# Ingham's Theorem on Convergent Dirichlet Series

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## Theorem

Let $\sequence {a_n} \le 1$

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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 $\map \Re z > 1$.

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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 $\map \Re z \ge 0$.

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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 $R \ge 1$.

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

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

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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.

- $\ds 2 \pi i \map F w = \oint_\Gamma \map F {z + w} N^z \paren {\frac 1 z + \frac z {R^2} } \rd z$

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:

\(\ds \) | \(\) | \(\ds 2 \pi i \paren {\map F w - \map {s_N} w}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int_\Gamma \map F {z + w} N^z \paren {\frac 1 z + \frac z {R^2} } \rd z - \int_A \map {s_N} {z + w} N^z \paren {\frac 1 z + \frac z {R^2} } \rd z - \int_A \map {s_N} {w - z} N^{-z} \paren {\frac 1 z + \frac z {R^2} } \rd z\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int_A \map F {z + w} N^z \paren {\frac 1 z + \frac z {R^2} } \rd z + \int_B \map F {z + w} N^z \paren {\frac 1 z + \frac z {R^2} } \rd z\) | ||||||||||||

\(\ds \) | \(\) | \(\, \ds - \, \) | \(\ds \int_A \map {s_N} {z + w} N^z \paren {\frac 1 z + \frac z {R^2} } \rd z - \int_A \map {s_N} {w - z} N^{-z} \paren {\frac 1 z + \frac z {R^2} } \rd z\) | |||||||||||

\(\ds \) | \(=\) | \(\ds \int_A \paren {\map {r_N} {z + w} N^z - \map {s_N} {w - z} N^{-z} } \paren {\frac 1 z + \frac z {R^2} } \rd z + \int_B \map F {z + w} N^z \paren {\frac 1 z + \frac z {R^2} } \rd z\) |

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

So:

\(\ds \frac 1 z + \frac z {R^2}\) | \(=\) | \(\ds \frac {\overline z} {\cmod z^2} + \frac z {R^2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {x - i y} {R^2} + \frac {x + i y} {R^2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {2 x} {R^2}\) |

and on $B$:

\(\ds \size {\frac 1 z + \frac z {R^2} }\) | \(=\) | \(\ds \size {\frac 1 z \paren {1 + \paren {\frac z R}^2} }\) | ||||||||||||

\(\ds \) | \(\le\) | \(\ds \size {\frac 1 \delta \paren {1 + 1} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac 2 \delta\) |

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

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

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## Source of Name

This entry was named for Albert Edward Ingham.