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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.

For a primitive Dirichlet character whose conductor of a Dirichlet character is $q>1$ and whose character parity is $a$, Poisson summation gives ${\theta }_{\chi }(x)={\epsilon }_{\chi }{x}^{-a-1/2}{\theta }_{\overline{\chi }}(1/x)$, where ${\epsilon }_{\chi }=\tau (\chi )/(i^a\sqrt{q})$. Its rapid decay at both ends yields the entire Mellin representation of the completed Dirichlet L-function. The case of conductor of a Dirichlet character equal to one uses the ordinary Jacobi theta function and a subtracted constant term.

The inducing primitive Dirichlet character has conductor of a Dirichlet character $q_0$. Increasing the modulus removes Euler factors only at new prime divisors. The correction creates zeros on the imaginary axis; raising the exponents of primes already dividing the conductor of a Dirichlet character changes no Euler factor. Primitive gamma factors continue to use $q_0$. For real-even characters this includes principal primitive conductor of a Dirichlet character one, with its Riemann zeta function pole treated separately.

A primitive Dirichlet character modulo $q$ is a Dirichlet character which is not induced from a character of a proper divisor of $q$. To determine the inducing primitive Dirichlet character, use the Chinese remainder theorem to decomposeOn each factor choose the least exponent $c_p\in \{0,\dots ,k\}$ through whose reduction the restricted character factors. Exponent zero means the trivial unit group modulo one. Put $q_0={\prod }_p{p}^{c_p}$ and define ${\chi }_0$ on the units modulo $q_0$ by these descended factors, extending by zero off the units. Every reduction of unit groups is surjective, so the descended character is unique. Its local exponents cannot be decreased, hence it is primitive. The original character is $\chi (n)={\chi }_0(n)$ when $(n,q)=1$, and zero otherwise.

Any other inducing modulus must have exponent at least $c_p$ at every prime, by restriction to the corresponding local factor. Therefore $q_0$ is the unique minimal modulus, the conductor of a Dirichlet character, and ${\chi }_0$ is the unique primitive Dirichlet character inducing $\chi$. The argument also explains why removing extra prime factors can change values at integers that were nonunits for $q$.

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.

The completed Dirichlet L-function isFor nonprincipal primitive Dirichlet characters, termwise Mellin transformation initially in $\Re s>1$ givesThe Dirichlet character theta function decays exponentially at infinity; its transformation makes it decay faster than any power at zero. Hence the integral is entire in $s$. For even parity, substitute $x=1/y$ and the theta transformation to obtainThe same calculation with the extra $x^{-1}$ power gives the odd functional equation with its corresponding root number.

The gamma function has no zeros and has simple poles at nonpositive integers. Thus the nontrivial zeros of $L$ and $\Lambda$ coincide with multiplicities. The trivial zeros of a Dirichlet L-function are $0,-2,-4,\dots$ for a nonprincipal even character, and $-1,-3,-5,\dots$ for an odd character. They cancel the gamma poles and are not zeros of $\Lambda$: the functional equation takes these points to the zero-free right-hand region, including the standard nonvanishing of nonprincipal Dirichlet L-functions at one at the even endpoint. The canceled zeros are simple.

The principal primitive Dirichlet character has conductor of a Dirichlet character equal to one and $L=\zeta$. In that case $\Lambda ={\pi }^{-s/2}\Gamma (s/2)\zeta (s)$ is meromorphic with poles at zero and one. Its canceled trivial zeros begin at $-2$, while $\zeta (0)=-1/2$ is not zero. Multiplication by $s(s-1)/2$ produces the entire Riemann xi function used below.

Let $q_0$ be the conductor of a Dirichlet character and ${\chi }_0$ the inducing primitive even character. Removing the Euler factors absent from $L(s,\chi )$ gives the imprimitive Dirichlet L-function Euler correctionThe primitive functional equation therefore givesEquivalently, replace the final primitive function by $L(1-s,\overline{\chi })/P_{\overline{\chi }}(1-s)$, interpreted as a meromorphic identity with removable values handled by continuation. It is the conductor of a Dirichlet character $q_0$, rather than the possibly inflated modulus $q$, that enters the gamma factor and root number. The complex conjugation bar in the original PDF is lost in the converted TeX.

The zeros are those of $L(s,{\chi }_0)$ together with the zeros of the finite Euler product, and multiplicities add. Since $|{\chi }_0(p)|=1$ at each extra prime, an extra factor vanishes at the imaginary points determined byEach extra prime creates infinitely many such points. Its nonzero-imaginary points are not zeros of the primitive function: the functional equation and nonvanishing of Dirichlet L-functions on the line one exclude them. Thus the zero sets are identical precisely when every prime dividing $q$ already divides $q_0$, making $P=1$. Increasing prime-power exponents alone can make a character imprimitive without changing its L-function. If $q_0=1$, the primitive function is zeta; the same Euler correction applies, with its pole at one retained.

Put $D_0=|D|$. The reference to part (c) in the printed hint is a reference to the positive-coefficient function from part (b). For $\sigma >1$ close to one, the preceding positivity and the supplied partial-fraction expansion giveAll omitted zero terms have nonnegative real parts because their real parts are at most one. The $O(1)$ zeta-pole remainder is included in $C\log D_0$, increasing the absolute constant if needed; a nonprincipal primitive real conductor of a Dirichlet character is at least three.

Suppose there were two real zeros, counted with multiplicity, with $\beta \ge 1-c/\log D_0$. Set $\sigma =1+a/\log D_0$. Division by $\log D_0$ givesChoose $C\ge 1$, $a=1/(4C)$ and $c=a/4$. The right side is strictly negative. ThusThis is the uniqueness of a possible exceptional real Dirichlet zero. It proves uniqueness, rather than existence of such a zero.