Mellin transform at $0$

Question

Let $f$ be a smooth complex valued function with support on $[0,2]$ and with Mellin transform $\tilde{f}$. 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 $0$?

Answer 1

By definition, $\tilde{f}(0)$ exists if and only if $\int_0^\infty f(x)/x\,dx$ exists. As $f(x)$ is smooth and supported on $[0,2]$, we have that that $$|f(x)|=\left|\int_0^x f'(t)\,dt\right|\leq x\sup_{0\leq t\leq x}|f'(t)|,\qquad x\geq 0.$$ Hence $f(x)/x$ is bounded, so $\tilde{f}(0)$ exists (and can be bounded easily in terms of $f'$).

More generally, since all the derivatives of $f$ vanish at $0$, repeated integration by parts shows that (cf. Taylor's theorem) $$|f(x)|=\left|\int_0^x \frac{(x-t)^{k-1}}{(k-1)!}f^{(k)}(t)\,dt\right|\leq\frac{x^k}{k!}\sup_{0\leq t\leq x}|f^{(k)}(t)|,\quad x\geq 0,\quad k\in\mathbb{Z}_{\geq 1}.$$ So $f(x)$ decays very rapidly as $x\to 0+$, and (in the same way) also as $x\to 2-$. Of course the same is true of every derivative of $f(x)$. Hence in fact $\tilde f(w)$ converges absolutely for every $w\in\mathbb{C}$, and it decays rapidly in every vertical strip.