Let $F$ be an L-function in the sense of the Selberg class --> http://en.wikipedia.org/wiki/Selberg_class.

We are observing the integral $$\frac{1}{2\pi i}\int_{(c)}F(s+w)\Gamma(w)z^w dw$$ for $\Re(s)>1$ and $c>0$. Why can we shift the line of integration to the left, say, to $(-R)$, where $R$ is some positive real number, non-integer, the difference coming only from the residues in between?

This is equivalent to asking why does the integral of $F(s+w)\Gamma(w)z^w$ over the horizontal line segments vanish when we push them to $\pm i\infty$?

The thing is that by doing so we are moving the line of integration over the critical stripe of $F$. Outside the critical stripe we can combine the Stirling's approximation of $\Gamma(w)$ with the fact that $F(\sigma+ it)$ is $o(t)$, when $t\to\infty$, as a Dirichlet series, plus Stirling's approximation for the Gamma-factors of the functional equation, when we observe the asymptotics of $F$ left from the critical stripe. However, is there a simple way to obtain some basic "vertical asymptotics" of $F$ over the critical stripe that would allow us the aforementioned move?

My only idea is to use the fact that $(s-1)^m F(s)$ ($m$ is the multiplicity of the pole of $F$ in $s=1$) is an entire function of finite order. But unless the order is $1$, it would actually appear to overweight any Stirling's approximation. I guess, I am missing something obvious in the whole story.

The reason why I am asking this is because the above argument appears to be pretty standard throughout various materials conerning L-functions, starting with the more elementary Dedekind-L-functions and going through automorphic L-functions and similar. Thus is seems to be a general argument that is not closely related to the specifics of each of these L-functions, and I would like to understand the principle behind it. To be honest, it has been bugging me for a few weeks now...

Thanks in advance for any help!


1 Answer 1


I have just realized how simple this is, so I decided to answer my own question. I hope no one minds it.

The trick is the following: take $\Re(s)>R-1$ or $\Re(s)<R+1$. Then shifting the integration line is justified by the asymptotics of the Gamma-factor of the functional equation of $F$ plus the fact that $F(\sigma+it+w)=o(t)$ whenever $F$ admits the representation as Dirichlet-series.

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    $\begingroup$ perhaps it would be helpful to the novice to mention that $\Gamma$ has rapid decay in every vertical strip; i.e. for $s=\sigma+it$, $|\Gamma(s)|\ll|s|^A e^{-t}$ as $t\rightarrow\infty$ and $\sigma\in[a,b]$ for constants $A,a,b\in\mathbf{R}$. this assures that as long as everything else in your line integral (i.e. the other terms) has (have) finite order (i.e. $O(|s|^B)$ for some real constant $B$) in vertical strips, you can move the line wherever you want in the plane. $\endgroup$
    – Tomo
    Jun 5, 2014 at 22:49

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