Theorem: Ingham's Theorem on Convergent Dirichlet Series

Let $\sequence {a_n} \le 1$
{{explain|What exactly is $a_n$ in this context?}}
For a complex number $z \in \C$, let $\map \Re z$ denote the real part of $z$. 
Form the series $\ds \sum_{n \mathop = 1}^\infty a_n n^{-z}$ which converges to an analytic function $\map F z$ for $\map \Re z > 1$.
{{explain|We have "$\map \Re z > 1$" used here and below to mean $\set {z \in \C: \map \Re z > 1}$? In which case, rather than just call it "$\map \Re z > 1$", which is unwieldy and suboptimal, it might be better and clearer to give it a single-symbol identifier.}}
Let $\map F z$ be analytic throughout $\map \Re z \ge 1$.
Then $\ds \sum_{n \mathop = 1}^\infty a_n n^{-z}$ converges throughout $\map \Re z \ge 1$.

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