Cauchy formula for repeated integration

A proof is given by induction. Since f is continuous, the base case follows from the fundamental theorem of calculus:

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}f^{(-1)}(x)={\frac {\mathrm {d} }{\mathrm {d} x}}\int _{a}^{x}f(t)\,\mathrm {d} t=f(x)}$;

where

${\displaystyle f^{(-1)}(a)=\int _{a}^{a}f(t)\,\mathrm {d} t=0}$.

Now, suppose this is true for n, and let us prove it for n+1. Firstly, using the Leibniz integral rule, note that

${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\left[{\frac {1}{n!}}\int _{a}^{x}\left(x-t\right)^{n}f(t)\,\mathrm {d} t\right]={\frac {1}{(n-1)!}}\int _{a}^{x}\left(x-t\right)^{n-1}f(t)\,\mathrm {d} t}$.

Then, applying the induction hypothesis,

${\displaystyle {\begin{aligned}f^{-(n+1)}(x)&=\int _{a}^{x}\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1})\,\mathrm {d} \sigma _{n+1}\cdots \,\mathrm {d} \sigma _{2}\,\mathrm {d} \sigma _{1}\\&=\int _{a}^{x}{\frac {1}{(n-1)!}}\int _{a}^{\sigma _{1}}\left(\sigma _{1}-t\right)^{n-1}f(t)\,\mathrm {d} t\,\mathrm {d} \sigma _{1}\\&=\int _{a}^{x}{\frac {\mathrm {d} }{\mathrm {d} \sigma _{1}}}\left[{\frac {1}{n!}}\int _{a}^{\sigma _{1}}\left(\sigma _{1}-t\right)^{n}f(t)\,\mathrm {d} t\right]\,\mathrm {d} \sigma _{1}\\&={\frac {1}{n!}}\int _{a}^{x}\left(x-t\right)^{n}f(t)\,\mathrm {d} t\end{aligned}}}$

This completes the proof.