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Fix a modulus $q$ and a residue class $a$ with $(a,q)=1$. The Dirichlet characters modulo $q$ are the one-dimensional characters of a representation of $(\mathbb{Z}/q\mathbb{Z}{)}^{\times }$, extended by zero on nonunits. Let ${\chi }_0$ be the principal Dirichlet character.

For every nonprincipal Dirichlet character the partial sums ${\sum }_{n\le u}\chi (n)$ are bounded: the sum over a full period is zero, and a remaining incomplete period has bounded length. Partial summation therefore continues its Dirichlet L-function holomorphically to $\Re s>0$ byThis shows in particular that it has no pole at one. By contrasthas a simple pole there. We use the stipulated nonvanishing $L(1,\chi )\ne 0$ for every real nonprincipal Dirichlet character, and establish the remaining nonvanishing as follows.

For real $\sigma >1$, take the product over all Dirichlet characters. The Euler product logarithms and Orthogonality of Dirichlet characters giveThus this product is real and at least one. If a nonreal Dirichlet character had $L(1,\chi )=0$, complex conjugation would also give $L(1,\overline{\chi })=0$. These are distinct factors. Two zeros and only one possible simple pole force the whole product to tend to zero as $\sigma \downarrow 1$, contradicting the lower bound. Hence all nonprincipal Dirichlet characters have $L(1,\chi )\ne 0$, under the given real-character assumption.

For $\sigma >1$, absolute convergence of each Euler product giveswith bounded remainder as $\sigma \downarrow 1$: the terms of prime-power exponent at least two have total modulus at most ${\sum }_{n\ge 2}1/(n(n-1))$. Since a nonprincipal $L$ defines a holomorphic function that is nonzero near one, its Euler-product logarithm differs from a local holomorphic logarithm by a fixed integer multiple of $2\pi i$ on a short real interval. It is therefore bounded there. For the principal Dirichlet character the pole instead givesApply the Dirichlet character orthogonality relation at each prime. The prime-character sum near one yieldsThis tends to infinity. A finite set of primes would have a bounded sum as $\sigma \downarrow 1$, so every coprime residue class contains infinitely many primes. This proves the Dirichlet theorem on primes in arithmetic progressions from the specified assumption. In fact it proves divergence of the corresponding reciprocal-prime series and gives Dirichlet density $1/\phi (q)$.

Finally, for real $s>1$ each prime satisfiesThe integrands are nonnegative, so Tonelli theorem permits summation inside the integral. At a fixed $t$, exactly $\pi (t,q,a)$ primes from the residue class have entered. ThereforeBoth sides are finite, since they are bounded respectively by ${\sum }_{n\ge 2}{n}^{-s}$ and $s{\int }_1^{\infty }{t}^{-s}dt$. The values of a counting function at its isolated jump endpoints do not change this integral.

For a positive integer $N$, a Dirichlet character $\psi$ modulo $N$ is a group homomorphism $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }\to {\mathbb{C}}^{\times }$, extended periodically to the integers and set equal to zero on integers not coprime to $N$. Its nonzero values are roots of unity. For $\mathrm{Re}s>1$, its Dirichlet L-function is the absolutely convergent Dirichlet seriesComplete multiplicativity of $\psi$ proves the Euler product by expanding each factor as a geometric series and applying unique integer factorization; absolute convergence justifies these rearrangements.

Suppose $\psi$ is a nonprincipal Dirichlet character. Choose a unit $u$ with $\psi (u)\ne 1$. Multiplication by $u$ permutes a complete residue system, so ${\sum }_{a=1}^N\psi (a)=\psi (u){\sum }_{a=1}^N\psi (a)$, and the sum is zero. Consequently $A(x)={\sum }_{n\le x}\psi (n)$ is bounded: complete periods contribute zero and the remaining incomplete period has at most $N$ terms. The Abel summation formula gives, initially for $\mathrm{Re}s>1$,For $\mathrm{Re}s>0$ the integral converges, locally uniformly together with its derivatives in $s$, because on a compact set with $\mathrm{Re}s\ge \delta >0$ the differentiated integrands are bounded by constants times $x^{-1-\delta }(1+\log x{)}^k$. Thus holomorphy of a Dirichlet series from bounded partial sums proves

For a nonreal $\psi$, the character ${\psi }^2$ is also nonprincipal. The supplied logarithmic Euler product yields, for real $\sigma >1$,At a prime dividing $N$ the numerator is $3$. At every other prime put $\psi (p{)}^k=e^{i\theta }$; then the numerator is $3+4\cos \theta +\cos 2\theta =2(1+\cos \theta {)}^2\ge 0$. This Euler product positivity for L-function nonvanishing proves that the product inside the logarithm is at least one. If $L_N(\psi ,1)=0$, its holomorphy would give $L_N(\psi ,\sigma )=O(\sigma -1)$. Meanwhile $L_N({\psi }^2,\sigma )$ is bounded near one and the Riemann zeta function has a simple pole there. The product would therefore be $O(\sigma -1)$ and tend to zero, a contradiction.

