Ingham's Theorem on Convergent Dirichlet Series

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

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

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

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

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

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

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

By the residue theorem,

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

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

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

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

$\displaystyle \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

$\displaystyle 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 \ $

$\displaystyle = \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 \ $

$\displaystyle = \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 $\displaystyle z=x+iy \ $ and observe that on $\displaystyle A, |z|=R, \ $ so

$\displaystyle \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 $\displaystyle B \ $, we have

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

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

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