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Let's talk about the function ##f(x)=x^n##.

It's derivative of ##k^{th}## order can be expressed by the formula:

$$\frac{d^k}{dx^k}=\frac{n!}{(n-k)!}x^{n-k}$$

Similarly, the ##k^{th}## integral (integral operator applied ##k## times) can be expressed as:

$$\frac{n!}{(n+k)!}x^{n+k}$$

According the the Wikipedia article https://en.wikipedia.org/wiki/Fractional_calculus, we can replace the factorial with the Gamma function to get derivatives of fractional order.

So, applying the derivative of half order twice to ##\frac{x^{n+1}}{n+1}+C##, should get us to ##x^n##.

Applying the half-ordered derivative once gives:

$$\frac{d^{1/2}}{{dx^{1/2}}}\left(\frac{x^{n+1}}{n+1}+Cx^0\right)=\frac{1}{n+1}\frac{\Pi(n+1)}{\Pi(n+1/2)}x^{n+1/2}+C\frac{1}{\Pi(-1/2)}x^{-1/2}$$

where ##\Pi(x)## is the generalization of the factorial function, and ##\Pi(x)=\Gamma(1+x)##.

Again, applying the half-ordered derivative gives:

$$\frac{1}{n+1}\frac{\Pi(n+1)}{\Pi(n)}x^n+\frac{C}{\Pi(-1)}x^{-1}=x^n$$

which works fine because ##\frac{C}{\Pi(-1)}\rightarrow 0##. So, the derivative works good but that's not the case with fractional-ordered integration.

Applying the half-ordered integral operator twice to ##x^n## should give us ##\frac{x^{n+1}}{n+1}+C##. Applying the half-ordered integral once means finding a function whose half-ordered derivative is ##x^n##. So, applying it once gives:

$$\frac{\Pi(n)}{\Pi{(n+1/2)}}x^{n+1/2}+C\frac{1}{\Pi(-1/2)}x^{-1/2}$$

Again, applying the half ordered derivative to this function should give a function whose half-ordered derivative is this function. So, again applying the half-integral operator gives:

$$\frac{x^{n+1}}{n+1}+C+C'\frac{1}{\Pi(-1/2)}x^{-1/2}\neq \frac{x^{n+1}}{n+1}+C$$

where ##C'## is another constant. So, why this additional term containing ##C'## gets introduced? Is the theory of fractional derivatives flawed? Is there any way to get a single constant ##C## in the end by applying the half-integral operator two times?

It's derivative of ##k^{th}## order can be expressed by the formula:

$$\frac{d^k}{dx^k}=\frac{n!}{(n-k)!}x^{n-k}$$

Similarly, the ##k^{th}## integral (integral operator applied ##k## times) can be expressed as:

$$\frac{n!}{(n+k)!}x^{n+k}$$

According the the Wikipedia article https://en.wikipedia.org/wiki/Fractional_calculus, we can replace the factorial with the Gamma function to get derivatives of fractional order.

So, applying the derivative of half order twice to ##\frac{x^{n+1}}{n+1}+C##, should get us to ##x^n##.

Applying the half-ordered derivative once gives:

$$\frac{d^{1/2}}{{dx^{1/2}}}\left(\frac{x^{n+1}}{n+1}+Cx^0\right)=\frac{1}{n+1}\frac{\Pi(n+1)}{\Pi(n+1/2)}x^{n+1/2}+C\frac{1}{\Pi(-1/2)}x^{-1/2}$$

where ##\Pi(x)## is the generalization of the factorial function, and ##\Pi(x)=\Gamma(1+x)##.

Again, applying the half-ordered derivative gives:

$$\frac{1}{n+1}\frac{\Pi(n+1)}{\Pi(n)}x^n+\frac{C}{\Pi(-1)}x^{-1}=x^n$$

which works fine because ##\frac{C}{\Pi(-1)}\rightarrow 0##. So, the derivative works good but that's not the case with fractional-ordered integration.

Applying the half-ordered integral operator twice to ##x^n## should give us ##\frac{x^{n+1}}{n+1}+C##. Applying the half-ordered integral once means finding a function whose half-ordered derivative is ##x^n##. So, applying it once gives:

$$\frac{\Pi(n)}{\Pi{(n+1/2)}}x^{n+1/2}+C\frac{1}{\Pi(-1/2)}x^{-1/2}$$

Again, applying the half ordered derivative to this function should give a function whose half-ordered derivative is this function. So, again applying the half-integral operator gives:

$$\frac{x^{n+1}}{n+1}+C+C'\frac{1}{\Pi(-1/2)}x^{-1/2}\neq \frac{x^{n+1}}{n+1}+C$$

where ##C'## is another constant. So, why this additional term containing ##C'## gets introduced? Is the theory of fractional derivatives flawed? Is there any way to get a single constant ##C## in the end by applying the half-integral operator two times?

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