Suppose we are interested in $$I = \int_0^\infty \frac{x}{1+x^4} dx$$
and evaluate it by integrating $$f(z) = \frac{z}{1+z^4}$$
around a pizze slice contour with the horizontal side $\Gamma_1$ of the slice on the positive real axis and the slanted side $\Gamma_3$ parameterized by $z= \exp(2\pi i/4) t = \exp(\pi i/2) t = it$ with the two connected by a circular arc $\Gamma_2$ parameterized by $z= R \exp(it)$ with $0\le t\le \pi i/2.$ We let $R$ go to infinity.
The integral along $\Gamma_1$ is $I$ in the limit. Furthermore we have in the limit $$\int_{\Gamma_3} f(z) dz = - \int_0^\infty \frac{\exp(\pi i/2) t}{1+t^4 \exp(2\pi i)} \exp(\pi i/2) dt \\ = - \exp(\pi i) \int_0^\infty \frac{t}{1+t^4} dt = I.$$
Furthermore $$\int_{\Gamma_2} f(z) dz \rightarrow 0$$ by the ML bound which yields $$\lim_{R\rightarrow \infty} \pi i/2 R \frac{R}{R^4-1} = 0.$$
The four poles are at $$\rho_k = \exp(\pi i/4 + 2\pi i k/4).$$
Considering the one pole $\rho_0$ inside the slice ($\rho_1 = \exp(\pi i/4 + \pi i/2) = \exp(3\pi i/4)$ so it is not inside the contour) we thus obtain
$$2 I = 2\pi i \mathrm{Res}_{z=\rho_0} f(z)$$
or
$$I = \pi i \frac{\rho_0}{4\rho_0^3} = \pi i \frac{\rho_0^2}{4\rho_0^4} = -\frac{1}{4} \pi i \exp(\pi i/2) = \frac{\pi}{4}.$$
Remark. We see that the pizza slice is in fact a quarter slice and the parameterization may use the rotation $it$ by $\pi/2$ throughout. The four poles are centered on the diagonals of the four quadrants.