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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:

$$\mathcal{M}_x[\exp (-x)](s)=\Gamma (s)\neq \mathcal{M}_x[\int \exp (-x) \, dx](s)=-\Gamma (s) $$

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  • $\begingroup$ It is not well defined since you can not make a function of the integrand without limits like that. Maybe you intend $\mathcal M _x\left[\int_a^x f(\xi)d\xi\right]$ or $\mathcal M _x\left[\int_a^b f(x)dx\right]$ something of the sort? Well the second will be a constant unless $a$ and $b$ themselves are functions of $x$. But you should be very careful with which variable you integrate and which is in the limits. $\endgroup$ Dec 26 '17 at 18:43
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    $\begingroup$ If $F(s) = \int_0^\infty h(x) x^{s-1}dx,G(s) = \int_0^\infty h'(x) x^{s-1}dx$ both converge then integrating by parts $G(s) = h(x) x^{s-1}|_0^\infty- \int_0^\infty h(x) (s-1)x^{s-2}dx = (s-1) F(s-1)$. Now let $h(x) = \int_0^x f(y)dy$ and you get your answer. $\endgroup$
    – reuns
    Dec 27 '17 at 9:01
  • $\begingroup$ @reuns. thanks :) $\endgroup$ Dec 27 '17 at 9:20
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$$\mathcal{M}_x[f(x)](s)=-\frac{\mathcal{M}_x\left[\frac{\partial f(x)}{\partial x}\right](s+1)}{s}\tag{1}$$

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:

$$\begin{align*} &\mathcal{M}_x[\exp (-x)](s)=\color{red}{\Gamma (s)}\\ &=(1-s) \mathcal{M}_x[\int \exp (-x) \, dx](s-1)\\ &=(1-s) \mathcal{M}_x\left[-e^{-x}\right](s-1)\\ &=-(1-s) \Gamma (-1+s)\\ &=\color{red}{\Gamma (s)} \end{align*}$$

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