So we want to compute
$$\oint_C dz \frac{e^{i \pi z^2}}{\sin{\pi z}} $$
where $C = C_1+C_2+C_3+C_4$. Along $C_1$, $z=-1/2 + e^{i \pi/4} t$, $t \in [R,-R]$.
$$\begin{align}\int_{C_1} dz \frac{e^{i \pi z^2}}{\sin{\pi z}} &= -e^{i \pi/4} \int_{-R}^R dt \, \frac{e^{i \pi (-1/2+e^{i \pi/4} t)^2}}{\sin{\pi (-1/2+e^{i \pi/4} t)}}\\ &=i \int_{-R}^R dt \,\frac{e^{-i \pi e^{i \pi/4} t} e^{-\pi t^2}}{\cos{\left (\pi e^{i \pi/4} t\right )}} \end{align}$$
Along $C_3$, $z=1/2 + e^{i \pi/4} t$, $t \in [-R,R]$.
$$\begin{align}\int_{C_3} dz \frac{e^{i \pi z^2}}{\sin{\pi z}} &= e^{i \pi/4} \int_{-R}^R dt \, \frac{e^{i \pi (1/2+e^{i \pi/4} t)^2}}{\sin{\pi (1/2+e^{i \pi/4} t)}}\\ &=i \int_{-R}^R dt \,\frac{e^{i \pi e^{i \pi/4} t} e^{-\pi t^2}}{\cos{\left (\pi e^{i \pi/4} t\right )}} \end{align}$$
Thus,
$$\int_{C_1+C_3} dz \frac{e^{i \pi z^2}}{\sin{\pi z}} = i 2 \int_{-R}^R dt \, e^{-\pi t^2} $$
Along $C_2$, $z=x-i (R/\sqrt{2})$, $x \in \left [-1/2-R/\sqrt{2},1/2-R/\sqrt{2} \right ]$.
$$\begin{align}\int_{C_2}dz \frac{e^{i \pi z^2}}{\sin{\pi z}} &= \int_{-1/2-R/\sqrt{2}}^{1/2-R/\sqrt{2}} dx \, \frac{e^{i \pi (x-i R/\sqrt{2})^2}}{\sin{\left [\pi \left (x-i R/\sqrt{2} \right ) \right ]}}\\ &= \int_{-1/2}^{1/2} dx \frac{e^{i \pi (x-R e^{i \pi/4})^2}}{\sin{\left [\pi \left (x-R e^{i \pi/4} \right ) \right ]}} \end{align}$$
Along $C_4$, $z=x+i (R/\sqrt{2})$, $x \in \left [-1/2+R/\sqrt{2},1/2+R/\sqrt{2} \right ]$.
$$\begin{align}\int_{C_4}dz \frac{e^{i \pi z^2}}{\sin{\pi z}} &= -\int_{-1/2+R/\sqrt{2}}^{1/2+R/\sqrt{2}} dx \, \frac{e^{i \pi (x+i R/\sqrt{2})^2}}{\sin{\left [\pi \left (x+i R/\sqrt{2} \right ) \right ]}}\\ &= -\int_{-1/2}^{1/2} dx \frac{e^{i \pi (x+R e^{i \pi/4})^2}}{\sin{\left [\pi \left (x+R e^{i \pi/4} \right ) \right ]}} \end{align}$$
As $R \to \infty$, the integrals about $C_2$ and $C_4$ each vanish. Thus, in this limit,
$$\oint_C dz \frac{e^{i \pi z^2}}{\sin{\pi z}} = i 2 \int_{-\infty}^{\infty} dt \, e^{-\pi t^2}$$
By the residue theorem, the contour integral is equal to $i 2 \pi$ times the residue of the integrand at the pole at $z=0$, which is $1/\pi$. Thus
$$i 2 \int_{-\infty}^{\infty} dt \, e^{-\pi t^2} = i 2 \pi \frac1{\pi} = i 2$$
and the result follows.
PRIMER:
The complex logarithm function is a multi-valued function that is defined as
$$\log(z)=\log(|z|)+i\arg(z) \tag1$$
where $\arg(z)$ is the multivalued argument of $z$.
The function $f(z)=z^c$, where $c\in \mathbb{C}$, is defined as
$$f(z)=e^{c\log(z)} \tag2$$
Therefore, $f(z)$ is also multivalued when $c$ is not an integer.
BRANCH POINT
If $z_0$ is branch point of the multivalued function $f(z)$ then there is no open neighborhood $N(z_0)$ of $z_0$ on which $f$ is continuous. Loosely speaking, we cannot encircle $z_0$ without encountering a discontinuity.
We can see from $(1)$ that $z_0=0$ is a branch point of $\log(z)$. Let $z_0=e^{i\theta_0}$ be a point on the unit circle. Then $\log(z_0)=i\theta_0$.
We travel on the unit circle from $z_0$ by increasing $\arg(z)$ from $\theta_0$ to $\theta_0+2\pi$. While we have returned to $z_0$, the value of $\log(z)$ has jumped from $i\theta_0$ to $i(\theta_0+2\pi)$. (Note that we have tacitly cut the plane along the ray $\theta=\theta_0$).
Inasmuch as $(2)$ defines $z^c$, then for non-integer $c$, $z^c$ shares the branch point singularity of $\log(z)$.
To see the reason that $z=\infty$ is also a branch point, we let $w=1/z$. Since $\log(w)$ has a branch point at $w=0$, then $\log(z)=\log(1/w)$ has a branch point at $\infty$.
INTEGRATION OVER THE KEYHOLE CONTOUR
From the previous discussion, we know that $z^{1/3}$ has logarithmic branch points at $z=0$ and $z=\infty$. We choose to cut the plane along the positive real axis.
