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For a primitive nonprincipal Dirichlet character, the theta Mellin integral makes this an entire function satisfying $\Lambda (s,\chi )={\epsilon }_{\chi }\Lambda (1-s,\overline{\chi })$. The gamma poles cancel the trivial zeros of a Dirichlet L-function. For the principal Dirichlet character whose conductor of a Dirichlet character is one the completion is meromorphic, with poles at zero and one.

A nonprincipal Dirichlet character differs from the principal Dirichlet character. Equivalently, it is a nontrivial character on the unit group, so its sum over one full residue period is zero by character orthogonality. Its Dirichlet L-function is entire by Mellin continuation of a nonprincipal Dirichlet L-function.

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 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.

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.

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

For $\Re s>1$, the Euler products areThe Dirichlet character is completely multiplicative, and unique prime factorization expands each finite product into the terms of the corresponding Dirichlet series supported on those primes. Absolute convergence permits passage to all primes. More strongly, the absolutely convergent Euler product logarithm givessince the sum of absolute values is bounded by ${\sum }_{n\ge 2}{\sum }_{m\ge 1}{n}^{-m\Re s}/m<\infty$.

We prove nonvanishing of a nonprincipal Dirichlet L-function at one, including real Dirichlet characters. If $\chi$ is nonprincipal modulo $q$, periodicity and the given Orthogonality of Dirichlet characters bound $A_{\chi }(x)={\sum }_{n\le x}\chi (n)$ by $q$. Partial summation giveswhich is holomorphic for $\Re s>0$ by locally uniform convergence, including after differentiation. The principal Dirichlet character instead haswith a simple pole at one.

For a nonreal Dirichlet character, consider $P(s)={\prod }_{\chi \mathrm{mod}q}L(s,\chi )$. For real $\sigma >1$, its Euler product logarithm isThe given character inversion with $a=1$ makes each inner sum either $\phi (q)$ or zero, so $P(\sigma )\ge 1$. If $L(1,\chi )=0$ for a nonreal Dirichlet character, then $L(1,\overline{\chi })=0$ as well, since their values at real $s$ are conjugate. These are two distinct factors. All nonprincipal factors are holomorphic at one, and only the principal factor has a simple pole. Their two zeros would force $P(\sigma )\to 0$, a contradiction.

For a real nonprincipal Dirichlet character, suppose $L(1,\chi )=0$. Then $G(s)=\zeta (s)L(s,\chi )$ is holomorphic throughout $\Re s>0$: the supplied continuation of the Riemann zeta function has no other pole there, and the zero cancels the pole at one. On $\Re s>1$ its Dirichlet series has coefficientsTo check the sign explicitly, for a prime power $p^j$ the local sum is $j+1$ if $\chi (p)=1$, is $1$ for even $j$ and $0$ for odd $j$ if $\chi (p)=-1$, and is $1$ if $\chi (p)=0$. Multiplication over prime factors proves nonnegativity and $g_{m^2}\ge 1$. Also $g_n\le \tau (n)$, so termwise derivatives at $s=2$ are justified by absolute convergence:The disc $|s-2|<2$ lies in $\Re s>0$. Its Taylor series therefore converges at $s=1/2$, and Tonelli theorem givesThis contradicts holomorphy at $1/2$. It is the positive-coefficient argument behind the Landau theorem for a Dirichlet series with nonnegative coefficients, proved here rather than invoked. Thus $L(1,\chi )\ne 0$ for every nonprincipal Dirichlet character.

It remains to detect the primes. For real $\sigma >1$, the Euler product logarithm satisfiesuniformly as $\sigma \downarrow 1$: terms with $m\ge 2$ are bounded in absolute value by ${\sum }_{n\ge 2}{\sum }_{m\ge 2}{n}^{-m}/m<\infty$. For a nonprincipal Dirichlet character, a holomorphic logarithm of $L$ exists near one by the nonvanishing just proved. On a short real interval it differs from the Euler product logarithm by a fixed multiple of $2\pi i$, so the prime sum stays bounded. For the principal Dirichlet character, the simple pole instead givesApplying the supplied Orthogonality of Dirichlet characters for a reduced residue class $a$ now yieldsThe right side tends to infinity, whereas a finite set of primes would give a bounded left side. Therefore every residue class coprime to $q$ contains infinitely many primes, proving the Dirichlet theorem on primes in arithmetic progressions.

