For $f(x)=x$, the half-derivative of $f$ is $$\frac{d^{\frac{1}{2}}}{dx^{\frac{1}{2}}} x = 2 \sqrt{\frac{x}{\pi}} \;.$$ Is there some geometric interpretation of (Q1) this specific derivative, and, (Q2) of the half-derivative more generally? I have read that fractional derivatives are nonlocal, but it seems strange to me that integral derivatives can be described in terms of local geometry only, while fractional derivatives cannot be so described. This would suggest an odd discontinuity between, say, $d^{1}$ and $d^{1.01}$. This seems especially at odds with the many applications of fractional derivatives, which (superficially) suggests continuity should reign.

I'd appreciate someone clearing up my elementary confusions—Thanks in advance!

Addendum (5Jan14). @AlexR. found this geometric interpretation of the fractional integral in Richard Herrmann's book, Fractional Calculus: An Introduction for Physicists,World Scientific, 2011:

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    $\begingroup$ Well, my view of fractional derivatives is from it's kernel (by using Fourier transform), though this has nothing to do with geometric interpretation. For integral derivatives,like $\frac{d^k}{dx^k}$, the Schwartz kernel is $\delta_0^{(k)}$, which measures only on a point, however, when considering the fractional derivative,like $\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}$ on $\mathbb{R}$, the respective kernel is $|x-y|^{-1-\frac{1}{2}}$, which is nonlocal anymore. $\endgroup$
    – Tomas
    Jan 4, 2014 at 2:43
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    $\begingroup$ I haven't found fractional powers of the first derivative $d/dx$, which is indefinite, to be of much use. Fractional powers $\Delta^\alpha$ of the (analyst's) Laplacian $\Delta = -\sum_{i=1}^n \frac{\partial}{\partial x_i^2}$, on the other hand, are much more useful (note that the Laplacian is positive definite, in contrast to the first order operator.) The square root of the Laplacian has a natural geometric interpretation as the Dirichlet-to-Neumann operator for the upper half-plane. $\endgroup$
    – Terry Tao
    Jan 4, 2014 at 2:46
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    $\begingroup$ Also, fractional derivatives depend continuously on the exponent $\alpha$ (at least in the distributional topology), as can be seen on the Fourier side. The non-locality disappears as $\alpha$ approaches a natural number due to denominators such as $\Gamma(-\alpha)$ that appear in the formulae for the kernel away from the origin (which is something like $\frac{1}{\Gamma(-\alpha)} |x-y|^{-1-\alpha}$ in one dimension). $\endgroup$
    – Terry Tao
    Jan 4, 2014 at 2:49
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    $\begingroup$ One obstruction to an obvious geometric interpretation is that the ordinary derivative can be regarded as generalizing to, for example, the exterior derivative, but I'm not aware of a generalization of the half-derivative this broad without extra structure (fractional powers of the Laplacian require a Riemannian metric, for example). $\endgroup$ Jan 4, 2014 at 6:00
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    $\begingroup$ A correction to my previous comment: the first order operator $d/dx$ is indefinite on the real line (the spectral variable $\xi$ can be either positive real or negative real), and so it is not natural to consider fractional powers of this operator (one has to arbitrarily choose a branch cut for $\xi^\alpha$). But if one is working on the half-line instead, then the spectral variable $\xi$ now naturally lives on the upper half-plane (Fourier-Laplace transform) and one now has a canonical interpretation of $\xi^\alpha$. So fractional powers of $d/dx$ are reasonable in half-line settings. $\endgroup$
    – Terry Tao
    Jan 4, 2014 at 16:54

2 Answers 2


A possible mechanical interpretation of the half-derivative can be given in terms of Abel's solution to a classical problem from the calculus of variations (the tautochrone problem).

Let there be a heavy particle which is constrained to slide without friction along the curve $y=y(t)$ in uniform gravity to its lowest point. Then, given a function $T(y)$ that specifies the total time of descent for a given starting height what is an equation of the curve that yields this result?

