mellin transform of a function related to the derivative of Riemann zeta function

Question

if we know that $$ \frac{\zeta(s)}{s}=\int_{0}^{\infty}dx \frac{[x]}{x^{s+1}}$$

then how could we get the function so

$$ \frac{d\zeta(s)}{ds}=s\int_{0}^{\infty}dx \frac{f(x)}{x^{s+1}}$$

the function whose mellin transform is the derivative of the riemann zeta function

Answer 1

Hint: Notice that $[x]$ is a constant on each interval between consecutive integers, and when we differentiate with regard to s inside the integral sign, a $\ln x$ pops out. To illustrate this more rigorously, break up the integral into an infinite sum over consecutive intervals, and replace $[x]$ with n, then, after differentiating inside the summation sign, join all those intervals back again into one.

Answer 2

Assume the following definitions for the unit-step staircase function $S(x)$ and the log-step staircase function $T(x)$:

(1) $\quad S[x]=\sum_{n=1}^{\lfloor x\rfloor}1=\lfloor x\rfloor$

(2) $\quad T[x]=\sum_{n=1}^{\lfloor x\rfloor}\log n$

The first-order derivatives of $S(x)$ and $T(x)$ are related as follows:

(3) $\quad T'(x)=\log(x)\,S'(x)$

The Riemann zeta function $\zeta(s)$ is related to $S(x)$ and $S'(x)$ as follows.

(4) $\quad\zeta(s)=s \int_0^\infty S(x)\,x^{-s-1}dx\,,\quad \Re(s)>1$

(5) $\quad\zeta(s)=\int_0^\infty S'(x)\,x^{-s}dx\,,\quad \Re(s)>1$

The Riemann zeta function $\zeta(s)$ is related to $T(x)$ and $T'(x)$ is as follows.

(6) $\quad\zeta'(s)=-s\int_0^\infty T(x)\,x^{-s-1}dx\,,\quad \Re(s)>1$

(7) $\quad\zeta'(s)=-\int_0^\infty T'(x)\,x^{-s}dx\,,\quad \Re(s)>1$

The relationship between $S(x)$ and $T(x)$ is analogous to the relationship between Riemann's prime-power counting function $J(x)$ and the second Chebyshev function $\psi(x)$.

The first-order derivatives of $J(x)$ and $\psi(x)$ are related as follows:

(8) $\quad \psi'(x)=\log(x)\,J'(x)$

The Riemann zeta function $\zeta(s)$ is related to $J(x)$ and $J'(x)$ as follows.

(9) $\quad\log\zeta(s)=s \int_0^\infty J(x)\,x^{-s-1}dx\,,\quad \Re(s)>1$

(10) $\quad\log\zeta(s)=\int_0^\infty J'(x)\,x^{-s}dx\,,\quad \Re(s)>1$

The Riemann zeta function $\zeta(s)$ is related to $\psi(x)$ and $\psi'(x)$ as follows.

(11) $\quad\frac{\zeta'(s)}{\zeta(s)}=-s\int_0^\infty \psi(x)\,x^{-s-1}dx\,,\quad \Re(s)>1$

(12) $\quad\frac{\zeta'(s)}{\zeta(s)}=-\int_0^\infty \psi'(x)\,x^{-s}dx\,,\quad \Re(s)>1$

Note that $\log\zeta(s)$ and $\frac{\zeta'(s)}{\zeta(s)}$ are related as follows.

(13) $\frac{\zeta'(s)}{\zeta(s)}=\frac{d\log\zeta(s)}{ds}$