Natural logarithm is defined as inverse function to exponent. This way defined it has the value of $0$ in $x=1$.
But if we define natural integral the following way
$$f^{(-1)}(x)=\frac{i}{2\pi}\int_{-\infty}^{+\infty} \frac{e^{- i \omega x}}{\omega} \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$
we will find that natural integral of $1/x$ is $\ln|x|+\gamma$.
Indeed, the natural differintegral of the function will be
$$D^s [1/x]=\frac{(-1)^s \Gamma(s+1)}{x^{s+1}}$$
This function diverges at $s=-1$, but its mean value at $s=-1$ in limit is
$$\lim_{h\to0}\frac12 \left( \frac{-\Gamma(h)}{x^{h}}+\frac{-\Gamma(-h)}{x^{-h}}\right)=\ln|x|+\gamma$$
Now if we consider natural antidifference of $1/x$, we get the following formula for discrete differintegral:
$$\Delta ^{s}[1/x]=\frac{(-1)^s \Gamma (s+1) \Gamma (x)}{(s+x)!}$$
which also diverges at $s=-1$ but its main value will be
$$\lim_{h\to0}\frac12\left(-\frac{\Gamma (x) \Gamma (-h+s+1)}{(-h+s+x)!}-\frac{\Gamma (x) \Gamma (h+s+1)}{(h+s+x)!}\right)=\psi(x)+\gamma$$
This is equal to the "harmonic number" function. It follows the rule that natural antidifference in point $x=1$ should have the value of natural integral minus Ramanujan's sum of $1/x$ (which is equal to $\gamma$).
That said I wonder, what is intuitive explanation why the natural integral and natural antidifference of $1/x$ are $\ln|x|+\gamma$ and $\psi(x)+\gamma$ rather than what would be expected more intuitive $\ln x$ (defined as inverse of exponent) and $\psi(x)$ (defined as logarithmic derivative of Gamma function $\psi(x)=\frac{\Gamma'(x)}{\Gamma(x)}$)