The principle of conservation of energy implies that the distance $S=S(t)$ travelled by the particle along the curve from the initial height $y_0$ satisfies the equation $$\left(\frac{dS}{dt}\right)^2=2g(y_0-y).$$ This is equivalent to the integral equation $$T(y_0)=\frac{1}{\sqrt{2g}}\int_0^{y_0}\frac{1}{(y_0-y)^{1/2}}\frac{dS}{dy}dy.$$

The r.h.s. of the latter equation is nothing else but the Riemann–Liouville fractional integral of $f=\pi^{1/2}(2g)^{-1/2}dS/dy$, i.e. $$D^{-\alpha}f(x)=\frac{1}{\Gamma(\alpha)}\int_0^x (x-y)^{\alpha-1}f(y)dy$$ of the order $\alpha=1/2$.

The solution to Abel's integral equation $dS/dy$ can be now interpreted (up to a constant factor) as the half-derivative of $T=T(y_0)$.

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    $\begingroup$ And what is the heuristic interpretation of this... interpretation? $\endgroup$
    – Qfwfq
    Jan 4, 2014 at 8:35
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    $\begingroup$ Beautiful!! Thanks for the lucid lesson! $\endgroup$ Jan 4, 2014 at 13:31

A simple perspective for all fractional integro-derivative operators (FID) of this type is that they satisfy the group power sum property (law of exponents)

$$D_x^{\alpha}D_x^{\beta} = D_x^{\alpha+\beta}$$

and that their actions satisfy the dual translations

$$D_x^{\beta} \; H(x) \; \frac{x^{\alpha}}{\alpha!} = H(x) \; \frac{x^{\alpha-\beta}}{(\alpha-\beta)!} = D_x^{-\alpha-1} \; H(x) \; \frac{x^{-\beta-1}}{(-\beta-1)!} ,$$

where $H(x)$ is the Heaviside step function. This is consistent with the rep of this action as the convolutional integral rep of the Euler beta function analytically continued via the Pochhammer contour or the methods of generalized functions/distributions. See below and my response to the MO-Q "What's the matrix of logarithm of derivative operator (lnD)? What is the role of this operator in various math fields?" and the numerous links therein.

The Euler beta integral has a variety of physical, geometric, and probabilistic interpretations. See the MO-Q "Connections to physics, geometry, geometric probability theory of Euler's beta integral (function)".

Sinc function / cardinal series interpolation and a circular Fourier transform:

Let me summarize some old MO and MSE responses and blog notes for easy reference to relate the FIDs here to a Fourier transform on the circle and to hopefully clear up some confusion of this family of FIDs in fractional calculus with pseudo-diff ops/symbols related to Fourier transforms of continuous functions over the whole real line.

As mentioned above one (of several reps) for the action of the FIDs is the Euler beta integral rep

$$D_x^{\beta} \; H(x)\frac{x^{\alpha}}{\alpha!}= H(x)\int_{0}^{x}\frac{z^{\alpha}}{\alpha!} \; \frac{(x-z)^{-\beta-1}}{(-\beta-1)!}\; dz $$

$$= [ \; \int_{0}^{1}\frac{t^{\alpha}}{\alpha!} \; \frac{(1-t)^{-\beta-1}}{(-\beta-1)!} \; dt \;] \; H(x) \; x^{\alpha-\beta} .$$

Focusing on the Euler beta integral,

$$ \int_{0}^{1}\frac{t^{\alpha}}{\alpha!} \; \frac{(1-t)^{-\beta-1}}{(-\beta-1)!} \; dt $$

$$= \int_{0}^{1}\frac{t^{\alpha}}{\alpha!} \; \sum_{n=0} (-1)^n \frac{1}{n!\;(-\beta-n-1)!} \; t^n \; dt$$

$$ =\frac{1}{a! (-\beta-1)!} \sum_{n=0} (-1)^n \binom{-\beta-1}{n} \; \frac{t^{n+\alpha+1}}{n+\alpha+1} \; |_{t=0}^1$$

$$ = \frac{(-\alpha-1)!}{(-\beta-1)!}\; \sum_{n \ge 0} \binom{-\beta-1}{n}\; \frac{\sin(\pi \; (n+\alpha+1))}{\pi(n+\alpha+1)}$$

$$ = \frac{(-\alpha-1)!}{(-\beta-1)!} \; \sum_{n \ge 0} \binom{-\beta-1}{n} \; \binom{0}{n+\alpha+1} = \frac{(-\alpha-1)!}{(-\beta-1)!} \; \binom{-\beta-1}{\alpha-\beta} = \frac{1}{(\alpha-\beta)!} $$

with the last two lines valid for all complex $\alpha$ and $Re(\beta) < 0$.

