Ingham's Theorem on Convergent Dirichlet Series

Theorem
Let $\left\vert{a_n}\right\vert \le 1$

For a complex number $z \in \C$, let $\Re \left({z}\right)$ denote the real part of $z$.

Form the series $\displaystyle \sum_{n \mathop = 1}^\infty a_n n^{-z}$ which converges to an analytic function $F \left({z}\right)$ for $\Re \left({z}\right) > 1$.

Let $F \left({z}\right)$ be analytic throughout $\Re \left({z}\right) \ge 1$.

Then $\displaystyle \sum_{n \mathop = 1}^\infty a_n n^{-z}$ converges throughout $\Re \left({z}\right) \ge 1$.

Proof
Fix a $w$ in $\Re \left({w}\right) \ge 1$.

Then $F \left({z + w}\right)$ is analytic in $\Re \left({z}\right) \ge 0$.

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 $R \ge 1$.

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 $\Re \left({z}\right) \ge -\delta, \left\vert{\Im \left({z}\right)}\right\vert \le R$.

We also choose an $M = M \left({R}\right)$ so that $F \left({z + w}\right)$ is bounded by $M$ in $-\delta \le \Re \left({z}\right), \left\vert{z}\right\vert \le R$.

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:

For what follows, allow $z = x + i y$ and observe that on $A, \left\vert{z}\right\vert = R$.

So:

and on $B$:

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


 * $\displaystyle \left\vert{\int_B F \left({z + w}\right) N^z \left({\frac 1 z + \frac z {R^2} }\right) \, \mathrm d z }\right\vert \le \int_{-R}^R M N^x \frac 2 \delta \, \mathrm d y + 2 M \int_{-\delta}^0 N^x \frac{2x}{R^2} \, \mathrm d x$