arithmetic progressions by showing that the prime Dirichlet sum in this residue class diverges as its real exponent decreases to one.

First justify the required Dirichlet L-function facts. If $\chi$ is a nonprincipal Dirichlet character modulo $q$, its sum over a full period is zero: choose a unit $b$ with $\chi (b)\ne 1$, and multiplication by $b$ permutes the residue classes, forcing the sum to equal $\chi (b)$ times itself. Therefore the partial sums $C_{\chi }(u)={\sum }_{n\le u}\chi (n)$ are bounded by $q$. Partial summation giveswhich supplies holomorphic continuation near one, with locally uniform convergence. For the principal Dirichlet character ${\chi }_0$,has a simple pole at one, with residue $\phi (q)/q$.

The assumption in the question gives nonvanishing at one for real nonprincipal characters. We must also prove it for nonreal characters. For a nonreal $\chi$, its square ${\chi }^2$ is nonprincipal. For real $\sigma >1$, logarithms of the Euler products giveFor a prime not dividing $q$, its logarithmic coefficient is $3+4\mathrm{Re}\chi (p{)}^k+\mathrm{Re}\chi (p{)}^{2k}=2(1+\cos \theta {)}^2\ge 0$, where $\chi (p{)}^k=e^{i\theta }$. Primes dividing $q$ contribute only the positive zeta term. If $L(1,\chi )$ vanished to order $m\ge 1$, the product would be $O((\sigma -1{)}^{4m-3})$ and tend to zero: the squared-character factor is bounded because ${\chi }^2$ is nonprincipal. This is a contradiction. HenceThis step uses the given real-character hypothesis and the Euler product positivity for L-function nonvanishing for all remaining characters.

In $\sigma >1$ choose the logarithm furnished by the absolutely convergent Euler product. Its prime-power expansion givesThe error from powers $k\ge 2$ is uniformly bounded as $\sigma \downarrow 1$, by ${\sum }_p{\sum }_{k\ge 2}1/(k{p}^{k\sigma })<\infty$. For nonprincipal $\chi$, nonvanishing and holomorphic continuation at one provide a bounded logarithm in a neighborhood of one. The Euler-product branch differs from that branch by a constant integral multiple of $2\pi i$ on the final real interval, so it too is bounded. ThusFor the principal character, its simple pole instead gives

The Orthogonality of Dirichlet characters gives, for primes not dividing $q$,Multiply by $p^{-\sigma }$ and sum. Absolute convergence allows the interchange for $\sigma >1$, soThe right side tends to infinity. If there were only finitely many primes in the progression, the left side would have a finite limit at one. This contradiction proves infinitely many primes occur in every reduced residue class. For $q=1$ or $2$ the character argument simply has no nonprincipal terms and gives the same conclusion.

For the final deduction the modulus $q$ is fixed. Let $A(t)=\pi (t;q,a)$. The given consequence of the Siegel–Walfisz theorem is $A(t)\sim t/(\phi (q)\log t)$. Partial summation, including the prime $2$ through the lower endpoint $2^{-}$ when appropriate, givesThe boundary is $O_q(1/\log x)$. Write $A(t)=t(1+\epsilon (t))/(\phi (q)\log t)$ with $\epsilon (t)\to 0$. The main integral is $(\log \log x-\log \log 2)/\phi (q)$. The error is $o(\log \log x)$: given $\eta >0$, choose $T$ such that $|\epsilon (t)|\le \eta$ for $t\ge T$. Its integral from $2$ to $T$ is a fixed constant, and its absolute value from $T$ to $x$ is at most $\eta \log \log x/\phi (q)$ plus a fixed constant. Divide by $\log \log x$ and then let $\eta \downarrow 0$. We obtain the reciprocal primes in a fixed arithmetic progression asymptoticThe argument uses the supplied prime-counting asymptotic in this last part; the preceding infinitude proof did not assume the Prime number theorem in arithmetic progressions.