A quick change of variables, reflecting the desirable property

$$D_x^{\beta} \; H(x) \; \frac{x^{\alpha}}{\alpha!} = D_x^{-\alpha-1} \; H(x) \; \frac{x^{-\beta-1}}{(-\beta-1)!} = H(x) \; \frac{x^{\alpha-\beta}}{(\alpha-\beta)!},$$


$$ \int_{0}^{1}\frac{t^{\alpha}}{\alpha!}\;\frac{(1-t)^{-\beta-1}}{(-\beta-1)!}\; dt = \int_{0}^{1}\frac{(1-t)^{\alpha}}{\alpha!}\;\frac{t^{-\beta-1}}{(-\beta-1)!}\; dt ,$$

with the binomial expansion and subsequent sinc function interpolation for the RHS valid for

all complex $\beta$ and $Re(\alpha) > -1.$

Since the domains of validity of the two different avenues of expansion overlap to give the same result at their intersection, we have via analytic continuation an expression valid for all $\alpha$ and $\beta$, real or complex,

$$D_x^{\beta} \; H(x) \; \frac{x^{\alpha}}{\alpha!} = D_x^{-\alpha-1} \; H(x) \; \frac{x^{-\beta-1}}{(-\beta-1)!} = H(x) \; \frac{x^{\alpha-\beta}}{(\alpha-\beta)!},$$

with the caveat (really an operator interpretation)

$$H(x) \frac{x^{-n-1}}{(-n-1)!} = (-1)^n \delta^{(n)}(x) $$

with the standard Dirac delta interpretation

$$H(x) \int_0^x f(x-t) \; \delta(t) \; dt = H(x) \int_0^x f(t) \; \delta(x-t) \; dt = H(x) f(x).$$

A Cauchy integral rep gives the blow-up from the real line of the Euler beta integral to the complex plane, and, with a bit of regularization,

$$D_x^{\beta} \;H(x)\frac{x^{\alpha}}{\alpha!}= H(x)\frac{1}{2\pi i} \; \oint_{|z-x|=|x|}\frac{z^{\alpha}}{\alpha!} \; \frac{\beta!}{(z-x)^{\beta+1}} \;dz$$

$$ = [\; \frac{1}{2\pi i} \; \oint_{|z-1|=1}\frac{z^{\alpha}}{\alpha!} \; \frac{\beta!}{(z-1)^{\beta+1}} \;dz \;] \; H(x) \; x^{\alpha-\beta}$$

$$ = [\;\frac{1}{2 \pi} \int_{-\pi}^{\pi} (1+e^{i \theta})^{\alpha} \; e^{-i \beta \theta} d\theta \; ] \; \frac{\beta!}{\alpha!} \; H(x) \; x^{\alpha-\beta}$$

$$ = [ \; \sum_{n \geq 0} \binom{\alpha}{n} \; \frac{\sin(\pi \; (n-\beta))}{\pi(n-\beta)} \;] \; \frac{\beta!}{\alpha!} \; H(x) \; x^{\alpha-\beta} = H(x) \; \frac{x^{\alpha-\beta}}{(\alpha-\beta)!}.$$

Here, we have that the FID operation is equivalent to the Fourier transform over a circle (or an average of the Fourier transform of a periodic, in general piecewise continuous function over the real line)

$$\frac{1}{2 \pi} \int_{-\pi}^{\pi} (1+e^{i \theta})^{\alpha} \; e^{-i \beta \theta} d\theta.$$

Added 2/3/21: As noted in other posts, this family of FIDs is also derived by Pincherle through an axiomatic approach, praised by Wiener, and presented in eqn. 4 of "The Role of Salvatore Pincherle in the Development of Fractional Calculus" by Mainardi and Pagnini. It is also consistent with Heaviside's operational calculus.

(I recall seeing eons ago that Ramanujan explored this last integral. Anyone have a ref?)

  • $\begingroup$ This is a reasonable justification for the Hadamard finite part regularization/analytic continuation of throwing out the lower limit of integration, when $t =0$, when it gives infinity. // The analysis can be couched also as convolution theorems for the Laplace or the Mellin transform. $\endgroup$ Feb 10, 2021 at 23:49
  • $\begingroup$ There is a universe of mathematics lying in between the complex differentiations and integrations. -- Heaviside $\endgroup$ Dec 1, 2021 at 8:51

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