Hello I have this formula for Mellin Transform :
$$\mathcal{M}_x[f(x)](s)=-\frac{\mathcal{M}_x\left[\frac{\partial f(x)}{\partial x}\right](s+1)}{s}$$
but I need formula like so (integral inside Mellin Transfrom):
$$\mathcal{M}_x[f(x)](s)=\mathcal{M}_x[\int f(x) \, dx](s) ?$$ or: $$\mathcal{M}_x[f(x)](s)=\frac{\mathcal{M}_x\left[\int_0^x f(u) \, du\right](s+1)}{s}?$$
Is there exist such a formula?
I checked on a few examples it seems dosen't work.
An Example:
$$\mathcal{M}_x[\exp (-x)](s)=\Gamma (s)\neq \mathcal{M}_x[\int \exp (-x) \, dx](s)=-\Gamma (s) $$
$$\mathcal{M}_x[f(x)](s)=-\frac{\mathcal{M}_x\left[\frac{\partial f(x)}{\partial x}\right](s+1)}{s}\tag{1}$$
If I integrate equation (1) from both sides and do algebraic manipulation I get:
$$\color{red}{\mathcal{M}_x[f(x)](s)=(1-s) \mathcal{M}_x[\int f(x) \, dx](s-1)}$$
Check:
$$\begin{align*} &\mathcal{M}_x[\exp (-x)](s)=\color{red}{\Gamma (s)}\\ &=(1-s) \mathcal{M}_x[\int \exp (-x) \, dx](s-1)\\ &=(1-s) \mathcal{M}_x\left[-e^{-x}\right](s-1)\\ &=-(1-s) \Gamma (-1+s)\\ &=\color{red}{\Gamma (s)} \end{align*}$$