About the remainder of Taylor expansion and Riemann-Liouville integral 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?
 A: Your first Question:
$$f(x)=f(a)+\int_a^x{f'(t)}dt$$
$$\int{U'(t)V(t)}dt=U(t)V(t)-\int{U(t)V'(t)}dt$$
$U(t)=t-x$ ,$V(t)=f'(t)$
$$f(x)=f(a)+\int_a^x{f'(t)}dt=$$$$=f(a)+ [(t-x)f'(t)]_{t=a}^{t=x}-\int_a^x{(t-x)f''(t)}dt=$$$$= f(a)- (a-x)f'(a)-\int_a^x{(t-x)f''(t)}dt=$$$$=f(a)+ (x-a)f'(a)-[\frac{(t-x)^2}{2}f''(t)]_{t=a}^{t=x}+\int_a^x{\frac{(t-x)^2}{2}f'''(t)}dt=$$$$=f(a)+ (x-a)f'(a)-[\frac{(t-x)^2}{2}f''(t)]_{t=a}^{t=x}+\int_a^x{\frac{(t-x)^2}{2}f'''(t)}dt=$$$$=f(a)+ (x-a)f'(a)+\frac{(x-a)^2}{2}f''(a)+\int_a^x{\frac{(t-x)^2}{2}f'''(t)}dt$$
If you continue in that way you can get final result
$$f(x)=\sum_{i=0}^k\frac{f^{(i)}(a)}{i!}(x-a)^i+\int_a^x\frac{f^{(k+1)}(t)}{k!}(x-t)^kdt$$
Riemann-Liouville integral is defined by $$I^{\alpha}[f(x)]=\frac{1}{\Gamma(\alpha)}\int_a^x{f(t)(x-t)^{(\alpha-1)}}dt$$    Equation (1)
Information about the integral from Wiki:
where Γ is the Gamma function and a is an arbitrary but fixed base point. The integral is well-defined provided ƒ is a locally integrable function, and α is a complex number in the half-plane re(α) > 0. The dependence on the base-point a is often suppressed, and represents a freedom in constant of integration. Clearly I1ƒ is an antiderivative of ƒ (of first order), and for positive integer values of α Iαƒ is an antiderivative of order α by Cauchy formula for repeated integration
$$\Gamma (\alpha)=\int_{0}^{\infty }t^{\alpha-1}e^{-t}dt\tag{0}$$
if $\alpha$  positive integer 
$$(\alpha-1)!=\Gamma (\alpha)=\int_{0}^{\infty }t^{\alpha-1}e^{-t}dt\tag{0}$$
if you replace $\alpha=k+1$   and  $f(x)-->f^{(k+1)}(x)$   in Equation (1)
$$I^{k+1}[f^{(k+1)}(x)]=\frac{1}{\Gamma(k+1)}\int_a^x{f^{(k+1)}(t)(x-t)^{k}}dt=\frac{1}{k!}\int_a^x{f^{(k+1)}(t)(x-t)^{k}}dt$$   
Your second Question:
$$f(x)=\sum_{i=0}^k\frac{f^{(i)}(a)}{i!}(x-a)^i+\int_a^x\frac{f^{(k+1)}(t)}{k!}(x-t)^kdt=\sum_{i=0}^k \frac{f^{(i)}(a)}{i!}(x-a)^i+I^{k+1}[f^{k+1}(x)]$$
$$I^{k+1}[f^{k+1}(x)]=f(x)-\sum_{i=0}^k \frac{f^{(i)}(a)}{i!}(x-a)^i$$
