Perron's formula in the region of conditional convergence

Question

I am a bit confused about the proof of Perron's formula. It states that for a Dirichlet series $f(s) = \sum_{n\geq 1} a_n n^{-s}$ and real numbers $c > 0$, $c > \sigma_c$, $x > 0$ we have

$$\sum\limits_{n\leq x}\kern-1pt^{'} a(n) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{f(s) x^s}{s}\,\text{d}s$$ where $\sigma_c$ is the abscissa of convergence of $f$. Now I understand the proof for the case where $c > \sigma_a$ (the abscissa of absolute convergence) perfectly well, but I don't understand how to extend it to $c > \sigma_c$.

Following some hints I found somewhere I tried to consider the integral $$\int_C \frac{f(s)x^s}{s}\,\text{d}s$$ where $C$ denotes a rectangular path with corners $c - iT$ and $c' + iT$ where $c'$ is some sufficiently big number (in particular $c' > \sigma_a$). If the integrals along the horizontal lines of the rectangle vanish for $T\to\infty$, this solves our problem since then the integral $\int_{c-i\infty}^{c+i\infty}$ is equal to the integral $\int_{c'-i\infty}^{c'+i\infty}$. But I don't see why those integrals should vanish.

The best I can come up with is $$\left|\int_{c+iT}^{c'+iT} \frac{f(\sigma+iT)x^{\sigma+iT}}{\sigma+iT}\,\text{d}\sigma \right| \leq (c' - c)\cdot|f(\sigma + iT)|\cdot\max(x^c, x^{c'}) / T$$ but I don't see how that gets me anywhere unless I could also show that $\lim_{T\to\pm\infty} f(\sigma + i T)/T = 0$ uniformly for all $\sigma\in[c,c']$. Is that the case? If so, why?

Answer 1

What you hope is the case is indeed the case.

Theorem. If $\sum a_n/n^s$ converges for ${\rm Re}(s) > \sigma_0$ and $\sigma_0 < r < \sigma_0 + 1$, then $\sum a_n/n^{\sigma + it} = o(|t|)$ as $|t| \to \infty$, uniformly for $\sigma \geq r$.

By replacing $a_n$ with $a_n/n^{\sigma_0}$, we can assume $\sigma_0 = 0$.

I don't have a reference to give you right now for a proof, but perhaps knowing this result is true will inspire you to look around some more to find a proof if you're not able to work it out on your own.

And you're right that to prove Perron on a line $\sigma = c$ outside the half-plane of absolute convergence for the Dirichlet series, you want to link it by a rectangle to Perron on a line in the half-plane of absolute convergence: pick $c' > c+1$, so convergence on $\sigma = c$ implies absolute convergence on $\sigma = c'$. Then you deduce Perron on $\sigma = c$ from $\sigma = c'$ by a contour shift.