For the contour you describe in your text, you have to indent about the poles at $z=0$ and $z=1$. In that case, the contour integral
$$\oint_C dz \frac{e^{i a z}}{e^{2 \pi z}-1}$$
is split into $6$ segments:
$$\int_{\epsilon}^R dx \frac{e^{i a x}}{e^{2 \pi x}-1} + i \int_{\epsilon}^{1-\epsilon} dy \frac{e^{i a R} e^{-a y}}{e^{2 \pi R} e^{i 2 \pi y}-1} + \int_R^{\epsilon} dx \frac{e^{-a} e^{i a x}}{e^{2 \pi x}-1} \\+ i \int_{1-\epsilon}^{\epsilon} dy \frac{ e^{-a y}}{e^{i 2 \pi y}-1} + i \epsilon \int_{\pi/2}^0 d\phi \:e^{i \phi} \frac{e^{i a \epsilon e^{i \phi}}}{e^{2 \pi \epsilon e^{i \phi}}-1}+ i \epsilon \int_{2\pi}^{3 \pi/2} d\phi\: e^{-a} e^{i \phi} \frac{e^{i a \epsilon e^{i \phi}}}{e^{2 \pi \epsilon e^{i \phi}}-1}$$
The first integral is on the real axis, away from the indent at the origin. The second integral is along the right vertical segment. The third is on the horizontal upper segment. The fourth is on the left vertical segment. The fifth is around the lower indent (about the origin), and the sixth is around the upper indent, about $z=i$.
We will be interested in the limits as $R \rightarrow \infty$ and $\epsilon \rightarrow 0$. The first and third integrals combine to form, in this limit,
$$(1-e^{-a}) \int_0^{\infty} dx \frac{e^{i a x}}{e^{2 \pi x}-1}$$
The fifth and sixth integrals combine to form, as $\epsilon \rightarrow 0$:
$$\frac{i \epsilon}{2 \pi \epsilon} \left ( -\frac{\pi}{2}\right) + e^{-a} \frac{i \epsilon}{2 \pi \epsilon} \left ( -\frac{\pi}{2}\right) = -\frac{i}{4} (1+e^{-a}) $$
The second integral vanishes as $R \rightarrow \infty$. The fourth integral, however, does not, and must be evaluated, at least partially. We rewrite it, as $\epsilon \rightarrow 0$:
$$-\frac{1}{2} \int_0^1 dy \frac{e^{-a y} e^{- i \pi y}}{\sin{\pi y}} = -\frac{1}{2} PV\int_0^1 dy \: e^{-a y} \cot{\pi y} + \frac{i}{2} \frac{1-e^{-a}}{a}$$
where
$$PV\int_0^1 dy \: e^{-a y} \cot{\pi y} = \lim_{\epsilon \to 0} \int_{\epsilon}^{1-\epsilon} dy \: e^{-a y} \cot{\pi y}$$
is the Cauchy principal value of that integral. By Cauchy's theorem, the contour integral is zero because there are no poles within the contour. Thus,
$$(1-e^{-a}) \int_0^{\infty} dx \frac{e^{i a x}}{e^{2 \pi x}-1} -\frac{i}{4} (1+e^{-a}) -\frac{1}{2} PV \int_0^1 dy \: e^{-a y} \cot{\pi y} + \frac{i}{2} \frac{1-e^{-a}}{a}=0$$
Now take the imaginary part of the above equation - note that the nasty Cauchy PV integral drops out - and get
$$(1-e^{-a}) \int_0^{\infty} dx \frac{\sin{ a x}}{e^{2 \pi x}-1} = \frac{1}{4} (1+e^{-a})-\frac{1}{2} \frac{1-e^{-a}}{a}$$
or, after a little algebra and simplifying things, we get:
$$\int_0^{\infty} dx \frac{\sin{ a x}}{e^{2 \pi x}-1} = \frac14 \coth{\left (\frac{a}{2}\right )} - \frac{1}{2 a}$$
There are a number of ways to attack this using the residue theorem. One way is to consider the following contour integral:
$$\oint_C dz \frac{\log{z}}{1+z^4} $$
where $C$ is a quarter circle of radius $R$ in the upper-right quadrant, modified by a small quarter-circle of radius $\epsilon$ so as to avoid the branch point at $z=0$. Thus the contour integral is equal to
$$\int_{\epsilon}^R dx \frac{\log{x}}{1+x^4} + i R \int_0^{\pi/2} d\theta \, e^{i \theta} \frac{\log{(R e^{i \theta})}}{1+R^4 e^{i 4 \theta}} \\ + i \int_R^{\epsilon} dy \, \frac{\log{y} + i \pi/2}{1+y^4} + i \epsilon \int_{\pi/2}^0 d\phi \, e^{i \phi} \frac{\log{(\epsilon e^{i \phi})}}{1+\epsilon^4 e^{i 4 \phi}}$$
