Abscissa of convergence for a Dirichlet series Let $\alpha \in \mathbb{Z}$ and $f(n) = n^{i \alpha n}$. What is the abscissa of convergence, $\sigma_c$, for the associated Dirichlet series, $\sum_{n=1}^{\infty} \frac{f(n)}{n^s}$? Since $|f(n)| = 1$, it follows that $\sigma_a = 1$ (where $\sigma_a$ is the abscissa of absolute convergence), and we may conclude from general theory of Dirichlet series that $\sigma_c \in [0,1]$.
My feeling is that there should be a bit of cancellation, resulting in $\sigma_c$ being smaller than 1, though I haven't been able to quantify this. 
 A: For $\alpha = 0$ we trivially have $\sigma_c = \sigma_a = 1$. For $\alpha \neq 0$ we have $\sigma_c = \frac{1}{2}$. We can of course assume $\alpha > 0$ since replacing $\alpha$ with $-\alpha$ just conjugates the coefficients.
First I prove $\sigma_c \geqslant \frac{1}{2}$, by showing that there is a $c > 0$ such that $\bigl\lvert \sum_{n = 1}^{N} f(n)\bigr\rvert \geqslant c\sqrt{N}$ holds for infinitely many $N$.
Let $m$ be a large integer and put $N = \bigl\lfloor \exp \bigl(\frac{2\pi m}{\alpha} - 1\bigr)\bigr\rfloor$. Then
\begin{gather}
2\pi m - \frac{\alpha}{N} < \alpha(1 + \log N) \leqslant 2\pi m\,, \\
\alpha(1 + \log N) = 2\pi m - \theta_1 \frac{\alpha}{N}\,,
\end{gather}
where $0 \leqslant \theta_1 \leqslant 1$ ($\theta_2, \theta_3$ below satisfy the same inequalities, $\lvert \theta_4\rvert \leqslant 1$). Put further
$$r = \biggl\lfloor \sqrt{\frac{\pi N}{2\alpha}}\biggr\rfloor\,.$$
For $0 \leqslant k \leqslant r$ we have
\begin{align}
(N+k)\log (N+k) &{}- N\log N \\
&= k\log N + N\log\biggl(1 + \frac{k}{N}\biggr) + k\log\biggl(1 + \frac{k}{N}\biggr) \\
&= k\log N + N\biggl[\frac{k}{N} - \frac{k^2}{2N^2} + \theta_2\frac{k^3}{3N^3}\biggr]
+ k\biggl[\frac{k}{N} - \theta_3\frac{k^2}{2N^2}\biggr] \\
&= k\bigl(1 + \log N\bigr) + \frac{k^2}{2N} + \theta_4 \frac{k^3}{2N^2}
\end{align}
and hence
$$\alpha\bigl((N+k)\log(N+k) - N\log N\bigr) = 2\pi m k + \frac{\alpha k^2}{2N} - \theta_1 \frac{\alpha k}{N} + \theta_4 \frac{\alpha k^3}{2N^2}\,,$$
whence
$$\cos \bigl(\alpha\bigl((N+k)\log(N+k) - N\log N\bigr)\bigr) = \cos \biggl(\frac{\alpha k^2}{2N} - \theta_1 \frac{\alpha k}{N} + \theta_4 \frac{\alpha k^3}{2N^2}\biggr) > \frac{1}{2}$$
provided that $m$ was chosen so large that
$$\frac{\alpha k}{N} + \frac{\alpha k^3}{2N^2} < \frac{\pi}{12}\,.$$
(Since the left hand side is $O(N^{-1/2})$ that is possible.)
Then we have
\begin{align}
\Biggl\lvert \sum_{k = 0}^{r} f(N+k)\Biggr\rvert
&= \Biggl\lvert \sum_{k = 0}^{r} \frac{f(N+k)}{f(N)}\Biggr\rvert \\
&\geqslant \sum_{k = 0}^{r} \cos \bigl(\alpha\bigl((N+k)\log(N+k) - N\log N\bigr)\bigr) \\
&> \frac{r+1}{2} \\
&> \sqrt{\frac{\pi N}{8\alpha}}\,.
