#3**+6 **

To compute: \(\displaystyle\int^{\infty}_{0}\dfrac{\ln^2\left(z\right)dz}{z^4+z^2+1}\).

I think this does come out by Cauchy, but it's quite heavy.

First you need the poles created by the zeros of \(z^4+z^2+1=0\): first put \(Z=z^2\) so \(Z^2+Z+1=0.\) The roots of this equation are the complex cube roots of unity \(\omega_1={1\over2}+i\sqrt{3\over2}\) and \(\omega_2=\overline\omega_1=\omega_1^2\).

Therefore the roots of \(z^4+z^2+1=0\) are the square roots of \(\omega_1,\ \omega_2\). But since \(\omega_2^2=\omega_1\)and \(\omega_1^2=\omega_2\) these four square roots are just \(z=\pm\omega_2\) and \(z=\pm\omega_1\).

You then have to compute the residues at these 4 poles: it helps that \(\omega_1=\exp(2\pi i/3)\) being a cube root of unity has absolute value 1 which simplifies the computation of \(\ln(z)\).

Having got the residues you then do the usual contour integration trick round two circles radius R and r joined by a cut up the negative real axis. This very much follows the example given in Wikipedia for the case where the denominator is just \((1+z^2)^2\) given here:

https://en.wikipedia.org/wiki/Contour_integration#Example_5_%E2%80%93_the_square_of_the_logarithm

although I think they make rather a meal of it; in particular I think you can ignore the \(i\varepsilon\) in the denominators throughout.

Does this help?

modusAug 23, 2018