**Theorem 4.**

$$(-\Delta)^{\frac{1}{2}}\arctan x = \frac{x}{1+x^2}.$$

**Proof.** Clearly, (arctan *x*) = 1 1 + *x*<sup>2</sup> is continuous on *R*, and

$$(\arctan x)' \* \delta\_{\mathbb{N}} = \left(\frac{1}{1 + t^2}, \delta\_{\mathbb{N}}(x - t)\right).$$

uniformly converges to 1/(1 + *x*<sup>2</sup>) on any compact subset of *R*. Therefore,

$$\begin{aligned} (-\Delta)^{\frac{1}{2}} \arctan x &= \lim\_{m \to \infty} -\frac{1}{\pi} \int\_0^\infty \Phi\_m(t) \frac{1}{t} \frac{1}{t + (\mathbf{x} + t)^2} dt \\ &= \frac{4x}{\pi} \lim\_{m \to \infty} \int\_0^\infty \Phi\_m(t) \frac{dt}{(1 + (\mathbf{x} - t)^2)(1 + (\mathbf{x} + t)^2)} \\ &= \frac{4x}{\pi} \int\_0^\infty \frac{dt}{(1 + (\mathbf{x} - t)^2)(1 + (\mathbf{x} + t)^2)} \end{aligned}$$

by noting that the integral:

$$\int\_0^\infty \frac{dt}{(1+(x-t)^2)(1+(x+t)^2)}$$

is well defined for every point *x* ∈ *R*. It remains to show that:

$$\int\_0^\infty \frac{dt}{(1+(x-t)^2)(1+(x+t)^2)} = \frac{\pi}{4+4x^2}$$

for all *x* ∈ *R*. First, we note that:

$$R(z) = \frac{1}{(1 + (\mathbf{x} - z)^2)(1 + (\mathbf{x} + z)^2)}$$

is even with respect to *z* and has two singular points *z*1 = *x* + *i* and *z*2 = *i* − *x* in the upper half-plane.Clearly, we have for *x* = 0 that:

$$\begin{aligned} \operatorname{Res}\{R(z), x+i\} &= \lim\_{z \to x+i} \frac{z-x-i}{(1+(x-z)^2)(1+(x+z)^2)} = \frac{1}{2i\left(1+(2x+i)^2\right)},\\ \operatorname{Res}\{R(z), i-x\} &= \lim\_{z \to i-x} \frac{z+x-i}{(1+(x-z)^2)(1+(x+z)^2)} = \frac{1}{2i\left(1+(2x-i)^2\right)}. \end{aligned}$$

By Cauchy's residue theorem, we get:

$$\begin{aligned} \int\_{-\infty}^{\infty} \frac{dt}{(1 + (\mathbf{x} - t)^2)(1 + (\mathbf{x} + t)^2)} &= \ 2\pi i [\operatorname{Res}\{R(z), \mathbf{x} + i\} + \operatorname{Res}\{R(z), i - \mathbf{x}\} ] \\ &= \ \frac{2\pi i}{2i \left(1 + (2\mathbf{x} + i)^2\right)} + \frac{2\pi i}{2i \left(1 + (2\mathbf{x} - i)^2\right)} \\ &= \ \pi \frac{8\mathbf{x}^2}{\left(1 + (2\mathbf{x} + i)^2\right)\left(1 + (2\mathbf{x} - i)^2\right)} \\ &= \ \frac{\pi}{2 + 2\mathbf{x}^2} \end{aligned}$$

using:

$$
\left(1 + (2\mathbf{x} + i)^2\right)\left(1 + (2\mathbf{x} - i)^2\right) = 16\mathbf{x}^2 + 16\mathbf{x}^4.
$$

This implies that:

$$\int\_0^\infty \frac{dt}{(1+(\mathbf{x}-t)^2)(1+(\mathbf{x}+t)^2)} = \frac{\pi}{4+4\pi^2}$$

for all nonzero *x*. Furthermore, we derive for *x* = 0 that:

$$\int\_0^\infty \frac{dt}{(1+t^2)^2} = \frac{\pi}{4}$$

by using the identity [31]:

$$\int\_{-\infty}^{\infty} \frac{dt}{(1+t^2)^n} = \frac{(2n-3)(2n-5)\cdots 1}{(2n-2)(2n-4)\cdots 2} \pi \,.$$

This completes the proof of Theorem 4.

