= Solution
The <residue> at $a$ is the coefficient $c_{-1}$ of $(z-a)^{-1}$ in the <Laurent series>. A <simple pole> means $f(z)=c_{-1}/(z-a)+h(z)$ near $a$, with $c_{-1}\neq0$ and $h$ holomorphic across $a$.
For the specified positively oriented arc, direct parametrization gives $\int_{\gamma_\epsilon}(z-a)^{-1}\,dz=i(\beta-\alpha)$. The remaining integral is bounded by $\epsilon(\beta-\alpha)\sup|h|$, which tends to zero because $h$ is bounded near $a$. Therefore
$$
\boxed{\int_{\gamma_\epsilon}f(z)\,dz\longrightarrow i(\beta-\alpha)\operatorname{res}(f;a).}
$$
The simple-pole hypothesis matters. For $f(z)=(z-a)^{-2}$, with $\alpha=0$, $\beta=\pi$, the <residue> is zero but
$$
\int_{\gamma_\epsilon}\frac{dz}{(z-a)^2}=\frac2\epsilon\longrightarrow\infty.
$$
For the particular function, expanding the exponential gives
$$
1+iz-e^{iz}=\frac{z^2}{2}+\frac{iz^3}{6}+O(z^4),\qquad
f(z)=\frac1{2z}+\frac i6+O(z).
$$
Thus $\boxed{\operatorname{res}(f;0)=1/2}$, and the pole is simple. On the large upper semicircle, $|e^{iz}|\leq1$, giving
$$
\left|\int_{\Gamma_R}f(z)\,dz\right|
\leq\pi R\frac{R+2}{R^3}=\frac{\pi(R+2)}{R^2}\longrightarrow0.
$$
Close the real-axis path from $-R$ to $-\epsilon$ and $\epsilon$ to $R$ by this upper arc and a clockwise upper indentation at zero. The function is holomorphic in the indented region, so the sum of all four integrals is zero. The clockwise indentation tends to $-i\pi/2$. For real $x$,
$$
f(x)=\frac{1-\cos x}{x^3}+i\frac{x-\sin x}{x^3}.
$$
The real part is odd and cancels on the symmetric real pieces; the imaginary part is even. Its integral converges since $(x-\sin x)/x^3\to1/6$ at zero and is $O(x^{-2})$ at infinity. Letting $\epsilon\to0$ and $R\to\infty$ therefore yields $2iI-i\pi/2=0$, where $I$ is the requested real integral. Hence
$$
\boxed{\int_0^\infty\frac{x-\sin x}{x^3}\,dx=\frac\pi4.}
$$
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