An even Dirichlet character satisfies $\chi (-1)=1$, while an odd Dirichlet character satisfies $\chi (-1)=-1$. Write $a=0$ or $1$ for its character parity and, for $x>0$, define the Dirichlet character theta functionFor a primitive Dirichlet character whose conductor of a Dirichlet character is $q>1$, the term at zero is zero. Put $\tau (\chi )={\sum }_{r\mathrm{mod}q}\chi (r)e(r/q)$, using the positive exponential, and ${\epsilon }_{\chi }=\tau (\chi )/(i^a\sqrt{q})$. The primitive Gauss sum of a Dirichlet character has magnitude $\sqrt{q}$, so $|{\epsilon }_{\chi }|=1$. The theta transformation isThus the powers are $x^{-1/2}$ in the even case and $x^{-3/2}$ in the odd case; the odd root number contains $1/i$. The conjugate character is necessary for a nonreal character. These formulas also follow by applying Poisson summation to the Gaussian function on each residue class, and to its derivative for odd parity. For the primitive principal Dirichlet character whose conductor of a Dirichlet character is one, use the ordinary Jacobi theta function with constant term one; its transformation has root number one.

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.

First establish rational conjugation of finite-index modular subgroups without assuming that $\Gamma$ is a congruence subgroup. Multiply $\gamma$ by a positive integer to obtain an integral matrix $A$, and let $D=\det A>0$. Conjugation is unchanged by this scalar. If $h=I+DB\in \Gamma (D)$, thenThus the principal congruence subgroup $\Gamma (D)$ is contained in $H=S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}S{L}_2(\mathbb{Z})\gamma$. It has finite index because reduction modulo $D$ has finite image. Inside $H$, pullback under conjugation of $\Gamma$ has relative index at most $[S{L}_2(\mathbb{Z}):\Gamma ]$. ConsequentlyThis argument does not assert that an arbitrary finite-index subgroup contains a principal congruence subgroup.

Use the determinant-normalized slash operatorThe positive real power of the determinant is used; on $S{L}_2(\mathbb{Z})$ this reduces to the usual slash operator for modular forms. The automorphy factor identity gives the right-action rule $(f{|}_k\gamma ){|}_k\eta =f{|}_k(\gamma \eta )$.

A modular form on a finite-index subgroup of integer weight $k$ is a holomorphic function on the complex upper half-plane, invariant under this weight-$k$ action of $\Gamma$, and holomorphic at a cusp at each of its cusps. A cusp of a modular group is a $\Gamma$ orbit in ${\mathbb{P}}^1(\mathbb{Q})$. If $\sigma \in S{L}_2(\mathbb{Z})$ carries infinity to its representative, choose a positive integer $h$ with $\sigma T^h{\sigma }^{-1}\in \Gamma$. Such $h$ exists by finite index. Then $f{|}_k\sigma$ is periodic and has a convergent expansion in $q_h=e^{2\pi iz/h}$ near zero; holomorphy means no negative exponents, and being a cusp form means zero constant term. Using an actual translation period avoids possible signs if a smaller width of a cusp is defined only modulo the center, particularly in odd weights.

For cusp holomorphy under rational slash operators, choose $\sigma \in S{L}_2(\mathbb{Z})$ with $\sigma \infty =\gamma \infty$, possible by completing a primitive integer pair to a determinant-one matrix. Then ${\sigma }^{-1}\gamma =\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$, with $a/d>0$. Up to a nonzero constant factor,The imaginary part of the argument tends to infinity with that of $z$, so this remains bounded by the cusp expansion of $f{|}_k\sigma$. It tends to zero if $f$ is a cusp form. Moreover $f{|}_k\gamma$ is invariant under ${\Gamma }^{\prime }$: for $\eta \in {\Gamma }^{\prime }$, $\gamma \eta {\gamma }^{-1}\in \Gamma$ and the right-action rule applies. Finite index gives a translation period for ${\Gamma }^{\prime }$, so boundedness is a removable singularity at zero in that periodic parameter. This proves holomorphy at infinity. For every other cusp, apply the same argument to the rational matrix $\gamma \rho$, with $\rho \in S{L}_2(\mathbb{Z})$. Thus all cusp conditions hold, and

