HP Prime CAS: RiemannLouiville Integral vs Taking the Indefinite Integral Twice
Introduction
The RiemannLouiville Integral takes the integral of the function f(x) of any positive order v. The integral is defined as:
c_D_x^(v) = 1 / Γ(v) * ∫( (x – t) * f(t) dt, t = c, t = x)
where:
c = a real constant, which can be zero
f(x) = function of x
t = dummy variable of integration
‘v = order where v >0
If v=1, this is the regular integral. However, the value of v can be a positive noninteger. If v=2, then the RiemannLouiville integral is a result if you integrate the function twice.
This is one of the formulas on determining indefinite integrals of various orders.
HP Prime CAS Function: dblint
Double Integral of f(x) which takes the integral of f(x) twice. The variable x is used in the function.
dblint(f):= ∫∫ f dx dx
HP Prime CAS Function: rlint
The RiemannLiouville Integral of f(x). The input has the variable x as the independent variable. The result of the function returns t as the independent variable.
rlint(f,c,v):=(∫(t–x)^(v–1)*f,x,c,t)) / Gamma(v)
Note that the variables x and t are switched to allow the input to be a function of x.
Notes
This was programmed on the CAS page in the format:
func(var) := function
I was not able to use the Program Editor mode at time of programming (July 30, 2024). (Beta Firmware 15048)
Examples

Double Integral: dblint 
RLI, v = 2: rlint with c = 0 
f(x) = x^m, m>0 
x^(m+2) / (m^2 + 3*m + 2) 
t^(m+2) / (t^2 + 3*t + 2) 
f(x) = a*x + b 
(a*x^3 + 3*b*x^2) / 6 
(a*t^3 + 3*b*t^2) / 6 
f(x) = e^x 
e^x 
t + e^t – 1 
f(x) = cos x 
cos x 
cos t + 1 
Taking the derivative twice will return us back to the original function. Note that the indefinite integral function assumes that the added constant is zero ( ∫ f(x) dx = F(x) + C ).
Source
Kimeu, Joseph M., "Fractional Calculus: Definitions and Applications" (2009).Masters Theses & Specialist Projects. Paper 115. http://digitalcommons.wku.edu/theses/115
Eddie
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