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$$ \int_{0}^{\infty}\frac{1}{1+x^{3}}{\mathrm{d} x} $$

I've been able to solve this integral by the residue theorem using a contour along a 1/3 circle and a line along the x-axis, but apparently it can be calculated by introducing a multivalued function, $\ln(z)$, and integrating along a branch cut. I'm not sure how to do that.

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    $\begingroup$ You can use a keyhole contour. $\endgroup$ – Henricus V. Nov 20 '15 at 0:51
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to use the method suggested by Henry

make the substitution $z=x^3$, giving: $$ I = \frac13 \int_0^{\infty}\frac{z^{-\frac23}}{1+z} dz $$ use a contour which goes towards infinity from a point near the origin along the real axis, followed by a large complete nticlockwise circle then a return along the real axis, stopping just before the origin. the closed contour is completed by a small clockwise circle around the origin.

the integrand is $O(|z|^{-\frac53})$ so becomes small at large radius.

on a small circle round the origin of radius $\rho$ the integrand is $O(\rho^{-\frac23})$ whilst the path length is $2\pi\rho$, so in the limit this reduces to zero

after the move round the large circle the integrand is multiplied by a factor $(e^{2\pi i})^{-\frac23}=\omega$, the cube root of unity with positive imaginary part. the direction of the path of integration is also reversed, so that we have: $$ (1-\omega)I = 2\pi i r $$ where $r$ is the residue at the pole at $-1$, which takes the value $\frac13(e^{i\pi})^{-\frac23}=\frac{\omega^2}3$

thus we have, after multiplying the equation by $\omega$ $$ (\omega-\omega^2)I = \frac{2\pi i}3 $$ giving: $$ I = \frac{2\pi}{3\sqrt{3}} $$

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