Dirichlet character chebyshev laplace transform

The linear theta bound and prime-power comparison give $\psi (x)=O(x)$, uniformly after enlarging the constant for bounded x. Since $\mathrm{\mid }\chi (n)\mathrm{\mid }\le 1$, $\mathrm{\mid }{\mathrm{\Psi }}_{\chi }(x)\mathrm{\mid }\le \psi (x)$; thus $f_{\chi }$ is bounded and locally integrable, with only finitely many jumps on each compact t-interval.

For nonprincipal chi, holomorphy on $\mathrm{Re}w>0$ and nonvanishing at w=1 and at every $1+it$, $t\mathrm{\ne }0$, show that the logarithmic derivative is holomorphic near every point of that line; the Euler product covers its right side. For the principal character, $L(w,{\chi }_0)=\zeta (w){\prod }_{p\mid q}(1-p^{-w})$ continues meromorphically, with a simple pole at one and no zero on $\mathrm{Re}w\ge 1$. The finite factors cannot vanish there because $\mathrm{\mid }{p}^{-w}\mathrm{\mid }<1$.

The Euler logarithm is normally absolutely convergent on $\mathrm{Re}w>1$; its differentiated series is dominated on each smaller half-plane by $\sum (\log n)n^{-\sigma }$. Differentiation gives $-L^{\mathrm{\prime }}\mathrm{/}L(w,\chi )=\sum \chi (n)\mathrm{\Lambda }(n)n^{-w}$. Apply the summatory integral at $w=s+1$ and substitute $x=e^t$, obtaining the displayed formula with the factor s+1 intact.

At s=0 in the principal case write $-L^{\mathrm{\prime }}\mathrm{/}L(1+s)=1\mathrm{/}s+h(s)$ with h holomorphic. Then $g_{{\chi }_0}(s)=-1\mathrm{/}(1+s)+h(s)\mathrm{/}(1+s)$ is holomorphic. Elsewhere shrink the pointwise neighborhoods to avoid s=-1. The union of these neighborhoods and the original half-plane is the required open set. For q=1 the finite product is empty and equals one.