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In fractional calculus, one usually tends to ignore the constant of "differ-integration" if you will, but when I attempted a problem with some fractional calculus, I found the result was somewhat off, which led me to the belief that I was missing the "simple" constant of integration.

After trying to put it to paper, I found this problem to be very annoying and confusing, as I came up with a few things.

Let $D^n_xf(x)$ represent the $n$th derivative of $f$ with respect to $x$.

$$D^n_xx^k=\frac{\Gamma(k+1)}{\Gamma(k+1-n)}x^{k-n}$$

For $k+1-n\in\mathbb Z$ and $k+1-n<0$, we get

$$D^n_xx^k=\frac{\Gamma(k+1)}{\pm\infty}x^{k-n}=0$$

So from this, we get

$$D^n_x0=\frac{\Gamma(k+1-n)}{\Gamma(k+1)}x^{k+n}$$

where $k\in\mathbb Z$ and $k\le0$.

I also run into other problems, like $D^n_xD^k_x0=D^{n+k}_x0$, and everything starts to fall apart. Is there a good defined way for representing the constant of differ-integration?

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  • $\begingroup$ you shouldn't use $n$ which is reserved for integers. and if $a = -2$, then $D^{a}_x x^k = \frac{x^{k+2}}{k (k+1)} + C_1 + C_2 x$ : i.e. the "constant of integration" is a degree $1$ polynomial $\endgroup$ – reuns Jun 4 '16 at 1:51
  • $\begingroup$ if you know convolution and Fourier transform/series, if $a < 0$ then $D^a_x$ is a filter equation : $g = D^a_x f \Leftrightarrow D^{-a}_x g = g \ast h^a_x = f$ where $h^a_x$ is some distribution representing the fractional derivative operator, and $\ast$ is the convolution. What is improtant is that since this is a convolution, it is a linear operator, and for understanding the integration constant, you can check what is the kernel of the linear operator $u \mapsto D^a_x u$. $\endgroup$ – reuns Jun 4 '16 at 1:55
  • $\begingroup$ if $n \in \mathbb{Z}$, you know that any polynomial $P$ of degree $n-1$ is mapped to $0$ by $D^{n}_x$ : $D^n_x P = 0$, hence the integration constant is a polynomial of degree $n-1$. Now from $D^a_x x^{k}=\dfrac{\Gamma(k+1)}{\Gamma(k+1-a)}x^{k-a}$ you can find what is the kernel of the $u \mapsto D^a_x u$ operator, and hence the form of the integration constant for $D^{-a}_x u$ $\endgroup$ – reuns Jun 4 '16 at 2:00
  • $\begingroup$ @user1952009 If we use the definition of your polynomial $P$, then $D^{1/2}_x0=C_1x^{-1/2}+C_2x^{-3/2}+C_3x^{-5/2}+\dots$, which, when differentiated $-1/2$ times, gives back $0$. And if $D^{1/2}_xD^{1/2}_xu=D^1_xu$, than your definition of $P$ is somewhat flawed, as after I get differintegrate the first time, I get $P_1$. After I differintegrate the second time, I get $D^{1/2}_xP_1=0$, $D^{1/2}_xu=P_2$, a second polynomial of integration, which goes against the normal $D^1_xu$ $\endgroup$ – Simply Beautiful Art Jun 4 '16 at 14:38
  • $\begingroup$ For any linear operator $T$ : $\ \ T0 = 0$, thus $D^{1/2} 0 = 0$, so I don't understand what you are trying to say. Let $a> 0$, did you get the "kernel of the linear operator $u \mapsto D^a u$ " idea for defining the integration constant in $u \mapsto D^{-a} u$ ? Hence the problem is finding a definition of $u \mapsto D^{a} u$ letting us find easily its kernel en.wikipedia.org/wiki/Kernel_(linear_algebra) $\endgroup$ – reuns Jun 4 '16 at 21:04
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You are attacking something complicated. Consider the functions of one real variable $x$, in my notations $a,b,c \in \mathbb{R}^{*+}$, $n,m \in \mathbb{N}^*$, $\ \ C_k,C$ are real constants, $u,v,w$ are differentiable/smooth functions of $x$.

