In the Laurent series, $${\displaystyle f(z)=\sum _{n=-\infty }^{\infty }a_{n}(z-c)^{n},}$$ the coefficients are the integrals $${\displaystyle a_{n}={\frac {1}{2\pi i}}\oint _{\gamma }{\frac {f(z)}{(z-c)^{n+1}}}\,dz.}$$ We know that the Taylor expansion and Laurent expansion, if both exist, are the same. However the coefficients in Taylor series are computed through derivatives, instead of integrals. Does this mean the derivative of $f(x)$ can be computed by some form of integral $\int \dots f(x)dx$?
From another point of view, in the language of functional analysis, or geralized funtions / distributions, a derivative is nothing but a linear operator. And a linear operator can always be viewed as an integral $\frac{d}{dx} f(x) = \int \beta(t, x)f(x) dx$ where $\beta(x)$ is a generalized function. However, this is just a formal rewriting which tells us nothing in essence. But the previous discussion suggesting that there might be an explicit relation between derivative and integral is what truly suprises me.