Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've been playing around with this equation:


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.

share|cite|improve this question
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
up vote 5 down vote accepted

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 $-i 2\pi i = 2\pi$.

share|cite|improve this answer
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
I know... he just asked for a way of doing it without using parameterizations. – Zarrax Nov 24 '10 at 0:06
Well, I 'fess up to be the one who made him look at formulating his integrals of ratios of exponentials as contour integrals... but this one is yet another nice way of sidestepping parametrizations. – J. M. Nov 24 '10 at 0:38
I should have mentioned my motivations...My mathematical background is a disjointed mess, and I'm currently focused on very complicated integrals (a simpler version being the one above). So I'm excited to learn as much as I can as fast as I can, and this seems to have potential. The power series expansions I've been working with are turning into a mess, and I was hoping for a quick and easy fix. As for J.M., he's been full of suggestions and I've been eager to try to learn about them all. I guess I'm more than a little disorganized, but I'm attempting to get a basic idea of things. – Matt Groff Nov 24 '10 at 0:50
Thinking about more difficult integrals, I'm having trouble picking out terms for the residue theorom. Are there a lot of options besides looking at residues? The reason I'm even integrating is to find the constant term. But again, I'd like to get a general feel for the basics. – Matt Groff Nov 24 '10 at 1:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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