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:

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

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

$$\mathcal{M}_x[f(x)](s)=-\frac{\mathcal{M}_x\left[\frac{\partial f(x)}{\partial x}\right](s+1)}{s}\tag{1}$$

If I integrate equation (1) from both sides and do algebraic manipulation I get:

$$\color{red}{\mathcal{M}_x[f(x)](s)=(1-s) \mathcal{M}_x[\int f(x) \, dx](s-1)}$$


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.