\end{align}
Thus
$$\limsup_{x \to \infty} \frac{1}{\sqrt{x}}\Biggl\lvert\sum_{n \leqslant x} f(n)\Biggr\rvert > 0$$
and $\sigma_c \geqslant \frac{1}{2}$ follows.
Now I show that there is a constant $C(\alpha)$ such that $\bigl\lvert \sum_{n = 1}^{N} f(n)\bigr\rvert \leqslant C(\alpha)(1 + \log N)^2\sqrt{N}$, which gives $\sigma_c \leqslant \frac{1}{2}$.
Lemma: For $0 \leqslant y < x$ and all $s \in \mathbb{R}$ we have
$$\biggl\lvert \int_y^x e^{iu(\log u - s)}\,du \biggr\rvert \leqslant 4\sqrt{2}\cdot \sqrt{x}\,.$$
Proof: For a fixed (but arbitrary) $s$, let $\psi(u) = u(\log u - s)$. Then $\psi'(u) = \log u - (s-1)$ and $\psi''(u) = \frac{1}{u}$. Thus $\psi$ is strictly convex, and attains its unique global minimum at $\xi = e^{s-1}$. With a small $\delta > 0$ (which will be chosen as $\sqrt{\frac{2}{x}}$) let $a = e^{-\delta}\xi$ and $b = e^{\delta}\xi$. Thus $-\psi'(a) = \psi'(b) = \delta$. If $[a,b] \cap [y,x]$ is nondegenerate, let $v = \max \{a,y\}$ and $w = \min \{b, x\}$. Then $v \geqslant a = e^{-2\delta}b \geqslant e^{-2\delta}w$, whence
$$w - v \leqslant (1 - e^{-2\delta})w \leqslant (1 - e^{-2\delta})x < 2\delta x\,.$$
Since the integrand has modulus $1$ it follows that the part between $a$ and $b$ contributes less than $2\delta x$ to the modulus of the integral. This clearly also holds if $[a,b]\cap [y,x]$ is empty or a degenerate interval.
If $[v,w]$ is an interval with $w \leqslant a$ or $v \geqslant b$, an integration by parts yields
\begin{align}
\int_v^w e^{i\psi(u)}\,du
&= \int_v^w \frac{i\psi'(u)e^{i\psi(u)}}{i\psi'(u)}\,du \\
&= \frac{e^{i\psi(w)}}{i\psi'(w)} - \frac{e^{i\psi(v)}}{i\psi'(v)} - i \int_v^w e^{i\psi(u)}\frac{\psi''(u)}{\psi'(u)^2}\,du\,.
\end{align}
Since $\psi''(u) > 0$ and $\psi'(u)$ has constant sign on $[v,w]$ it follows that
\begin{align}
\biggl\vert \int_v^w e^{i\psi(u)}\,du\biggr\rvert
&\leqslant \frac{1}{\lvert \psi'(w)\rvert} + \frac{1}{\lvert\psi'(v)\rvert}
+ \int_v^w \frac{\psi''(u)}{\psi'(u)^2}\,du \\
&= \frac{1}{\lvert \psi'(w)\rvert} + \frac{1}{\lvert\psi'(v)\rvert}
+ \biggl\lvert \frac{1}{\psi'(w)} - \frac{1}{\psi'(v)}\biggr\rvert \\
&= \frac{2}{\min \{\lvert \psi'(v)\rvert, \lvert \psi'(w)\rvert\}} \\
&\leqslant \frac{2}{\delta}\,.
\end{align}
Thus, splitting the domain of the integral into three parts, $[0,a]\cap [y,x]$, $[a,b]\cap [y,x]$, and $[b,+\infty)\cap [y,x]$ (each of which might be empty or degenerate) we find
$$\biggl\lvert \int_y^x e^{i\psi(u)}\,du\biggr\rvert \leqslant \frac{2}{\delta} + 2\delta x + \frac{2}{\delta} = \frac{4}{\delta} + 2\delta x\,.$$
The right hand side is minimised for $\delta = \sqrt{\frac{2}{x}}$, and this gives the stated bound.