With this choice of branch cut, if we approach a point on the positive real axis along a contour in the first quadrant, then $\arg(z)$ approaches $0$. If we approach a point on the positive real axis along a contour in the fourth quadrant, then $\arg(z)$ approaches $2\pi$.
Referring to the diagram in the OP, we can formally parameterize the green (red) segments as $z=x \pm i\epsilon$, $x\in [\sqrt{\nu^2-\epsilon^2},\sqrt{R^2-\epsilon^2})$, where $\nu>0$ is the radius of the blue-colored circular arc centered at the origin and $R$ is the radius of the gray-colored circular arc. Then, we have
\begin{align}
\lim_{\epsilon\to 0}\int_{\sqrt{\nu^2-\epsilon^2}}^{\sqrt{R^2-\epsilon^2}}\frac{(x+ i\epsilon)^{1/3}}{(x+ i\epsilon+1)^2}\,dx&=\int_\nu^R \frac{x^{1/3}}{(x+1)^2}\,dx\\\\
\lim_{\epsilon\to 0}\int_{\sqrt{\nu^2-\epsilon^2}}^{\sqrt{R^2-\epsilon^2}}\frac{(x- i\epsilon)^{1/3}}{(x- i\epsilon+1)^2}\,dx&=\int_\nu^R \frac{x^{1/3}e^{i2\pi/3}}{(x+1)^2}\,dx
\end{align}
FINISHING IT UP
It can be shown that as $\nu\to 0$ and $R\to \infty$, the contributions from the integrals around the circular arcs vanish. This leaves
$$\begin{align}
(1-e^{i2\pi/3})\int_0^\infty \frac{x^{1/3}}{(x+1)^2}\,dx&=2\pi i \text{Res}\left(\frac{z^{1/3}}{(z+1)^2},z=-1\right)\\\\
&=2\pi i\lim_{z\to -1}\frac13z^{-2/3}\\\\
&=\frac{2\pi i}3 e^{-i2\pi/3}
\end{align}$$
Solving for
$$\int_0^\infty\frac{x^{1/3}}{(x+1)^2}\,dx=\frac{2\pi}{3\sqrt{3}}$$
Best Answer
First, the integral $I=\int_0^\infty \frac{\log(x)}{x^4-16}\,dx$ diverges. However, its Principal Value, $$\text{PV}\left(\int_0^\infty \frac{\log(x)}{x^4-16}\,dx\right)=\lim_{\varepsilon\to 0^+}\left(\int_0^{2-\varepsilon} \frac{\log(x)}{x^4-16}\,dx+\int_{2+\varepsilon}^\infty \frac{\log(x)}{x^4-16}\,dx\right)$$ converges. We shall interpret $I$ in terms of the Principal Value integral.
We simplify the problem by enforcing the substitution $x\mapsto 2x$ and find that
$$\begin{align} I&=\frac18 \text{PV}\left(\int_0^\infty \frac{\log(2x)}{x^4-1}\,dx\right)\\\\ &=\frac18 \text{PV}\left(\int_0^\infty \frac{\log(x)}{x^4-1}\,dx\right)-\frac{\pi\log(2)}{32}\\\\ \end{align}$$
Next, we let $f(z)=\frac{\log^2(z)}{z^4-1}$. Then, we cut the plane along the non-negative real axis and integrate aound the classical keyhole contour with deformations around the pole at $z=1$.
Proceeding, we have for $\varepsilon\in (0,1)$ and $R>1$
$$\begin{align} \oint_{C_{\varepsilon,R}} f(z)\,dz&=\int_0^{1-\varepsilon}\frac{\log^2(x)-(\log(x)+i2\pi)^2}{x^4-1}\,dx+\int_{1+\varepsilon}^R \frac{\log^2(x)-(\log(x)+i2\pi)^2}{x^4-1}\,dx\\\\ &+\underbrace{\int_\pi^0 \frac{\log^2(1+\varepsilon e^{i\phi})}{(1+\varepsilon e^{i\phi})^4-1}\,i\varepsilon e^{i\phi}\,d\phi}_{\to 0\,\,\text{as}\,\varepsilon\to 0^+}+\underbrace{\int_{2\pi}^\pi \frac{\log^2(1+\varepsilon e^{i\phi})}{(1+\varepsilon e^{i\phi})^4-1}\,i\varepsilon e^{i\phi}\,d\phi}_{\to i\pi^3\,\,\text{as}\,\,\varepsilon\to 0^+}\\\\ &+\underbrace{\int_0^{2\pi}\frac{\log^2(Re^{i\phi})}{(Re^{i\phi})^4-1}\,iRe^{i\phi}\,d\phi}_{\to 0\,\,\text{as}\,\,R\to \infty}\\\\ &=2\pi i \left(\text{Res}\left(f(z), z=- 1\right)+\text{Res}\left(f(z), z=i\right)+\text{Res}\left(f(z), z=- i\right)\right) \end{align}$$
where we used the residue theorem to arrive at the last line.
Noting that
$$\begin{align} 2\pi i \left(\text{Res}\left(f(z), z=- 1\right)+\text{Res}\left(f(z), z=i\right)+\text{Res}\left(f(z), z=- i\right)\right)&=i \pi^3/2-\pi^3 \end{align}$$
and letting $\varepsilon\to 0^+$ and $R\to \infty$ reveals
$$\begin{align} -i4\pi \text{PV}\int_0^\infty \frac{\log(x)}{x^4-1}\,dx+4\pi^2 \underbrace{\text{PV}\int_0^\infty \frac{1}{x^4-1}\,dx}_{=-\pi/4}+i\pi^3=i \pi^3/2-\pi^3 \end{align}$$
Finally, putting it all together, we find that
$$I=\frac{\pi^2}{64}-\frac{\pi\log(2)}{32}$$