Elementary Fractional Calculus

In this note we define the fractional derivative \(\frac{\mathrm{d}^\alpha f}{\mathrm{d}x^\alpha}\) of a function \(f\), where \(\alpha\) is allowed to be an arbitrary complex parameter. Notice that, heuristically speaking,

\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}x} \int f = f = \int \frac{\mathrm{d}}{\mathrm{d}x} f. \end{equation*}

So, in a sense that will be made more precise, we have

\begin{equation*} \int = \left(\frac{\mathrm{d}}{\mathrm{d}x}\right)^{-1} = \frac{\mathrm{d}^{-1}}{\mathrm{d}x^{-1}}. \end{equation*}

This means that by extending the derivative we also obtain a generalized integration in “fractional dimension”. In fact, it turns out that fractional integration makes a good starting point for fractional calculus theory in general. To keep things clear we disregard all convergence issues in what follows.

Fractional Integration Approach

We begin by defining fractional integration. Since

\begin{equation*} \frac{\mathrm{d}}{\mathrm{d}x} \int_{c}^{x} f(t) ~\text{d} t = f(x) \end{equation*}

we can think of the operator \(f \longmapsto \int_{c}^{x} f(t) ~\text{d} t\) as an inverse to differentiation. We express this by writing

\begin{equation*} {}_{c}D_x^{-1}f(x) = \int_{c}^{x} f(t) ~\text{d} t \end{equation*}

Here we think of \(D\) as a generalized differentation/integration operator. The superscript denotes the order to which we are differentiating. The fact that integration is the inverse to differentation translates to a superscript of \(-1\). Notice that these generalized derivatives depend on a choice of base point \(c\).

If we repeat this process, we see that it makes sense to define

\begin{equation*} {}_{c}D_{x}^{-2} = {}_{c}D_{x}^{-1} \circ {}_{c}D_{x}^{-1} \end{equation*}

Let us compute what this means explicitly:

\begin{align*} {}_{c}D_x^{-2}f(x) &= \int_{c}^{x} {}_{c}D_{t_0}^{-1}f(t_0) ~\text{d} t_0 = \int_{c}^{x} \int_{c}^{t_0} f(t_1) ~\text{d} t_1 ~\text{d} t_0 \\ &= \int_{c}^{x} f(t_1) \int_{t_1}^{x} ~\text{d} t_0 ~\text{d} t_1 \\ &= \int_{c}^{x} f(t_1)(x-t_1) ~\text{d} t_1 \end{align*}

Applying this procedure recursively we arrive at a definition for \({} _ {c}D _ {x}^{-n}\):

\begin{equation*} {}_{c}D_{x}^{-n}f(x) = \frac{1}{(n-1)!} \int_{c}^{x} f(t)(x-t)^{n-1} ~\text{d} t \end{equation*}

To keep notation standard, we replace \(-n\) with \(n\) in the formula above to obtain

\begin{equation*} {}_{c}D_{x}^{n}f(x) = \frac{1}{(-n-1)!} \int_{c}^{x} \frac{f(t)}{(x-t)^{n+1}} ~\text{d} t \end{equation*}

valid for negative integers \(n\). In order to extend \(D\) to arbitrary complex order, we simply replace the factorials with gamma functions, then swap out the negative integers \(n\) for arbitrary complex \(\alpha\). In this way we arrive at the following formula for the generalized \(\alpha^{th}\)-derivative of \(f\):

\begin{equation} \label{eq:def1} {}_{c}D_{x}^{\alpha}f(x) = \frac{1}{\Gamma(-\alpha)} \int_{c}^{x} \frac{f(t)}{(x-t)^{\alpha+1}} ~\text{d} t \end{equation}

Fourier Transform Approach

In this section we attempt to define a fractional derivative using Fourier analtyic methods. Recall that the Fourier transform is the automorphism of \(L^2(\mathbb{R})\) defined by

\begin{equation*} \mathcal{F}[f](\xi) = \int_{\mathbb{R}}^{} f(x)e^{-2\pi i\xi x} ~\text{d}x \end{equation*}

Let \(S\) be an operator defined on \(L^2\) functions, and let \(S\) satisfy the following Fourier transform identity:

\begin{equation*} \sigma(\xi) \cdot \mathcal{F}[f](\xi) = \mathcal{F}[Sf](\xi) \end{equation*}

