About fractional differentiation under the integral sign

Question

$1.$ Does $\dfrac{d^n}{dx^n}\int_a^bf(x,t)~dt=\int_a^b\dfrac{\partial^n}{\partial x^n}f(x,t)~dt$ correct when $n$ is a positive real number?

$2.$ How about $\dfrac{d^n}{dx^n}\int_{a(x)}^{b(x)}f(x,t)~dt$ when $n$ is a positive real number?

Answer 1

The first rule of Leibniz you mention will still hold for the fractional case, as the definitions of fractional derivatives are equivalent to whole-order differentiation of an integral which is operating orthogonal to the integral in another variable. Symbolically the proof for 1 using the traditional differintegral:

$$ D^n_x = \frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^xf(X,t)(X-x)^{\lceil n \rceil - n -1}dX $$

so: $$ D^n_x\int_a^bf(x,t)dt = \frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \int_a^b f(X,t)dt(X-x)^{\lceil n \rceil - n -1}dX $$ by Fubini's theorem: $$ = \frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_a^b \int_\alpha^x f(X,t) (X-x)^{\lceil n \rceil - n -1}dXdt $$ By Leibniz theorem (for integer integrals and derivatives): $$ = \int_a^b\frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x (X-x)^{\lceil n \rceil - n -1} f(X,t)dXdt $$ $$ = \int_a^bD_x^nf(x,t)dt $$

For Caputo's fractional differential the differential is inside of the integral with respect to x, so the use of Leibniz for integers is used before Fubini's theorem (the steps are just reversed).

For part 2: $$ D^n_x\int_{a(x)}^{b(x)}f(x,t)dt = \frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \int_{a(X)}^{b(X)} f(X,t)dt(X-x)^{\lceil n \rceil - n -1}dX $$ Fubini's theorem applies here but it must be noted that you're operating over an entire measure space simultaneously (the symbolic idea of reversing the integrals just doesn't make sense here): $$ =\frac{d^{\lceil n \rceil}}{dx^{\lceil n \rceil}}\frac{1}{\Gamma(\lceil n \rceil - n)}\int_{[a(X),b(X)]\times [\alpha,x]} f(X,t)(X-x)^{\lceil n \rceil - n -1}d(t,X) $$

This is where things get a bit dicey for me, as the integral in this case involves a bit more measure theory than I'm entirely comfortable with (due to the interdependence of the product that forms the integral). My inclination is that this fails to make sense, as you wind up with differentiation of a variable which is caught in the bounds of both directions in the integral.

if, however the Caputo differ-integral were used:

$$ D^n_x\int_{a(x)}^{b(x)}f(x,t)dt = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \frac{d^{\lceil n \rceil}}{dX^{\lceil n \rceil}}\int_{a(X)}^{b(X)} f(X,t)dt(X-x)^{\lceil n \rceil - n -1}dX $$

applying Leibniz here makes more sense since the derivative variable matches the variables in the limits of integration (for 0 < n < 1): $$ D^n_x\int_{a(x)}^{b(x)}f(x,t)dt = $$ $$ \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \frac{d^{\lceil n \rceil}b(X)}{dX^{\lceil n \rceil}}f(X,b(X))(X-x)^{\lceil n \rceil - n -1} -\frac{d^{\lceil n \rceil}a(X)}{dX^{\lceil n \rceil}}f(X,a(X))(X-x)^{\lceil n \rceil - n -1} + \int_{a(X)}^{b(X)} \frac{d^{\lceil n \rceil} f(X,t)(X-x)^{\lceil n \rceil - n -1}}{dX}dtdX $$ At this point some breakdown can occur into three integrals: $$ = B(x)- A(x) + C(x) $$ By substituting the appropriate integral values for f(x,a(x)) and f(x,b(x)) so that they would cancel in the product of derivatives: $$ A(x) = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \frac{d^{\lceil n \rceil}a(X)}{dX^{\lceil n \rceil}}f(X,a(X))(X-x)^{\lceil n \rceil - n -1} dX $$ $$ B(x) = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x \frac{d^{\lceil n \rceil}b(X)}{dX^{\lceil n \rceil}}f(X,b(X))(X-x)^{\lceil n \rceil - n -1} dX $$ $$ C(x) = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x\int_{a(X)}^{b(X)} \frac{d^{\lceil n \rceil} f(X,t)(X-x)^{\lceil n \rceil - n -1}}{dX}dtdX $$

By the product rule for fractional differentiation we can then say: $$ A(x) = \sum_{i=1}^\infty \binom {n}{i} D_x^j(a(x))D_x^{n-j}(\int_\alpha^xf(X,b(X))(X-x)^{|n-\lceil n \rceil| -1} dX) $$ $$ B(x) = \sum_{i=1}^\infty \binom {n}{i} D_x^j(b(x))D_x^{n-j}(\int_\alpha^xf(X,b(X))(X-x)^{|n-\lceil n \rceil| -1} dX) $$

The term for C(x) then solvable by substitution and the fundamental theorem: $$ C(x) = \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x f(X,b(X))(X-x)^{\lceil n \rceil - n -1}dX $$ $$ - \frac{1}{\Gamma(\lceil n \rceil - n)}\int_\alpha^x f(X,a(X))(X-x)^{\lceil n \rceil - n -1}dX $$

Which implies, by substituting the operators back in: $$ C(x) = D_x^nf(x,b(x)) - D^n_xf(x,a(x)) $$

So the total equation would then be:

$$ D^n_x\int_{a(x)}^{b(x)}f(x,t)dt = \sum_{i=1}^\infty \binom {n}{i} D_x^j(b(x))D_x^{n-j}(\int_\alpha^xf(X,b(X))(X-x)^{|n-\lceil n \rceil| -1} dX) $$ $$ - \sum_{i=1}^\infty \binom {n}{i} D_x^j(a(x))D_x^{n-j}(\int_\alpha^xf(X,b(X))(X-x)^{|n-\lceil n \rceil| -1} dX) $$ $$ + D_x^nf(x,b(x)) - D^n_xf(x,a(x)) $$

So I would be confident in saying that there is a general formula for the fractional derivative, but it doesn't necessarily match the Leibniz rule for normal integration and differentiation. I've already noted my issues trying to find an expression for the normal fractional derivative but after this I'd guess one exists.

It's also a bit late so there might be some typos in my work I can't see at the moment so I'll double check it tomorrow just to be sure.