I was trying to follow along with the proof on Wikipedia for Cauchy's formula for repeated integrals and I'm stuck on the last step. How do you go from$$\int_a^x \frac {d}{dy} \left( \int_a^y (y-t)^nf(t)dt\right) dy$$to$$\int_a^x (x-t)^nf(t) dt$$
$\endgroup$ 02 Answers
$\begingroup$We always have$$\int_a^b\frac{\mathrm d\varphi}{\mathrm dy}\,\mathrm dy=\varphi(b)-\varphi(a),$$by the Fundamental Theorem of Calculus. So,$$\int_a^x\frac{\mathrm d}{\mathrm dy}\left(\int_a^y(y-t)^nf(t)\,\mathrm dt\right)\,\mathrm dy=\int_a^x(x-t)^nf(t)\,\mathrm dt-\overbrace{\int_a^a(a-t)^nf(t)\,\mathrm dt}^{\phantom{0}=0}.$$
$\endgroup$ 3 $\begingroup$I´m going to show a sketch of a proof for Cauchy´s formula for repeated integrals.
The formula says:$$f^{(-n)}(x)=\frac{1}{(n-1)!}\int_{a}^{x} (x-t)^{n-1}f(t)dt$$
Let start with $n=3$, and later on we try to generalize it.
Then:$$f^{(-3)}(x)=\int_{a}^{x} \int_{a}^{u} \int_{a}^{s}f(t)dtdsdu $$$$\int_{a}^{x} \int_{a}^{u} \int_{a}^{s}f(t)dtdsdu=\int_{a}^{x}\left[ \int_{a}^{u} \int_{a}^{s}f(t)dtds\right]du $$Now concentrate in the double integral inside the parenthesis. Here is a sketch of it´s region of integration.
We can swap the order the integration, according to the following picture:
If we do so, we get the following expression for the inner double integral$$ \int_{a}^{u} \int_{a}^{s}f(t)dtds=\int_{a}^{u} \int_{t}^{u} f(t)dsdt$$$$\int_{a}^{u} \int_{t}^{u} f(t)dsdt=\int_{a}^{u}f(t)dt \int_{t}^{u}ds=\int_{a}^{u}(u-t)f(t)dt$$
$$ \Rightarrow f^{(-3)}(x)=\int_{a}^{x} \int_{a}^{u} \int_{a}^{s}f(t)dtdsdu =\int_{a}^{x} \left[\int_{a}^{u}(u-t)f(t)dt\right]du$$
Proceeding in the same way we did before (changing the order of integration), we get$$\int_{a}^{x} \int_{a}^{u}(u-t)f(t)dtdu=\int_{a}^{x} \int_{t}^{x}(u-t)f(t)dudt$$$$\int_{a}^{x} \int_{t}^{x}(u-t)f(t)dudt=\int_{a}^{x}f(t)dt \int_{t}^{x}(u-t)du$$The second integral on the right hand side of the equation can be solved using a substitution: let $s=u-t \Rightarrow ds=du$$$\int_{a}^{x}f(t)dt \int_{t}^{x}(u-t)du=\int_{a}^{x}f(t)dt\int_{0}^{x-t} sds$$And finally we get:$$f^{(-3)}(x)=\int_{a}^{x} \int_{a}^{u} \int_{a}^{s}f(t)dtdsdu =\int_{a}^{x} \frac{(x-t)^{2}}{2!}f(t)dt$$It´s starting to emerge a pattern, if we keep integrating further, we may prove by induction Cauchy´s formula.
$\endgroup$
