= Accumulating asymptotic exponents obstruct Mellin continuation
Strict increase alone does not suffice in the preceding theorem. Let $\sigma_j=1-1/j$, $c_j=2^{-j}$ and $f(y)=\eta(y)\sum_{j\geq1}2^{-j}y^{1-1/j}$, where $\eta$ is continuous, equals one on $(0,1]$ and vanishes on $[2,\infty)$. After division by the next power, every finite remainder extends continuously to zero. Nevertheless the <Mellin transform> has genuine poles at $-\sigma_j$ accumulating at $-1$, so it cannot be a <meromorphic function> on the whole plane.
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