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 $$\tilde{f}(w)=\int_0^{\infty}f(x)x^{w-1}dx.$$ The Mellin inversion formula states that $$f(x)=\frac{1}{2\pi i}\int_{(c)}\tilde{f}(w)x^{-w}dw.$$ 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)^N}{w(w+1)\dotsb(w+N-1)}\int_0^2f^{(N)}(x)x^{w+N-1}dx. \end{equation} Since $f^{(N)}(x)$ is bounded on $[0,2]$ and $x^{w+N-1}$ is bounded (due to the compact support as $x\le 2$), 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 $0$?