N-th derivative of n-fold integral

Question

I want to justify that the n-th derivative of an n-fold integral gives the original function. In other words that $$ \frac{d^n}{dx^n}\frac{1}{(n-1)!}\int_{0}^{x}(x-s)^{n-1}f(s)ds=f(x) $$ If I substitute n=1, then is clear that the equation holds. But If I substitute n=2, then: $$ \frac{d^2}{dx^2}\frac{1}{(2-1)!}\int_{0}^{x}(x-s)^{2-1}f(s)ds=f(x) $$ which is: $$ \frac{d^2}{dx^2}\int_{0}^{x}(x-s)f(s)ds=f(x) $$ Now I can split this like this: $$ \frac{d^2}{dx^2}\int_{0}^{x}xf(s)ds- \frac{d^2}{dx^2}\int_{0}^{x}sf(s)ds $$ Which wolfram alpha says that it is: $$ xf'(x)+2f(x)-(xf'(x)+f(x)) $$ That clearly gives f(x). But how do I prove that the second derivatives of such integrals in fact gives those results?

Answer 1

I think I've got the answer. We have to use the Leibniz's rule integral form.

So for example

$$ \frac{d^2}{dx^2}\int_{0}^{x}xf(s)ds=\frac{d}{dx}(xf(x)+\int_{0}^{x}f(s)ds) $$ which is equal to: $$ =xf'(x)+f(x)+f(x)+\int_{0}^{x}0ds=xf'(x)+2f(x) $$ As wolfram alpha said.

Answer 2

I think I've got the answer, using the mathematical induction. We have verify that the formula is valid for n=1 and n=2, also n=3 (but not shown here).

Now we make our hypothesis:

I.H. : $$ \frac{1}{(n-1)!}\frac{d^n}{dx^n}\int_{0}^{x}(x-s)^{n-1}f(s)ds=f(x) $$

Now we make the inductive step, which is that for n+1 we will got f(x) also, so make n+1 in the above formula:

$$ \frac{1}{(n)!}\frac{d^{n+1}}{dx^{n+1}}\int_{0}^{x}(x-s)^{n}f(s)ds=f(x) $$ Which is equal to: $$ \frac{1}{(n)!}\frac{d^n}{dx^n}(\frac{d}{dx}\int_{0}^{x}(x-s)^{n}f(s)ds)= $$ Using the Leibniz Rule in its integral form we have: $$ \frac{1}{(n)!}\frac{d^n}{dx^n}((x-x)^{n}f(x)+\int_{0}^{x}n(x-s)^{n-1}f(s)ds)= $$ $$ \frac{n}{(n)!}\frac{d^n}{dx^n}(\int_{0}^{x}(x-s)^{n-1}f(s)ds)= $$ $$ \frac{n}{1*2*3...*(n-1)*n}\frac{d^n}{dx^n}(\int_{0}^{x}(x-s)^{n-1}f(s)ds)= $$ $$ \frac{1}{(n-1)!}\frac{d^n}{dx^n}(\int_{0}^{x}(x-s)^{n-1}f(s)ds) $$ Using the I.H. we observe that this is equal to f(x):

$$ \frac{1}{(n-1)!}\frac{d^n}{dx^n}(\int_{0}^{x}(x-s)^{n-1}f(s)ds)=f(x) \\ \square $$