#### **3. The Implicit Approach to (***−***<sup>Δ</sup>) 12** *u*

It seems infeasible to calculate directly the fractional Laplacian operator of some functions or distributions by Definition 2. For example,

$$\begin{aligned} (-\Delta)^{\frac{1}{2}}e^{\mathbf{x}} &= \lim\_{m \to \infty} \lim\_{n \to \infty} -\frac{1}{\pi} \int\_0^\infty \phi\_m(t) \frac{u\_n'(\mathbf{x}+t) - u\_n'(\mathbf{x}-t)}{t} dt \\ &= \lim\_{m \to \infty} -\frac{1}{\pi} \int\_0^\infty \phi\_m(t) \frac{e^{\mathbf{x}+t} - e^{\mathbf{x}-t}}{t} dt \\ &= -\frac{e^x}{\pi} \lim\_{m \to \infty} \int\_0^\infty \phi\_m(t) \frac{e^t - e^{-t}}{t} dt \end{aligned}$$

where:

$$u\_n(\mathbf{x}) = (e^t, \delta\_n(\mathbf{x} - t))$$

uniformly converges to *e<sup>x</sup>* as it is continuous on *R*. Clearly, the right-hand side of the above integral is divergent as:

$$\lim\_{t \to \infty} \frac{e^t - e^{-t}}{t} = \infty.$$

In this section, we are going to provide another definition for dealing with (−<sup>Δ</sup>) 12 *u*(*x*) efficiently, based on a testing function with compact support. This definition is implicit and only used to define the meaning of:

$$((-\Delta)^{\frac{1}{2}}u(x), \phi(x)),$$

rather than finding an explicit function of *x*. It clearly makes sense in the distribution as we regard (−<sup>Δ</sup>) 12 *u*(*x*) as a functional (not a function) on the Schwartz testing space <sup>D</sup>(*R*). We must point out that this implicit-definition, using a different generalization, is independent of the explicit one provided in Section 2.

**Definition 3.** *Let u* ∈ D(*R*) *and un* = *u* ∗ *δn* = (*u*(*t*), *<sup>δ</sup>n*(*<sup>x</sup>* − *t*)) *for n* = 1, 2, ··· *. We define the half-order Laplacian operator* (−<sup>Δ</sup>) 1 2 *i on* D(*R*) *(adding the index i to distinguish from* (−<sup>Δ</sup>) 12 *) for φ* ∈ D(*R*) *as:*

$$\begin{aligned} \langle ( - \Delta)\_{\text{i}}^{\frac{1}{2}} u(\mathbf{x}), \phi(\mathbf{x}) \rangle &= -\frac{1}{\pi} \lim\_{n \to \infty} \left( P.V. \frac{1}{\mathbf{f}}, \phi(t) u'\_n(\mathbf{x} + t) \right) \\ &= -\frac{1}{\pi} \lim\_{n \to \infty} \int\_0^\infty \frac{u'\_n(\mathbf{x} + t) \phi(t) - u'\_n(\mathbf{x} - t) \phi(-t)}{t} dt \end{aligned} \tag{9}$$

*if it exists.*

> Clearly, the integral:

$$\int\_0^\infty \frac{\mu\_n'(\mathbf{x} + t)\phi(t) - \mu\_n'(\mathbf{x} - t)\phi(-t)}{t} dt$$

is well defined, as *φ*(*t*) has a bounded support and:

$$\lim\_{t \to 0^{+}} \frac{u\_{n}'(\mathbf{x} + t)\phi(t) - u\_{n}'(\mathbf{x} - t)\phi(-t)}{t} = 2u\_{n}''(\mathbf{x})\phi(0) + 2u\_{n}'(\mathbf{x})\phi'(0).$$

> It follows from Definition 3 that:

$$((-\Delta)^{\frac{1}{2}}\_i e^{\mathfrak{x}}, \phi(\mathfrak{x})) = -\frac{e^{\mathfrak{x}}}{\pi} (\text{P.V.} \frac{1}{\mathfrak{x}'}, e^{\mathfrak{x}} \phi(\mathfrak{x})) $$

by noting that:

$$(\text{P.V.} \frac{1}{\mathfrak{x}'} \,\, \mathfrak{e}^x \mathfrak{o}(\mathfrak{x})) = \int\_0^\infty \frac{\mathfrak{e}^x \mathfrak{o}(\mathfrak{x}) - \mathfrak{e}^{-x} \mathfrak{o}(-\mathfrak{x})}{\mathfrak{x}} d\mathfrak{x}'$$

is well defined and is a number for each *φ*, since *e<sup>x</sup>φ*(*x*) is also a testing function in the Schwartz space. Furthermore,

$$((-\Delta)^{\frac{1}{2}}\_{i}\cosh x,\,\phi(x)) = -\frac{e^{x}}{2\pi}(\text{P.V.}\frac{1}{\text{x}}\,\,\,\epsilon^{x}\phi(x)) + \frac{e^{-x}}{2\pi}(\text{P.V.}\frac{1}{\text{x}}\,\,\,\,\epsilon^{-x}\phi(x)).$$

**Theorem 5.** *Let u*(*x*) *be an infinitely-differentiable function satisfying:*

$$u(\mathbf{x}) = \sum\_{k=0}^{\infty} \frac{u^{(k)}(0)}{k!} \mathbf{x}^k.$$

*Then,*

$$\rho\left((-\Delta)\_{\rm i}^{\frac{1}{2}}u(\mathbf{x}),\,\phi(\mathbf{x})\right) = -\frac{u'(\mathbf{x})}{\pi}(\text{P.V.}\_{\mathbf{x}'}^{1},\,\phi(\mathbf{x})) - \frac{1}{\pi}\left(\frac{u'(\mathbf{x}+t) - u'(\mathbf{x})}{t},\,\phi(t)\right). \tag{10}$$