For a real nonprincipal $\psi$, suppose instead that $L_N(\psi ,1)=0$. The product $H(s)=\zeta (s)L_N(\psi ,s)$ would then be holomorphic on the whole half-plane $\mathrm{Re}s>0$, because the zero cancels the only zeta pole there. In $\mathrm{Re}s>1$ it has coefficientsThese are nonnegative zeta-times-real-L coefficients. They are multiplicative, and their prime-power values are $k+1$ if $\psi (p)=1$, one or zero according as $k$ is even or odd if $\psi (p)=-1$, and one if $\psi (p)=0$. In particular $b_n\ge 0$ and $b_{m^2}\ge 1$ for every $m$.

The Taylor series of the assumed holomorphic $H$ about $2$ converges in the disk $|s-2|<2$. Termwise differentiation of its absolutely convergent Dirichlet series at $2$ gives $(-1{)}^k{H}^{(k)}(2)={\sum }_n{b}_n(\log n{)}^k/n^2$. Evaluate the Taylor series at $s=1/2$ and apply the Tonelli theorem to the resulting nonnegative terms:This contradicts holomorphy at $1/2$. The argument is the positive-coefficient mechanism of the Landau theorem for a Dirichlet series with nonnegative coefficients, with the needed step proved explicitly. Combining the real and nonreal cases gives the nonvanishing of a nonprincipal Dirichlet L-function at one:

Now let $a$ be coprime to $N$. The Dirichlet theorem on primes in arithmetic progressions asserts that infinitely many prime numbers satisfy $p\equiv a(\mathrm{mod}N)$. To prove it, separate the first-power terms in the given logarithmic Euler product:For $\sigma \ge 1$, the remainder is bounded uniformly byFor a nonprincipal character, holomorphy and nonvanishing near one give a local holomorphic logarithm. Along a short real interval to the right of one, this logarithm and the supplied continuous logarithm differ by a fixed integer multiple of $2\pi i$. Hence the prime-character sum near one is bounded:For the principal Dirichlet character ${\psi }_0$,The latter follows from the simple zeta pole and the bounded remainder. Finally Orthogonality of Dirichlet characters givesHere $\phi$ is the Euler totient function. The right-hand side tends to infinity as $\sigma \downarrow 1$, whereas a finite set of primes would give a bounded left-hand side. Therefore

Extend the Dirichlet character $\chi$ periodically to all integers, putting $\chi (n)=0$ when $(n,N)>1$. Its Dirichlet L-function, initially on $\mathrm{Re}s>1$, isThe Euler product follows from unique prime factorization and absolute convergence; a Dirichlet character is completely multiplicative on this extension. Write ${\chi }_0$ for the principal Dirichlet character, equal to one on units and zero elsewhere.

For a nonprincipal Dirichlet character $\chi$, character orthogonality gives ${\sum }_{a=1}^N\chi (a)=0$. Explicitly, multiplication by a unit $u$ with $\chi (u)\ne 1$ permutes the unit residues and multiplies this sum by $\chi (u)$, forcing it to vanish. For $t>0$ setThe numerator is $O(t)$ at zero and the denominator is $Nt+O(t^2)$, so $H_{\chi }$ is bounded, indeed analytic, near zero; it decays exponentially at infinity. Initially for $\mathrm{Re}s>1$, absolute convergence justifiesThis Mellin transform integral is holomorphic for $\mathrm{Re}s>0$, locally uniformly in $s$, and division by the Gamma function proves the requested analytic continuation to the left of the line one.

In fact, the same argument proves Mellin continuation of a nonprincipal Dirichlet L-function to the entire plane. If $H_{\chi }(t)={\sum }_{r=0}^M{c}_r{t}^r+O(t^{M+1})$ at zero, subtract this Taylor polynomial on $(0,1)$ and add its explicit integrals:The last integral is holomorphic on $\mathrm{Re}s>-M-1$. Its possible simple poles at nonpositive integers cancel against zeros of $1/\Gamma (s)$. Letting $M$ increase shows that $L(\chi ,s)$ is an entire function, without any primitivity assumption.

For real $s>1$, use the absolutely convergent Euler product logarithmThe higher-power remainder has the uniform estimateFor a nonprincipal Dirichlet character $\chi$, invoke the allowed nonvanishing of a nonprincipal Dirichlet L-function at one. Its holomorphy and nonvanishing give a holomorphic logarithm on a small disk about one. On the connected real interval $1<s<1+\eta$, $B_{\chi }(s)$ differs from this logarithm by a fixed element of $2\pi i\mathbb{Z}$: the difference is continuous with exponential one. Thus the prime-character sum near one is bounded. This branch argument is needed for complex-valued Dirichlet characters.

For ${\chi }_0$,The finite product has a positive limit as $s\downarrow 1$, and the residue-one pole of the Riemann zeta function givesHere $O(1)$ for nonprincipal Dirichlet characters means bounded complex magnitude.

Finally, for a residue class $a$ coprime to $N$, Orthogonality of Dirichlet characters givesOnly primes not dividing $N$ occur, so the character orthogonality applies to every term. This sum diverges as $s\downarrow 1$. A finite collection of primes would give a bounded sum, a contradiction. Every reduced residue class contains infinitely many primes. This is the Dirichlet theorem on primes in arithmetic progressions; the coprimality hypothesis is essential.