Ingham's Theorem on Convergent Dirichlet Series

Theorem
Let $\sequence {a_n} \le 1$

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 $\map \Re z > 1$.

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 $\map \Re z \ge 0$.

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 $R \ge 1$.

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 $\map \Re z \ge -\delta, \size {\map \Im z} \le R$.

We also choose an $M = \map M R$ so that $\map F {z + w}$ is bounded by $M$ in $-\delta \le \map \Re z, \cmod z \le R$.

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 the 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 the 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:

For what follows, allow $z = x + i y$ and observe that on $A, \cmod z = R$.

So:

and on $B$:

Already we can place an upper bound on one of these integrals:


 * $\ds \size {\int_B \map F {z + w} N^z \paren {\frac 1 z + \frac z {R^2} } \rd z} \le \int_{-R}^R M N^x \frac 2 \delta \rd y + 2 M \int_{-\delta}^0 N^x \frac {2 x} {R^2} \rd x$