Cauchy formula for repeated integration

Let f be a continuous function on the real line. Then the nth repeated integral of f with base-point a, $${\displaystyle f^{(-n)}(x)=\int _{a}^{x}\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n-1}}f(\sigma _{n})\,\mathrm {d} \sigma _{n}\cdots \,\mathrm {d} \sigma _{2}\,\mathrm {d} \sigma _{1},}$$ is given by single integration $${\displaystyle f^{(-n)}(x)={\frac {1}{(n-1)!}}\int _{a}^{x}\left(x-t\right)^{n-1}f(t)\,\mathrm {d} t.}$$

A proof is given by induction. The base case with n = 1 is trivial, since it is equivalent to $${\displaystyle f^{(-1)}(x)={\frac {1}{0!}}\int _{a}^{x}{(x-t)^{0}}f(t)\,\mathrm {d} t=\int _{a}^{x}f(t)\,\mathrm {d} t.}$$

Now, suppose this is true for n, and let us prove it for n + 1. Firstly, using the Leibniz integral rule, note that $${\displaystyle {\frac {\mathrm {d} }{\mathrm {d} x}}\left[{\frac {1}{n!}}\int _{a}^{x}(x-t)^{n}f(t)\,\mathrm {d} t\right]={\frac {1}{(n-1)!}}\int _{a}^{x}(x-t)^{n-1}f(t)\,\mathrm {d} t.}$$ Then, applying the induction hypothesis, $${\displaystyle {\begin{aligned}f^{-(n+1)}(x)&=\int _{a}^{x}\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1})\,\mathrm {d} \sigma _{n+1}\cdots \,\mathrm {d} \sigma _{2}\,\mathrm {d} \sigma _{1}\\&=\int _{a}^{x}\left[\int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1})\,\mathrm {d} \sigma _{n+1}\cdots \,\mathrm {d} \sigma _{2}\right]\,\mathrm {d} \sigma _{1}.\end{aligned}}}$$ Note that the term within square bracket has n-times successive integration, and upper limit of outermost integral inside the square bracket is ${\displaystyle \sigma _{1}}$. Thus, comparing with the case for n = n and replacing ${\displaystyle x,\sigma _{1},\cdots ,\sigma _{n}}$ of the formula at induction step n = n with ${\displaystyle \sigma _{1},\sigma _{2},\cdots ,\sigma _{n+1}}$ respectively leads to $${\displaystyle \int _{a}^{\sigma _{1}}\cdots \int _{a}^{\sigma _{n}}f(\sigma _{n+1})\,\mathrm {d} \sigma _{n+1}\cdots \,\mathrm {d} \sigma _{2}={\frac {1}{(n-1)!}}\int _{a}^{\sigma _{1}}(\sigma _{1}-t)^{n-1}f(t)\,\mathrm {d} t.}$$ Putting this expression inside the square bracket results in $${\displaystyle {\begin{aligned}&=\int _{a}^{x}{\frac {1}{(n-1)!}}\int _{a}^{\sigma _{1}}(\sigma _{1}-t)^{n-1}f(t)\,\mathrm {d} t\,\mathrm {d} \sigma _{1}\\&=\int _{a}^{x}{\frac {\mathrm {d} }{\mathrm {d} \sigma _{1}}}\left[{\frac {1}{n!}}\int _{a}^{\sigma _{1}}(\sigma _{1}-t)^{n}f(t)\,\mathrm {d} t\right]\,\mathrm {d} \sigma _{1}\\&={\frac {1}{n!}}\int _{a}^{x}(x-t)^{n}f(t)\,\mathrm {d} t.\end{aligned}}}$$

• It has been shown that this statement holds true for the base case ${\displaystyle n=1}$.
• If the statement is true for ${\displaystyle n=k}$, then it has been shown that the statement holds true for ${\displaystyle n=k+1}$.
• Thus this statement has been proven true for all positive integers.

This completes the proof.