Integral form of Taylor expansion looks like this:$$f(x)=\sum_{i=0}^k\frac{f^{(i)}(a)}{i!}(x-a)^i+\int_a^x\frac{f^{(k+1)}(t)}{k!}(x-t)^kdt$$ Riemann-Liouville integral is $$I^{\alpha}f=\frac{1}{\Gamma(\alpha)}\int_a^x{f(t)(x-t)^{(\alpha-1)}}dt$$ Q1: The integral form remainder of Taylor expansion is exactly $I^{k+1}f^{(k+1)}$. Why is that?
Q2: As far as I know, Riemann-Liouville integral is basically $\alpha$th antiderivative of $f(x).$ Shouldn't $f(x)=I^{k+1}f^{(k+1)}$? Is that right?