Define the Fourier transform
$$
\widehat{u}(y) = \mathcal F(u)(y) = \int_{\mathbb R} e^{-2iπ \,x\cdot y} \,u(x)\,\mathrm d x.
$$
One can then define the fractional Laplacian of order $s > -1$ by
$$
\Lambda^{s} u(x) = \mathcal F^{-1}(|2\pi y|^{s}\,\widehat{u}(y)).
$$
(so that in particular $\Lambda^{2} = -\Delta$, $\Lambda^{0} = \mathrm{Id}_{\mathbb R}$ and $\Lambda^{1} = \Lambda$). With this definition it is easy to see that (if $a>-1$, $b>-1$ and $a+b>-1$)
$$
\Lambda^{a+b} u(x) = \mathcal F^{-1}(|2\pi y|^{a+b}\,\widehat{u}(y)) = \mathcal F^{-1}(|2\pi y|^{a}\mathcal F\mathcal F^{-1}(|2\pi y|^{b}\,\widehat{u}(y)) = \Lambda^{a}\Lambda^{b} u(x).
$$
However, it is also possible to prove$^{1}$ that for $s\in(0,1)$
$$
\Lambda^{s} u(x) = c_s \int_{\mathbb R} \frac{u(x)-u(y)}{|x-y|^{1+s}}\, \mathrm d y
\tag{1}\label{1}
$$
for some constant $c_s = (2\pi)^s \frac{|\omega_{-s}|}{\omega_{d+s}} > 0$ where $\omega_a = \frac{2\,\pi^{a/2}}{\Gamma(a/2)}$. Therefore, defining $u(x) = \theta(t,x)$, in your case you get
$$
\begin{align*}
\int_0^{\infty} \frac{(\Lambda^{\gamma}u)(x)-(\Lambda^{\gamma}u)(0)}{x^{1+\delta}}\,\mathrm dx &= \frac{1}{2}\int_{\mathbb R} \frac{(\Lambda^{\gamma}u)(y)-(\Lambda^{\gamma}u)(0)}{|y|^{1+\delta}}\,\mathrm dy
\\
&= -\frac{1}{2\,c_\delta}\,\Lambda^{\delta}(\Lambda^{\gamma}u)(0)
\\
&= -\frac{1}{2\,c_\delta}\,\Lambda^{\gamma+\delta}u(0)
\\
&= \frac{c_{\delta+\gamma}}{2\,c_\delta}\,\int_{\mathbb R} \frac{u(y)-u(0)}{|y|^{1+\delta}}\,\mathrm dy
\\
&= \frac{c_{\delta+\gamma}}{c_\delta}\,\int_{0}^\infty \frac{u(y)-u(0)}{|y|^{1+\delta}}\,\mathrm dy
\end{align*}
$$
So your equality is almost right, except that there is a constant $\frac{c_{\delta+\gamma}}{c_\delta} = (2\pi)^\gamma \,\frac{\omega_{d+\delta}\,\omega_{-\delta-\gamma}}{\omega_{d+\delta+\gamma}\,\omega_{-\delta}}$ in front of the right-hand side.
$~$
$^1$ A way to understand and prove the integral formula for the fractional Laplacian and the constant in it is as follows. One can prove by a simple scaling argument (see e.g. here) that
$$
\mathcal F\!\left(\frac{1}{\omega_c\,|x|^{c}}\right) = \frac{1}{\omega_{1-c}\, |x|^{1-c}}
$$
for $c∈ (0,1)$. When $c = -s\in(-1,0)$, then the formula is still quite the same but since $1/|x|^{1+s}$ is not locally integrable, it has to be replaced by the distribution $\mathrm P(1/|x|^{1+s})$ defined by
$$
\int_{\mathbb R} \mathrm P\!\left(\frac{1}{|x|^{1+s}}\right)\varphi(x)\,\mathrm d x := \int_{\mathbb R} \frac{\varphi(x)-\varphi(0)}{|x|^{1+s}}\,\mathrm d x.
$$
hence you get
$$
\mathcal F\!\left(\frac{1}{\omega_{-s}\,|x|^{-s}}\right) = \frac{1}{\omega_{1+s}}\mathrm P\!\left(\frac{1}{|x|^{1+s}}\right)
$$
which can be written
$$
\mathcal F\!\left(|x|^s\right) = \frac{\omega_{-s}}{\omega_{1+s}}\,\mathrm P\!\left(\frac{1}{|x|^{1+s}}\right)
$$
Now, since the Fourier transform of a product is the convolution of the Fourier transforms, the fractional Laplacian is
$$
\begin{align*}
\Lambda^{s} u(x) &= \mathcal F^{-1}(|2\pi y|^{s}\,\widehat{u}(y)) = (2\pi)^s \, \mathcal F_y(|y|^s) * u
\\
&= \frac{(2\pi)^s\,\omega_{-s}}{\omega_{1+s}}\,\mathrm P\!\left(\frac{1}{|y|^{1+s}}\right) * u
\\
&= \frac{(2\pi)^s\,\omega_{-s}}{\omega_{1+s}}\,\int_{\mathbb R}\mathrm P\!\left(\frac{1}{|y|^{1+s}}\right) u(x-y)\,\mathrm d y
\\
&= \frac{(2\pi)^s\,\omega_{-s}}{\omega_{1+s}}\,\int_{\mathbb R}\frac{u(x-y)-u(x-0)}{|y|^{1+s}} \,\mathrm d y
\\
&= \frac{(2\pi)^s\,\omega_{-s}}{\omega_{1+s}}\,\int_{\mathbb R}\frac{u(z)-u(x)}{|x-z|^{1+s}} \,\mathrm d y
\end{align*}
$$
where the last line just follows from the change of variable $y=x-z$. Since $\omega_{-s} = -|\omega_{-s}|$
$$
\begin{align*}
\Lambda^{s} u(x) &= \frac{(2\pi)^s\,|\omega_{-s}|}{\omega_{1+s}}\,\int_{\mathbb R}\frac{u(x)-u(z)}{|x-z|^{1+s}} \,\mathrm d y
\end{align*}
$$
which is Formula $\eqref{1}$.
