Ingham's Theorem on Convergent Dirichlet Series

Theorem
Suppose $$|a_n| \leq 1 \ $$, and form the series $$\sum_{n=1}^\infty a_n n^{-z} \ $$ which converges to an analytic function $$F(z) \ $$ for $$\Re \left({z}\right) > 1 \ $$, where $$\Re \left({z}\right)$$ is the real part of $$z \ $$.

If $$F(z) \ $$ in analytic throughout $$\Re \left({z}\right) \geq 1 \ $$, then $$\sum_{n=1}^\infty a_n n^{-z} \ $$ converges throughout $$\Re \left({z}\right) \geq 1 \ $$.

Proof
Fix a $$w \ $$ in $$\Re \left({w}\right) \geq 1 \ $$. Then $$F(z+w) \ $$ is analytic in $$\Re \left({z}\right) \geq 0 \ $$.

We note that since $$F(z+w) \ $$ is analytic on $$\Re(z)=0 \ $$, it must be analytic on an open set containing $$\Re(z)=0 \ $$.

Choose an $$R \geq 1 \ $$. Then, since $$F(z+w) \ $$ is analytic on such an open set, we can determine $$\delta = \delta (R) > 0, \delta \leq \tfrac{1}{2} \ $$, such that $$F(z+w) \ $$ is analytic in $$\Re(z) \geq - \delta, | \Im (z) | \leq R \ $$.

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

Now form the counterclockwise contour $$\Gamma \ $$ as the arc $$|z|=R, \Re \left({z}\right) > - \delta \ $$ and the segment $$\Re \left({z}\right) = -\delta, |z| \leq R \ $$. We denote by $$A, B \ $$ respectively, the parts of $$\Gamma \ $$ in the right and left half-planes.

By the residue theorem,

$$2\pi i F(w) = \oint_{\Gamma} F(z+w) N^z \left({ \frac{1}{z} + \frac{z}{R^2} }\right) dz \ $$

Since $$F(z+w) \ $$ converges to its series on $$A \ $$, we may split it into the partial sum and remainder after $$N \ $$ terms, $$s_N(z+w), r_N(z+w) \ $$ respectively, and find that, again, by the residue theorem,

$$\int_A s_N(z+w) N^z \left({ \frac{1}{z}+\frac{z}{R^2} }\right) dz = 2\pi i s_N(w) - \int_{-A} s_N(z+w) N^z \left({ \frac{1}{z}+\frac{z}{R^2} }\right) dz \ $$

where $$-A \ $$ is the reflection of $$A \ $$ through the origin.

Changing $$z \to -z \ $$, we have

$$\int_A s_N(z+w) N^z \left({ \frac{1}{z}+\frac{z}{R^2} }\right) dz = 2\pi i s_N(w) - \int_{A} s_N(w-z) N^{-z} \left({ \frac{1}{z}+\frac{z}{R^2} }\right) dz \ $$

Combining these results gives

$$2\pi i \left({F(w) - s_N(w) }\right) = \int_\Gamma F(z+w)N^z \left({\frac{1}{z}+\frac{z}{R^2} }\right) dz - \int_A s_N(z+w) N^z \left({ \frac{1}{z}+\frac{z}{R^2} }\right) dz - \int_{A} s_N(w-z) N^{-z} \left({ \frac{1}{z}+\frac{z}{R^2} }\right) dz \ $$

$$= \int_A  F(z+w)N^z \left({\frac{1}{z}+\frac{z}{R^2} }\right) dz   + \int_B F(z+w)N^z \left({\frac{1}{z}+\frac{z}{R^2} }\right) dz- \int_A s_N(z+w) N^z \left({ \frac{1}{z}+\frac{z}{R^2} }\right) dz - \int_{A} s_N(w-z) N^{-z} \left({ \frac{1}{z}+\frac{z}{R^2} }\right) dz \ $$

$$= \int_A \left({ r_N(z+w)N^z -s_N(w-z)N^{-z} }\right) \left({ \frac{1}{z}+\frac{z}{R^2} }\right) dz + \int_B F(z+w)N^z \left({\frac{1}{z}+\frac{z}{R^2} }\right) dz \ $$

For what follows, allow $$z=x+iy \ $$ and observe that on $$A, |z|=R, \ $$ so

$$\frac{1}{z}+\frac{z}{R^2} = \frac{\overline{z}}{|z|^2} + \frac{z}{R^2} = \frac{x-iy}{R^2} + \frac{x+iy}{R^2} = \frac{2x}{R^2} \ $$

and on $$B \ $$, we have

$$\left|{ \frac{1}{z}+\frac{z}{R^2} }\right| = \left|{ \frac{1}{z} \left({ 1+ \left({\frac{z}{R} }\right)^2 }\right) }\right| \leq \left|{ \frac{1}{\delta} \left({ 1 + 1 }\right) }\right| = \frac{2}{\delta}$$

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

$$\left|{ \int_B F(z+w)N^z \left({\frac{1}{z}+\frac{z}{R^2} }\right) dz }\right| \leq \int_{-R}^R M N^x \frac{2}{\delta}dy  + 2M \int_{-\delta}^0 N^x \frac{2x}{R^2}dx$$