Then we call the function \(\sigma\) the symbol of the operator \(S\). We shall be particularly interested in the symbol of the differentiation operator \(\frac{\mathrm{d}}{\mathrm{d}x}\). We know that the derivative transforms like

\begin{equation*} (2\pi i \xi) \cdot \mathcal{F}[f](\xi) = \mathcal{F} \left[ \frac{\mathrm{d}f}{\mathrm{d}x}\right](\xi) \end{equation*}

so \(\frac{\mathrm{d}}{\mathrm{d}x}\) has symbol \(2\pi i \xi\). In general \(\frac{\mathrm{d}^n}{\mathrm{d}x^n}\) has symbol \((2\pi i\xi)^n\) since

\begin{equation*} (2\pi i \xi)^n \cdot \mathcal{F}[f](\xi) = \mathcal{F} \left[ \frac{\mathrm{d}^nf}{\mathrm{d}x^n}\right](\xi) \end{equation*}

If we invert the Fourier transform in the equations above, we obtain a new expression for the derivative of \(f\), namely

\begin{align*} \frac{\mathrm{d}^nf}{\mathrm{d}x^n}(x) &= \mathcal{F}^{-1}[(2\pi i\xi)^n \cdot \mathcal{F}[f](\xi)] \\ &= \mathcal{F}^{-1}[(2\pi i\xi)^n] \ast f(x) \end{align*}

Here we must interpret the Fourier transform in the sense of Schwartz distributions. This line of reasoning suggests we define the generalized \(\alpha^{th}\) derivative by

\begin{equation} \label{eq:def2} D^\alpha f(x) = \mathcal{F} ^{-1}[(2\pi i \xi)^\alpha]\ast f(x) \end{equation}

Let us try to reconcile this definition with the one given in equation \ref{eq:def1}. We will show that the so-called Weyl fractional derivative \({} _ {-\infty}D _ {x}^\alpha f\) with base point \(-\infty\) has symbol \((2\pi i\xi)^\alpha\), meaning that equation \ref{eq:def2} defines the same operator as \ref{eq:def1} when \(c = -\infty\). To this end we compute

\begin{align*} \mathcal{F}[{} _ {-\infty} D _ {x}^\alpha f](\xi) &= \int _ {\mathbb{R}}^{} {} _ {-\infty}D _ {x}^\alpha f(x) e^{-2\pi i \xi x} ~\text{d}x \\ &= \frac{1}{\Gamma(-\alpha)} \int_{\mathbb{R}}^{} \int_{-\infty}^{x} \frac{f(t)}{(x-t)^{\alpha + 1}} ~\text{d} t ~ e^{-2\pi i \xi x} ~\text{d} x \\ &= \frac{1}{\Gamma(-\alpha)} \int_{\mathbb{R}} \int_{t}^{\infty} \frac{f(t)}{(x-t)^{\alpha + 1}} e^{-2\pi i \xi x} ~\text{d} x ~\text{d} t \\ &= \frac{1}{\Gamma(-\alpha)} \int_{\mathbb{R}} f(t)e^{-2\pi i \xi t} \int_{0}^{\infty} x^{-\alpha - 1} e^{-2\pi i \xi x} ~\text{d} x ~\text{d} t \\ &= (2\pi i \xi)^{\alpha} \int_{\mathbb{R}}^{} f(t) e^{-2\pi i \xi x} ~\text{d} x \\ &= (2\pi i \xi)^{\alpha} \mathcal{F}[f](\xi) \end{align*}

In order to complete the last two steps of the above computation, we use the defintion of the \(\Gamma\)-function, which is

\begin{equation*} \Gamma(\alpha) = \int_{0}^{\infty} t^{\alpha} e^{-\alpha} \frac{~\text{d}t}{t} \end{equation*}

along with the change of variables \(s = 2\pi i\xi s\) with \(\mathrm{d}s = 2\pi i\xi \mathrm{d}t\), in order to derive

\begin{equation*} (2\pi i \xi)^{-\alpha} = \frac{1}{\Gamma(\alpha)} \int_{0}^{\infty} s^{\alpha}e^{-2\pi i\xi s} \frac{~\text{d} s}{s} \end{equation*}

Or, what is the same,

\begin{equation*} (2\pi i \xi)^{\alpha} = \frac{1}{\Gamma(-\alpha)} \int_{0}^{\infty} t^{-\alpha}e^{-2\pi i\xi t} \frac{~\text{d} t}{t} \end{equation*}

which is exactly the identity we used in the computation above. So to recap, we have shown that the Fourier transform definition of the fractional deriviative corresponds to the previous definition, provided we choose the base point \(c = -\infty\).

Cauchy Integral Formula Approach

In this section we take a different approach to defining fractional differentiation, this time based on the Cauchy integral formula. As a result, we temporarily require \(f\) to be analytic in a neighbourhood of \(\mathbb{R}\). As we shall see, this approach ends up agreeing with the one in the previous section.