As I Wrote $$u \mapsto D^a u$$

is a linear operator. That is : $$D^a (u+v) =D^a u+D^a v, \qquad \qquad D^a (C u) = C D^au$$

When you write $D^{-a}u$, you are in fact considering the solutions of the linear equation $D^a v = u$ where $v$ is the unknown, and what you call the "integration constant" means how to define this set of solutions.

As every linear equation, once we have a solution $D^a v = u$, we can find all the other solutions from the kernel of $D^a$ : if $D^a w = 0$, then $v+w$ is also a solution of $D^a v = u$, and all the solutions are of this form.

You know that $D^1 x^0 = 0$, and from $D^{n+m} =D^{n}D^{m}$ you get that the kernel of $D^n$ are the degree $n-1$ polynomials. This is why you can write : $$D^n v = u \quad \implies \quad v = J^n u + \sum_{k=0}^{n-1} C_k x^k$$

Where "$J^n$ is the integrate $n$ times linear operator", i.e. $J^1 u (x) = \int_0^x u(t) dt$ and by induction $J^n = J^1 J^{n-1}$. Clearly $D^n J^n u = u$, so that $J^n$ is the right-inverse of $D^n$, but not the left-inverse, and there is no such, since the kernel of $D^n$ is non-empty, and hence not invertible.

Then you have to prove that with $\gamma(a,c) = \frac{\Gamma(c+1)}{\Gamma(c+1-a)}$ and when everything is well-defined (see below) : $$D^a x^c = \gamma(a,c) x^{c-a}, \qquad\qquad D^{a}D^{b} x^c = D^{a+b} x^c$$

How do you explain then that $D^1 x^0 = 0$ ? In fact $D^1 x^0 = \gamma(1,0) x^{-1}$ and this is $= 0$ only because $\gamma(1,0) =0$.

(Since every differentiable function can be written as a linear combination of powers of $x$, we can look only at those powers of $x$. )

And the kernel of $D^a$ is obtained from the set of values $c$ for which $\gamma(a,c) = 0$. It is a theorem that the $\Gamma$ function is nowhere zero, and has poles of order $1$ at negative integers. hence $$\gamma(a,c) = \frac{\Gamma(c+1)}{\Gamma(c+1-a)} = 0 \Leftrightarrow \Gamma(c+1)\ne \infty, \Gamma(c+1-a)= \infty \Leftrightarrow \ -(c+1) \not\in \mathbb{N}, \ -(c+1-a) \in \mathbb{N}$$

And this is the answer to your question. If $a$ is not an integer, then the kernel of $D^a$ is $$D^a w = 0 \qquad \Leftrightarrow \qquad w = \sum_{k=1}^\infty C_k x^{a-k} $$

And when $n$ is an integer $$D^n w = 0 \qquad \Leftrightarrow \qquad w = \sum_{k=1}^n C_k x^{n-k} $$

Now there is another problem : what happens if $\gamma(a,c) = \infty$ ? Because yes it can happen, when $-(c+1) \in \mathbb{N}, \ -(c+1-a) \not\in \mathbb{N}$.

This is how you get your problem, when trying to relate the kernel of $D^1$ with the kernel of $D^{1/2}$. Indeed $D^{1/2} u = 0 $ doesn't imply in general that $ D^1 u = D^{1/2}(D^{1/2} u) = 0$.

This is why you can't write that $D^{a+b} = D^a D^b$.

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  • $\begingroup$ So we try to avoid $D^{a+b}=D^aD^b$ simply because we can't do it? Thank you for the nice explanation. $\endgroup$ – Simply Beautiful Art Jun 5 '16 at 15:53
  • $\begingroup$ So is this kernel concept how fractional differ-integrals of the power functions get "past" the singularity that is the constant function? Asking for a friend=>. math.stackexchange.com/questions/2698895/… $\endgroup$ – No Name Mar 21 '18 at 0:55

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