Bounds for vertical integrals and Mellin transforms

Question

I am trying to bound Mellin transforms. They look like $$I(x) = \int_{(c)} f(w) \frac{x^w}{w} dw$$ where the integral is on the vertical line of abscissa $c$. Let's say that $f$ has vertical growth controlled by $$f(w) \ll (1+\Im(w))^k$$ for a certain real number $k$. How can I bound $I(x)$ in terms of $x$, in general? I tried with some explicitly truncated versions of the integral, but I keep stuck with the remaining pieces of the integral...

Answer 1

This question isn't quite a model for the classical application of an integral transform like Perron's formula to Dirichlet series.

The bound $f(w) \ll (1 + \lvert \textrm{Im}(w) \rvert)^k$ would be unusual in that there is no dependence on the real part of $w$. If $k \geq 0$, it's not obvious that the integral converges anywhere. The only information we are given here is to say that the integral is bounded by

$$ \int_{(c)} (1 + \lvert \textrm{Im} w \rvert)^k \frac{\lvert X^w \rvert}{\lvert w \rvert} dw \asymp \int_{(c)} \lvert X^w \rvert \lvert w \rvert^{k - 1} dw.$$ This doesn't say very much.

Applying Perron's formula directly can be a bit subtle and often requires some knowledge about the underlying Dirichlet series. A better abstract model of the Dirichlet series situation would be to suppose that $f(w)$ satisfies a set of bounds like $$f(w) \ll \begin{cases} 1 & \mathrm{Re}(w) > 1,\\ (1 + \lvert w \rvert)^{\sigma_1(\mathrm{Re}(w))} & 0 < \mathrm{Re}(w) \leq 1, \\ (1 + \lvert w \rvert)^{\sigma_2(\mathrm{Re}(w))} & \mathrm{Re}(w) \leq 0, \end{cases}$$ where $\sigma_1(x)$ describes the line segment from $(0, \alpha)$ to $(1, 0)$ and where $\sigma_2(x)$ describes the line segment with slope $-2\alpha$ that goes through $(0, \alpha)$. The trivial bounds for every $L$-function take this form. (Here, by trivial, I mean that we use the trivial bound in the region of absolute convergence, the trivial bound in the reflection under $w \mapsto 1 - w$ in the region $\mathrm{Re}(w) < 0$, and the Phragmen-Lindelof convexity bound in the middle).

In practice, Perron's formula needs to be truncated at some height $T$ to be used in situations like this. Then the game is to relate the truncation points $T$ to the growths in $f$ where one shifts the line of integration. As a rule of thumb, shifting the line of integration left forces one to have worse bounds on $f$ and thus larger truncation costs in $T$, but saves powers of $X$. I suggest looking at the book of Montgomery and Vaughan for their treatment of truncated Perron and its applications to see this game played.

I will also note that an alternate game is to instead use various forms of smoothing. That is, instead of integrating $f$ against $X^w / w$, you integrate against $X^w V(w)$ for some weight function $w$ that has better convergence properties. If $f(w) = \sum_{n \geq 1} \frac{a(n)}{n^w}$, then the general Mellin transform pair states that $$ \sum_{n \geq 1} a(n) v(n/X) = \int_{(c)} f(w) V(w) X^w dw, $$ where $v$ is the Mellin tranform of $V$. Choosing nice weight functions $V$ (and thus $v$) simplify the analysis at the cost of giving a weighted sum on the left.