Let $f$ be a smooth complex valued function with support on $[0,2]$ and with Mellin transform $\tilde{f}(s)$. Recall that the Mellin transform is defined as $$\tilde{f}(w)=\int_0^{\infty}f(w)x^{w-1}dw.$$ The Mellin inversion formula states that $$f(u)=\frac{1}{2\pi i}\int_{(c)}\tilde{f}(w)u^{-w}dw.$$ What can be said about $\tilde{f}(0)$?

I note that such an $f$ decays rapidly; integrating by parts $N$ times we have \begin{equation} \tilde{f}(w)=\frac{1}{w^N}\int_0^2f^{(N)}(u)u^{w+N-1}du. \end{equation} Since $f^{(N)}(u)$ is bounded on $[0,2]$ and $u^{w+N-1}$ is bounded (due to the compact support as $u\le 2$), the integral is bounded for all $w$. We conclude that $\tilde{f}(w)$ decays faster than any polynomial in $|w|$. With any function I try, it seems as though $\tilde{f}(0)=\infty$, so I wonder if it's always the case that it diverges, or are there functions satisfying the properties which has a convergent Mellin transform at $0$?