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

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.

  • 2
    $\begingroup$ You can use a keyhole contour. $\endgroup$ – Henricus V. Nov 20 '15 at 0:51

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: $$ I = \frac{2\pi}{3\sqrt{3}} $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.