The $\psi(x)$ is the second Chebyshev function.
The $\zeta(s)$ is the Riemann zeta function.
Mellin inversion theorem:
$$\varphi(s)=\int_0^\infty f(x)x^{s-1}\,dx$$ $$f(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\varphi(s) x^{-s}\,ds$$
Starting with: $$\tag{1}-\frac{\zeta'(s)}{\zeta (s)} \frac{1}{s}=\int_0^{\infty } \frac{\psi (x)}{x^{s+1}} \, dx$$ Using substitutution $s=-z$: $$\tag{2}-\frac{\zeta'(-z)}{\zeta (-z)} \frac{1}{-z}=\int_0^{\infty } \psi (x) x^{z-1} \, dx$$ Applying inverse Mellin transform: $$\tag{3}\psi(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\left(-\frac{\zeta'(-z)}{\zeta (-z)} \frac{1}{-z}\right) x^{-z}\,dz$$ Reversing substitution $z=-s$ and $\frac{dz}{ds}=-1$: $$\tag{4}\psi(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\left(-\frac{\zeta'(s)}{\zeta (s)} \frac{1}{s}\right) x^{s}\,(-1) ds$$ This simplifies to: $$\tag{5}\psi(x)=\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\zeta'(s)}{\zeta (s)} \frac{x^{s}}{s}\,ds$$
But the correct formula should be: $$\tag{6}\psi(x)=-\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\zeta'(s)}{\zeta (s)} \frac{x^{s}}{s}\,ds$$
I cannot spot a mistake. Can you tell me where I went wrong?