As $R\to\infty$, the second integral vanishes as $\log{R}/R^3$; as $\epsilon \to 0$, the fourth integra. vanishes as $\epsilon \log{\epsilon}$. In these limits, the contour integral becomes
$$(1-i) \int_0^{\infty} dx \frac{\log{x}}{1+x^4} + \frac{\pi}{2} \int_0^{\infty} \frac{dx}{1+x^4}$$
By the residue theorem, the contour integral is $i 2 \pi$ times the residue of the pole inside $C$, namely $e^{i \pi/4}$, so that we have
$$(1-i) \int_0^{\infty} dx \frac{\log{x}}{1+x^4} + \frac{\pi}{2} \int_0^{\infty} \frac{dx}{1+x^4} = i 2 \pi \frac{i \pi/4}{4 e^{i 3 \pi/4}} = \frac{\pi^2}{8} e^{i \pi/4} $$
By equating imaginary parts, we find that
$$\int_0^{\infty} dx \frac{\log{x}}{1+x^4} = -\frac{\sqrt{2}}{16} \pi^2 $$
We can also find the other integral as well from the real part:
$$\int_0^{\infty} \frac{dx}{1+x^4} = \frac{\sqrt{2}}{4} \pi $$
Best Answer
You have found, correctly, that $$\begin{eqnarray*} I &\equiv& \int_0^\infty\frac{x^{1/3}dx}{1+x^2} \\ &=& \frac{2}{1-e^{2\pi i/3}} \int_{-\infty}^\infty d z\, \frac{z^{5/3}}{1+z^4}. \end{eqnarray*}$$ The second integral has a branch cut at $z=0$ and singularities at the roots of $1+z^4$. Close the contour in the upper half-plane. Let $\gamma$ denote the contour. We will pick up residues at $e^{i\pi/4}$ and $e^{i 3\pi/4}$.
We deform the contour slightly near $z=0$ to avoid the cut. Letting $z=\epsilon e^{i\theta}$ we see that the contribution to the integral here goes like $\epsilon^{1+5/3} = \epsilon^{8/3}$ and so vanishes in the limit.
Letting $z = R e^{i\theta}$, we find the integral over the semicircle at infinity goes like $R^{1+5/3-4} = 1/R^{4/3}$. Thus, the contribution from this part of the contour is also zero. Therefore, $\int_{-\infty}^\infty d z\, z^{5/3}/(1+z^4) = \int_\gamma d z\, z^{5/3}/(1+z^4)$.
We find $$\begin{eqnarray*} I &=& \frac{2}{1-e^{2\pi i/3}} \int_\gamma d z\, \frac{z^{5/3}}{1+z^4} \\ &=& \frac{2}{1-e^{2\pi i/3}} 2\pi i \sum\mathrm{Res} \, \frac{z^{5/3}}{1+z^4} \\ &=& \frac{\pi}{\sqrt{3}} \end{eqnarray*}$$ where the residues are to be taken from the upper half-plane.
The residue of $f(z)$ at $z=z_0$ is just the coefficient of $(z-z_0)^{-1}$ in the Laurent series and, of course, $z^{5/3}/(1+z^4)$ has a Laurent series about the zeros of $1+z^4$. (It does not, however, have a Laurent series about $z=0$.)
A simpler solution
Recognize that the integral $I$ is already in the form $$\int_0^\infty d t\, t^\beta f(t^2).$$ But $$\begin{eqnarray*} \int_0^\infty d t\, t^\beta f(t^2) &=& \frac{1}{1+e^{\beta \pi i}} \int_{-\infty}^\infty d t\, t^\beta f(t^2) \\ &=& \frac{1}{1+e^{\beta \pi i}} \int_\gamma d t\, t^\beta f(t^2) \\ &=& \frac{1}{1+e^{\beta \pi i}} 2\pi i \sum \mathrm{Res} \, t^\beta f(t^2) \end{eqnarray*}$$ assuming the contributions from the contour around the branch cut at $z=0$ and the contour at infinity vanish. (Again, we have closed the contour in the upper half-plane and pick up only the residues residing there.) It is also important that the singularities of $f$ do not lie on the real line.
The contribution near $z=0$ goes like $\epsilon^{1+1/3} = \epsilon^{4/3}$. On the semicircle at infinity the integral goes like $R^{1+1/3-2} = R^{-2/3}$. Thus, $\int_{-\infty}^\infty d t\, t^{1/3}/(1+t^2) = \int_\gamma d t\, t^{1/3}/(1+t^2)$.
There is one residue, at $t=i=e^{i\pi/2}$, so we have halved our work. We find $$\begin{eqnarray*} I &=& \frac{1}{1+e^{\pi i/3}} 2\pi i \frac{e^{i\pi/6}}{2i} \\ &=& \frac{\pi}{2\cos \pi/6} \\ &=& \frac{\pi}{\sqrt{3}}. \end{eqnarray*}$$