Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $F$ be an L-function in the sense of the 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!

share|cite|improve this question
up vote 0 down vote accepted

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.

share|cite|improve this answer
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. – Owen Barrett Jun 5 '14 at 22:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.