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Let $L(s)$ be an L-function, given by its series expansion and admitting an Euler product, say, for $s$ of large enough real part, $$L(s) = \sum_n \frac{a_n}{n^s} = \prod_p \prod_{i=1}^k (1-\alpha_i p^{-s})^{-1}.$$

Assume that $L(s)$ admits a simple pole at $s=x$ with residue $R$, so that it admits a local expansion of the form $$L(s) = \frac{R}{s-x} + H(s),$$

where $H$ is an analytic function around $s=x$. My question is therefore:

Assuming we now explicitly the value of the residue $R$, is there any standard way of computing the value of $H(x)$?

(everything here is quite explicit, but I am still stuck when I try to do something with the cumbersome expression $L(s) - R/(s-x)$...)

NB: If it can help, I am interested in some kind of symmetric square L-functions.

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  • $\begingroup$ Define concretely your L-function and tell what you want to do with $H(x)$. You are supposed to know that L-functions have dozens of different representations and constructions, many of them are computationally intractable and based on conjectural properties. If we assume some of those properties then $L(s)-R \zeta(s-x+1)$ converges on $\Re(s) > x-b$ for some $b>0$ (which is difficult to find precisely, the Dirichlet divisor problem is the abscissa of convergence of $ ((1-2^{1-s})\zeta(s))^m$) and so does their ratio and its logarithm (the PNT for $L(s)$) $\endgroup$ – reuns Oct 30 '19 at 11:54
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The computation of the Taylor expansion of an $L$-function is implemented in Pari/GP. Here is an example for the Riemann zeta function at $s=1$:

Z=lfuncreate(1);
lfun(Z,1+x+O(x^4))

which returns

1.0000000000000000000000000000000000000*x^-1 + 0.57721566490153286060651209008240243104 + 0.072815845483676724860586375874901319140*x + O(x^2)

For the theory behind, you can look at Henri Cohen's course. The documentation for Pari/GP is available here.

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I don't claim my method is very clever, but let's look at this for the Riemann $\zeta$-function.

A standard trick is integrate around a symmetric rectangle, infinite in the vertical direction.

For instance with $\xi(s)=\Gamma(s/2)\zeta(s)/\pi^{s/2}$ consider $$\biggl(\int_{(2)}-\int_{(-1)}\biggr)(s-1/2)\xi(s){1\over s(s-1)}{ds\over 2\pi i}.$$ Here I've made there be a double pole at $s=0,1$, so that the desired term should fall out of a residue calculation.

On the one hand, Cauchy's Residue Theorem says this is twice (by symmetry) the residue at $s=1$ of $$((s-1)+1/2))({1\over s-1}+c_0+\cdots){1\over s-1}\sum_{n=0}^\infty (1-s)^n$$ and if my math is correct, this residue is $1+c_0/2-1/2$.

On the other hand, by symmetry this difference of integrals is twice the integral on the 2-line, where there is a Mellin expansion: $$\sum_{n=1}^\infty\int_{(2)}{\Gamma(s/2)(s-1/2)\over s(s-1)}{1\over\pi^{s/2}}{1\over n^s}{ds\over 2\pi i},$$ and this latter integral can be computed if desired to high precision for each $n$ (and by general theory the sum over $n$ converges). Perhaps one can be more clever, and actually get the result for $\zeta(s)$ theoretically, being able to compute the transform explicitly, and sum over $n$, but I don't know how offhand.

In general there will be coefficients depending on $n$, and likely one can't do more than just compute to some precision.

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