**Proof.** Clearly,

$$\begin{aligned} u(\mathbf{x} + t) &= \sum\_{k=0}^{\infty} \frac{u^{(k)}(\mathbf{x})}{k!} t^k, \\ u'(\mathbf{x} + t) &= \sum\_{k=0}^{\infty} \frac{u^{(k+1)}(\mathbf{x})}{k!} t^k. \end{aligned}$$

and:

$$u'\_n(\mathfrak{x}) = (u' \* \delta\_n)(\mathfrak{x}) = (u'(t), \delta\_n(\mathfrak{x} - t))$$

uniformly converges to *u*(*x*) on any compact subset of *R*. By Definition 3,

$$\begin{split} \left( (-\Delta)\_{\frac{t}{t}}^{\frac{1}{2}} u(x), \phi(x) \right) &= \lim\_{n \to \infty} -\frac{1}{\pi} \int\_{0}^{\infty} \frac{u'\_{n}(x+t)\phi(t) - u'\_{n}(x-t)\phi(-t)}{t} dt \\ &= -\frac{1}{\pi} \int\_{0}^{\infty} \frac{u'(x+t)\phi(t) - u'(x-t)\phi(-t)}{t} dt \\ &= -\frac{1}{\pi} \int\_{0}^{\infty} \frac{u^{(k+1)}(x)}{k!} t^{k} \phi(t) - \sum\_{k=0}^{\infty} \frac{u^{(k+1)}(x)}{k!} (-t)^{k} \phi(-t) dt \\ &= -\frac{1}{\pi} \int\_{0}^{\infty} \frac{\phi(t)u'(x) - \phi(-t)u'(x)}{t} dt \\ &- \frac{1}{\pi} \int\_{0}^{\infty} \frac{u^{(k+1)}(x)}{k!} t^{k} \phi(t) - \sum\_{k=1}^{\infty} \frac{u^{(k+1)}(x)}{k!} (-t)^{k} \phi(-t) dt \\ &= -\frac{u'(x)}{\pi} (\text{PV} \frac{1}{\pi}, \phi(x)) - \frac{1}{\pi} \left( \frac{u'(x+t) - u'(x)}{t}, \phi(t) \right) \end{split}$$

since

$$\sum\_{k=1}^{\infty} \frac{u^{(k+1)}(\mathbf{x})}{k!} t^k = u'(\mathbf{x} + t) - u'(\mathbf{x}).$$

> We should note that the term:

$$\left(\frac{\mu'(\mathbf{x} + t) - \mu'(\mathbf{x})}{t}, \phi(t)\right)$$

is well defined for every point *x* ∈ *R*, and it is indeed not Cauchy's principal value as:

$$\lim\_{t \to 0} \frac{\mu'(\mathbf{x} + t) - \mu'(\mathbf{x})}{t} = \mu''(\mathbf{x}).$$

This completes the proof of Theorem 5.

It follows from Theorem 5 that:

$$((-\Delta)^{\frac{1}{2}}\_{i}(a\mathfrak{x}+b),\,\phi(\mathfrak{x})) = -\frac{a}{\pi}(\text{P.V.}\frac{1}{\mathfrak{x}'},\,\phi(\mathfrak{x})) $$

which implies that:

$$(-\Delta)\_i^{\frac{1}{2}}(a\mathbf{x} + b) = -\frac{a}{\pi} \text{P.V.} \frac{1}{\mathbf{x}} \neq (-\Delta)^{\frac{1}{2}}(a\mathbf{x} + b) = 0.$$

Furthermore,

$$((-\Delta)^{\frac{1}{2}}\_{i}\sin x,\,\phi(x)) = -\frac{\cos x}{\pi}\left(\text{P.V.}\frac{\cos t}{t},\,\phi(t)\right) + \frac{\sin x}{\pi}\left(\frac{\sin t}{t},\,\phi(t)\right).$$

**Remark 1.** *To end this section, we must point out that if we replace φ*(*t*) *by φm*(*t*) *used in Section 2 and add the limit, then the two operators* (−<sup>Δ</sup>) 1 2 *iand* (−<sup>Δ</sup>) 12 *are identical for some functions. For instance,*

$$\lim\_{m \to \infty} \left( (-\Delta)\_i^{\frac{1}{2}} \sin x, \,\, \phi\_m(x) \right) = \frac{\sin x}{\pi} \int\_{-\infty}^{\infty} \frac{\sin t}{t} dt = \sin x.$$

$$\left( P.V. \frac{\cos t}{t}, \,\, \phi\_m(t) \right) = 0.$$

*as:*
