Fix a $w$ in $\Re \left({w}\right) \ge 1$. Then $F \left({z + w}\right)$ is analytic in $\Re \left({z}\right) \ge 0$. {{explain|The above needs to be proved. We know that $F \left({z}\right)$ is analytic throughout $\Re \left({z}\right) \ge 1$ but we are told nothing about what it's like on $\Re \left({z}\right) \ge 0$. Also see below where it is also stated that it is analytic on $\Re \left({z}\right) = 0$ as well. We also have further down an implicit assumption that $F \left({z}\right)$ is analytic on $\Re \left({z}\right) \ge -\delta$ where $-\delta$ appears to be negative (but this is also unclear). These also both need to be demonstrated.}} We note that since $F \left({z + w}\right)$ is analytic on $\Re \left({z}\right) = 0$, it must be analytic on an open set containing $\Re \left({z}\right) = 0$.  Choose some $R \ge 1$. {{explain|What is the domain of $R$?}} We have that $F \left({z + w}\right)$ is analytic on such an open set. Thus we can determine $\delta = \delta \left({R}\right) > 0, \delta \le \dfrac 1 2$ such that $F \left({z + w}\right)$ is analytic in $\Re \left({z}\right) \ge -\delta, \left\vert{\Im \left({z}\right)}\right\vert \le R$. {{explain|What is the domain and range of $\delta$?}} We also choose an $M = M \left({R}\right)$ so that $F \left({z + w}\right)$ is bounded by $M$ in $-\delta \le \Re \left({z}\right), \left\vert{z}\right\vert \le R$. {{explain|What are the domain and range of $M$?}} Now form the counterclockwise contour $\Gamma$ as the arc $\left\vert{z}\right\vert = R, \Re \left({z}\right) > - \delta$ and the segment $\Re \left({z}\right) = -\delta, \left\vert{z}\right\vert \le R$. We denote by $A, B$ respectively, the parts of $\Gamma$ in the right and left half-planes. By the Residue Theorem: :$\displaystyle 2 \pi i F \left({w}\right) = \oint_{\Gamma} F \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z$ Since $F \left({z + w}\right)$ converges to its series on $A$, we may split it into the partial sum and remainder after $N$ terms: :$s_N \left({z + w}\right), r_N \left({z + w}\right)$ respectively. Again, by the Residue Theorem: :$\displaystyle \int_A s_N \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z = 2 \pi i s_N \left({w}\right) - \int_{-A} s_N \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z$ where $-A$ is the reflection of $A$ through the origin. Changing $z \to -z$, we have: :$\displaystyle \int_A s_N \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z = 2 \pi i s_N \left({w}\right) - \int_A s_N \left({w - z}\right) N^{-z} \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z$ Combining these results gives: {{begin-eqn}} {{eqn | o =        | r = \displaystyle 2 \pi i \left({F \left({w}\right) - s_N \left({w}\right) }\right)       | c =  }} {{eqn | r = \int_\Gamma F \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z - \int_A s_N \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z - \int_A s_N (w-z) N^{-z} \left({ \frac 1 z + \frac z {R^2} }\right) \, \mathrm d z       | c =  }} {{eqn | r = \int_A F \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z + \int_B F \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z       | c =  }} {{eqn | o =        | ro= -       | r = \int_A s_N \left({z + w}\right) N^z \left({ \frac 1 z + \frac z {R^2} }\right) \, \mathrm d z - \int_A s_N(w-z) N^{-z} \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z       | c =  }} {{eqn | r = \int_A \left({r_N \left({z + w}\right) N^z - s_N(w-z)N^{-z} }\right) \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z + \int_B F \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z       | c =  }} {{end-eqn}} For what follows, allow $z = x + i y$ and observe that on $A, \left\vert{z}\right\vert = R$. So: {{begin-eqn}} {{eqn | l = \frac 1 z + \frac z {R^2}       | r = \frac {\overline z} {\left\vert{z}\right\vert^2} + \frac z {R^2}       | c =  }} {{eqn | r = \frac {x - i y} {R^2} + \frac {x + i y} {R^2}       | c =  }} {{eqn | r = \frac {2 x} {R^2}       | c =  }} {{end-eqn}} and on $B$: {{begin-eqn}} {{eqn | l = \left\vert{\frac 1 z + \frac z {R^2} }\right\vert       | r = \left\vert{\frac 1 z \left({1 + \left({\frac z R}\right)^2}\right)}\right\vert       | c =  }} {{eqn | o = \le       | r = \left\vert{\frac 1 \delta \left({1 + 1}\right)}\right\vert       | c =  }} {{eqn | r = \frac 2 \delta       | c =  }} {{end-eqn}} Already we can place an upper bound on one of these integrals: :$\displaystyle \left\vert{\int_B F \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z }\right\vert \le \int_{-R}^R M N^x \frac 2 \delta \, \mathrm d y + 2 M \int_{-\delta}^0 N^x \frac{2x}{R^2} \, \mathrm d x$ {{finish}} {{Namedfor|Albert Edward Ingham|cat = Ingham}} Category:Dirichlet Series nqq6d4mvtten96acay6c39hmy9946jc