Suppose \(C_R\) is the circle of radius \(R\) about the point \(x\in \mathbb{R}\). The generalized Cauchy integral formula states

\begin{equation*} D^n f(x) = \frac{n!}{2\pi i} \int_{C_R}^{} \frac{f(\zeta)}{(\zeta - x)^{n+1}} ~\text{d}\zeta \end{equation*}

We may attempt to replace the positive integer \(n\) in this formula with an arbitrary complex number to obtain

\begin{equation*} D^\alpha f(x) = \frac{\Gamma(\alpha + 1)}{2\pi i} \int_{C_R}^{} \frac{f(\zeta)}{(\zeta - x)^{\alpha + 1}} ~\text{d}\zeta \end{equation*}

However we have now introduced a branching singularity at \(x\). To get around this we:

  1. Choose \(r < R\), make a branch cut along the ray \((-\infty, x)\), and replace the contour \(C_R\) with the curve
\begin{equation*} \gamma = L_1 \cup C_r \cup L_2 \end{equation*}

where \(L_1\) is the line segment joining \(x - R\) to \(x - r\), and \(L_2\) is \(L_1\) with reversed orientation. Take the branch of the logarithm with argument ranging from \(-\pi\) to \(\pi\). Depending on which segment of \(\gamma\) contains \(\zeta\) we get the following representations for the function \((\zeta - x)^{-\alpha - 1}\):

  • On \(L_1\):
\begin{align*} (\zeta - x)^{-\alpha - 1} &= \exp[(-\alpha - 1)(\log(|\zeta - x|) - i\pi)[ \\ &= \exp[(-\alpha - 1)(\log(x - \zeta) - i\pi)] \end{align*}
  • On \(C_r\): (with angular variable \(-\pi \leq \theta \leq \pi\))
\begin{align*} (\zeta - x)^{-\alpha - 1} &= \exp[(-\alpha - 1)(\log(|\zeta - x|) + i\theta)] \\ &= \exp[(-\alpha - 1)(\log(r) + i\theta)] \end{align*}
  • On \(L_2\):
\begin{align*} (\zeta - x)^{-\alpha - 1} &= \exp[(-\alpha - 1)(\log(|\zeta - x|) + i\pi)] \\ &= \exp[(-\alpha - 1)(\log(x - \zeta) + i\pi)] \end{align*}

Therefore, we compute

\begin{align*} \int_{\gamma}^{} \frac{f(\zeta)}{(\zeta - x)^{\alpha + 1}} ~\text{d} \zeta &= e^{(\alpha + 1)i\pi}\int_{x-R}^{x-r} \frac{f(t)}{(x- t)^{\alpha + 1}} ~\text{d} t \\ &\quad+ \int_{C_r}^{} \frac{f(\zeta)}{(\zeta - x)^{\alpha + 1}} ~\text{d} \zeta \\ &\quad+ e^{-(\alpha + 1)i\pi} \int_{x-r}^{x-R} \frac{f(t)}{(x-t)^{\alpha + 1}} ~\text{d} t \end{align*}

We can estimate the portion of the integral over \(C_r\) via the inequality

\begin{align*} \left| \int_{C_r}^{} \frac{f(\zeta)}{(\zeta - x)^{\alpha + 1}} ~\text{d} \zeta \right| = \left| \int_{-\pi}^{\pi} \frac{f(x + re^{i\theta})}{r^{\alpha + 1}e^{(\alpha + 1)i\theta}} ire^{i\theta} ~\text{d} \theta \right| \leq r^{-\Re(\alpha)} \int_{-\pi}^{\pi} |f(x+re^{i\theta})| ~\text{d} \theta \end{align*}

It follows that once \(\Re(\alpha) < 0\) we can let \(r\to 0\) to obtain

\begin{align*} D^\alpha f(x) &= \frac{\Gamma(\alpha + 1)}{2\pi i} \int_{\gamma}^{} \frac{f(\zeta)}{(\zeta - x)^{\alpha + 1}} ~\text{d} \zeta \\ &= \frac{\Gamma(\alpha + 1)}{2\pi i} (e^{(\alpha + 1)i\pi} - e^{-(\alpha + 1)i\pi}) \int_{x-R}^{x} \frac{f(t)}{(x-t)^{\alpha + 1}} ~\text{d} t \\ &= \frac{\Gamma(\alpha + 1)\sin(\alpha + 1)\pi}{\pi} \int_{x-R}^{x} \frac{f(t)}{(x-t)^{\alpha + 1}} ~\text{d} t \end{align*}

Let us set \(c := x - R\) and employ the lovely reflection formula to rewrite the above as

\begin{align*} {}_{c}D_{x}^\alpha f(x) &= \frac{1}{\Gamma(-\alpha)} \int_{c}^{x} \frac{f(t)}{(x-t)^{\alpha + 1}} ~\text{d} t \end{align*}