Let $L>0$. Let $f(t)$ be a smooth step function that satisfies the following: $f(t) = 1$ if $t \leq 1$ and $f(t) = 0$ if $t \geq 1 + (1/L)$. Let $Z>0$. I am interested in obtaining an upper bound for the Mellin transform (of a related function) $$ \int_{0}^{\infty} f(x/Z) x^s \frac{1}{x} dx. $$
I am interested in learning what bound is available for this integral along with how I should choose such smooth step function or a good example of it. Thank you very much.
You meant a bound for $F(s) = \int_0^\infty f(t) t^{s-1}dt, \Re(s) > 0$. Then write $f(t) = 1_{t< 1} \ast \phi(t)= \int_0^\infty 1_{\tau < 1} \phi(t/\tau) \frac{d\tau}{\tau}=\int_0^1 \phi(t/\tau) \frac{d\tau}{\tau}$ where $\phi=f' \in C^\infty_c(0,\infty)$ so that by the (Mellin) convolution theorem $F(s) = \frac{1}{s}\Phi(s)$ where $\Phi(s) = \int_0^\infty \phi(t) t^{s-1}dt$ which is entire and Schwartz on every vertical line.
By Mellin/Fourier/Laplace inversion for $\sigma > 0$, $f(t) = \frac{1}{2i\pi}\int_{\sigma - i \infty}^{\sigma + i \infty} F(s) x^{-s} ds$ and for $\sigma < 0$ , $f(t)-1 = \frac{1}{2i\pi}\int_{\sigma - i \infty}^{\sigma + i \infty} F(s) x^{-s} ds$