i have a problem with 2 different functions that give the similar mellin transform, for example let be the floor function $ [x] $
then $ \int_{0}^{\infty}dt t^{s-1}[1/x]= \frac{\zeta(s)}{s} $
but also $ \int_{0}^{\infty}dt t^{s-1}frac(1/x)= -\frac{\zeta(s)}{s} $
sop they are the same except for a sign , what is wrong with this ?
which Mellin tarnsform should i use
For $\Re(s) > 0$, with $\mathcal{L}$ the bilateral Laplace transform :
$$\frac{1}{s} = \int_0^\infty e^{-st}dt =\mathcal{L}[1_{t > 0}](s)$$
But for $\Re(s) < 0$
$$\frac{1}{s} = -\int_{-\infty}^0 e^{-st}dt =\mathcal{L}[-1_{t < 0}](s)$$
$\displaystyle\frac{1}{s}$ is the (bilateral) Laplace transform of $1_{t > 0}$ and $-1_{t < 0}$, depending on the domain of convergence.
Here, it is the same since $$\mathcal{M}[f(t)](s) = \int_0^\infty f(x)x^{s-1}dx =\int_{-\infty}^\infty f(e^{-t})e^{-st}dt = \mathcal{L}[f(e^{-t})](s)$$ All you said is that $\displaystyle\frac{\zeta(s)}{s}$ is the bilateral Laplace transform of $\lfloor e^{t} \rfloor$ or $- \{e^t\}$ depending on the domain convergence $\Re(s) > 1$ or $\Re(s) \in (0,1)$.
