# Complex integrals and the possibility of avoiding parametric equations

I've been playing around with this equation:

$\displaystyle\int_{-\pi}^\pi{\displaystyle\frac{1-e^{3it}}{1-e^{it}}dt}$

Now it seems to me that we can (possibly) split the integral into four seperate integrals; one for each quadrant of the complex plane. Doing this, we can determine the complex part if we know the real part, and vice versa. So I therefore ponder that it might be possible to simplify the exponential expressions into expressions of variables. In other words, I wonder if we can rewrite the $e^{it}$s as $u$s.

I'm extremely interested in the potential of a variable substitution, but maybe I'm missing the obvious. I can't tell you exactly what I'm looking for, except for information concerning the possibility of this method. What is known? If there is a known way, I'd really enjoy seeing it. It would make many series expansions that I'm thinking about much easier, I'm almost certain. If no methods are known, I'd appreciate it very much if someone could take the time to elaborate on all of the reasons why this method is not known.

I'm very interested in going into detail on this matter, because it seems to decide issues concerning a great amount of potential.

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I'm wondering if we could get a list of potential options to avoid parametrizations. Feel free to construct your own example(s); you don't have to use mine. –  Matt Groff Nov 24 '10 at 1:26
Did you try to simplify the integrand? $1-t^3$ has a zero at $t=1$, hence you can write $1-t^3=(1-t)\cdot(....)$. –  AD. Nov 24 '10 at 8:29
No, but that's definitely an option; thanks! –  Matt Groff Nov 24 '10 at 13:04

I don't really see why you don't want to just divide and then integrate the three terms... but if you insist on not using parameterizations: you can rewrite your integral as $$-i\int_{-\pi}^{\pi} {1 - e^{3it} \over 1 - e^{it}} ie^{it} {1 \over e^{it}}\,dt$$ Then you do a complex variables version of the $u$-substitution you are trying. You let $z = e^{it}$ and $dz = i e^{it} dt$. Your integral becomes $$-i\int_{|z| = 1} {1 - z^3 \over z(1 - z)}\,dz$$ Dividing $(1 - z)$ into $(1 - z^3)$ turns this into $$-i \int_{|z| = 1} ({1 \over z} + 1 + z) \,dz$$ Using the residue theorem this immediately evaluates into $-i 2\pi i = 2\pi$.
Good answer. I don't really think we need the residue theorem for this (though it's always fun to shoot sparrows with cannons), as the original integrand can be directly calculated to be $1 + e^{it} + e^{2it}$. This expression can be integraded by hand. –  Gunnar Þór Magnússon Nov 23 '10 at 23:41