The substitution $u = \alpha t$ shows
$$\biggl\vert \int_y^x e^{i\alpha t(\log t- s)}\,dt\biggr\rvert \leqslant \frac{4\sqrt{2}}{\sqrt{\alpha}}\cdot \sqrt{x}$$
for $\alpha > 0$, $s \in \mathbb{R}$, and $0 \leqslant y < x$.
Now let $\varphi(t) = \alpha t\log t$, so that $f(t) = e^{i\varphi(t)}$. By the Euler–Maclaurin formula we have
$$\sum_{n = 1}^{N} f(n) = \int_1^N f(t)\,dt + \frac{f(1) + f(N)}{2} + \frac{f'(N) - f'(1)}{12} - \int_1^N p(t)f''(t)\,dt\,,$$
where
$$p(t) = \frac{\{t\}^2 - \{t\} + \frac{1}{6}}{2} = \sum_{k \neq 0} \frac{e^{2\pi i kt}}{(2\pi k)^2}\,.$$
Since $f(t)$ is improperly integrable over $[1,+\infty)$ and $\lvert f(t)\rvert = 1$, the first two terms are bounded independent of $N$ (the bound for the integral depends on $\alpha$ of course). The third term has order $\lvert f'(N)\rvert = \varphi'(N) = \alpha(1 + \log N)$. Since
$$f''(t) = \bigl(i\varphi'(t)f(t)\bigr)' = i\varphi''(t)f(t) - \varphi'(t)^2f(t) = \frac{i\alpha}{t}f(t) -\alpha^2(1+\log t)^2f(t)$$
we split the last integral into two parts.
$$\biggl\lvert \int_1^N p(t) \varphi''(t)f(t)\,dt\biggr\rvert \leqslant \frac{1}{12} \int_1^N \varphi''(t)\,dt = \frac{\alpha}{12}\log N$$
is immediate from $\lvert p(t)\rvert \leqslant \frac{1}{12}$. For the other part we use the Fourier series of $p$. This converges absolutely and uniformly, so we can interchange summation and integration:
$$\int_1^N p(t)\varphi'(t)^2f(t)\,dt = \sum_{k \neq 0} \frac{1}{(2\pi k)^2} \int_1^N \varphi'(t)^2e^{i\alpha t(\log t + 2\pi k/\alpha)}\,dt\,.$$
$\varphi'(t)^2$ is positive and increasing, hence by the second mean value theorem there are $\xi_k, \eta_k \in [1,N]$ with
$$\int_1^N \varphi'(t)^2 \cos\bigl(\alpha t(\log t + 2\pi k/\alpha)\bigr)\,dt = \varphi'(N)^2\int_{\xi_k}^N \cos\bigl(\alpha t(\log t + 2\pi k/\alpha)\bigr)\,dt$$
and
$$\int_1^N \varphi'(t)^2 \sin\bigl(\alpha t(\log t + 2\pi k/\alpha)\bigr)\,dt = \varphi'(N)^2\int_{\eta_k}^N \sin\bigl(\alpha t(\log t + 2\pi k/\alpha)\bigr)\,dt\,.$$
These are the real and imaginary parts respectively of integrals we estimated in the lemma, thus
$$\biggl\lvert \int_1^N \varphi'(t)^2 e^{i \alpha t(\log t + 2\pi k/\alpha)}\,dt \biggr\rvert \leqslant \frac{8}{\sqrt{\alpha}}\varphi'(N)^2\sqrt{N}$$
for all $k \in \mathbb{Z}\setminus \{0\}$, and hence
$$\biggl\lvert \int_1^N p(t)\varphi'(t)^2f(t)\,dt\biggr\rvert \leqslant \frac{2}{3}\alpha^{3/2}(1 + \log N)^2\sqrt{N}\,.$$
The other terms from the Euler–Maclaurin formula had smaller order of growth and can therefore be absorbed by enlarging the constant factor.
It is worth noting that the assumption $\alpha \in \mathbb{Z}$ played no role here, everything — result and proof — works the same for all positive real $\alpha$ (and for negative $\alpha$ only a few absolute values have to be inserted). It might be that for $\alpha \in \mathbb{Z}$ there is a simpler proof, but I don't see it.
