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# Help?

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Help with the following integral... I tried Cauchy's residue theorem but it didn't work(maybe I used it wrongly).

I know the answer, but I can't figure out the steps:

Answer to this integral is $$\dfrac{\pi^3}{9\sqrt3}$$.

Somebody help?

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

P.S.: The good thing is that I still know how to use $$\LaTeX$$ :)

Aug 23, 2018

### 3+0 Answers

#2
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Hi Max: Here is an online "Integral Calculator" that appears to be able to solve your Integral with many steps as well: https://www.integral-calculator.com/

Good luck.

Aug 23, 2018
#3
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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?

Aug 23, 2018
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Yes it helps. Thank you :D

MaxWong  Aug 27, 2018