For the character twist by rational translations of a cusp form, put $B_j=\left(\begin{array}{cc}1& j/N\\ 0& 1\end{array}\right)$ and $K={\Gamma }_1(N)\cap {\Gamma }_0(N^2)$. For $\eta =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in K$, direct conjugation givesIndeed $N^2\mid c$ and $a\equiv d\equiv 1(\mathrm{mod}N)$. Every $f{|}_k{B}_j=f(z+j/N)$ is therefore $K$-invariant and vanishes at all its cusps by the preceding rational-translate argument. Their finite weighted sum is a cusp form, for every Dirichlet character:

There is, however, a missing primitivity hypothesis in the printed final expansion claim. The exact Fourier expansion of a modular form is alwaysValues of a Dirichlet character on units have modulus one, so $\chi (j{)}^{-1}=\overline{\chi (j)}$. For unit $n$, substitution gives $S_{\overline{\chi }}(n)=\chi (n)g(\overline{\chi })$, where $g$ is the Gauss sum of a Dirichlet character. For nonunit $n$, this vanishing formula requires a primitive Dirichlet character.

Here is its proof in that case. Choose a prime $p\mid (n,N)$. Primitivity supplies a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$: otherwise the character would factor through the surjective reduction to units modulo $N/p$. Surjectivity follows by lifting a unit and, if needed, adjusting the lift to avoid the additional prime $p$, using the Chinese remainder theorem. Multiplication by $u$ fixes $e^{2\pi inj/N}$ because $p\mid n$, but multiplies the character factor by a nontrivial constant. Hence $S_{\overline{\chi }}(n)=0$. The finite Fourier transform of a primitive Dirichlet character now gives the corrected formulaThe constant is nonzero: finite exponential orthogonality gives ${\sum }_{n\mathrm{mod}N}|S_{\overline{\chi }}(n){|}^2=N\phi (N)$, whereas the proved formula makes this $\phi (N)|g(\overline{\chi }){|}^2$. Thus $|g(\overline{\chi })|=\sqrt{N}$.

For a concrete counterexample to the printed unrestricted claim, take $N=2$, the principal Dirichlet character, and $f=\Delta$. The translation sum is $\Delta (z+1/2)={\sum }_n(-1{)}^n\tau (n)q^n$, whose $q^2$ coefficient is $-24$. Any constant multiple of the proposed odd-index-only series has $q^2$ coefficient zero. Thus the general modularity conclusion is proved, while the claimed simplification is false without the stated extra hypothesis.

For real $s>1$, the logarithm defined by the Euler product of a Dirichlet L-function differs from $F_{\chi }(s)$ by a quantity of absolute value at most ${\sum }_{n\ge 2}1/(n(n-1))=1$. If $L(\chi ,1)\ne 0$, a local holomorphic logarithm differs from this continuous logarithm by one fixed multiple of $2\pi i$ on a short real interval, so $F_{\chi }(s)$ stays bounded as $s\downarrow 1$. For the principal Dirichlet character, the pole of the Riemann zeta function instead gives $F_{{\chi }_0}(s)=\log (1/(s-1))+O(1)$. Orthogonality of Dirichlet characters then makes the sum over any reduced residue class diverge, proving infinitude of its primes.

For a character of modulus $p^k$, a primitive Dirichlet character attains the bound by finite Plancherel theorem. An imprimitive character with $k\ge 2$ has periodic values modulo $p^{k-1}$, so its frequency-one sum vanishes. The principal Dirichlet character modulo $p$ has unnormalized sum $-1$. The sign convention of the exponential does not affect its magnitude.

The principal Dirichlet character modulo $N$ is one on integers coprime to $N$ and zero on all other integers. Its Dirichlet L-function is $L({\chi }_0,s)=\zeta (s){\prod }_{p\mid N}(1-p^{-s})$, so it has a simple pole at one. For $N>1$ it is induced from the character of modulus one and is not a primitive Dirichlet character.

For real $s>1$, Orthogonality of Dirichlet characters gives $\log {\mathcal{P}}_q(s)=\phi (q){\sum }_{p\nmid q}{\sum }_{j\ge 1,p^j\equiv 1(\mathrm{mod}q)}1/(j{p}^{js})\ge 0$. Thus ${\mathcal{P}}_q(s)\ge 1$. Only the principal Dirichlet character contributes a pole at one, and that pole is simple. A zero of a nonreal Dirichlet L-function at one brings a zero of its conjugate too; their two zeros overpower that pole and would make the product tend to zero. This proves nonvanishing for nonreal Dirichlet characters and is a useful complement to arguments for real Dirichlet characters.