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I would like to integrate $\int_{-a}^a {u\over 1+u+u^2} du$ as $a\to \infty$.

So I thought I might use the residue theorem. In the complex plane, the singularities occur at $z=e^{\pm i2\pi\over 3}$. So if we close the contour in the upper half plane we enclose $z=e^{i2\pi\over 3}$. This is a covert simple pole, so res$\left({z\over 1+z+z^2};e^{i2\pi\over 3}\right)={z\over 2z+1}|_{z=e^{i2\pi\over 3}}={1\over 2}+{i\over 2\sqrt3}$. So by the residue theorem, the integral is $2\pi i \left({1\over 2}+{i\over 2\sqrt3}\right)=\pi(i-{1\over\sqrt 3})$.

But this cannot be the answer since it contains $i$. Where have I gone wrong? Thanks.

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Cam answers your question below. But to evaluate the integral, I'd observe that you have an integrand $\frac{u+1/2}{1+u+u^2}-\frac{1/2}{1+u+u^2}$. The first term integrates to $\frac{1}{2}\ln(1+u+u^2)$ and the second can be integrated by completing the square and making a linear substitution. – alex.jordan Mar 6 '12 at 0:45
Thanks, @alex.jordan! – Herbert G Mar 6 '12 at 0:49
up vote 7 down vote accepted

Apologies in advance for the informal answer: I'm not in a convenient position for typesetting lots of integrals.

Your mistake is in assuming that the integral of a rational function automatically equals the sum of the residues. But remember how this argument goes: If you add on to the interval $[-a,a]$ the semi-circle of radius $a$ centered at the origin, then the integral over this whole contour is given by the sum of the residues. As $a\rightarrow\infty$, your region encloses all the poles, and so the residue theorem gives you the integral you're looking for provided that the integral over the semi-circular arc goes to zero. This can be shown to happen in many cases, and two famous integrands in particular: 1) a trig integral of the form rational times sine or cosine, where the denominator has degree at least one more than the numerator; and 2) a rational function with denominator of degree at least two more than the numerator. The point for these is that the function itself shrinks to zero faster than the arclength of the contour goes to infinity.

In your case, this doesn't hold: Your denominator has degree only one larger than the numerator, so as $u\rightarrow\infty$, your function shrinks at approximately the same rate as the arclength goes to infinity (they grow roughly like $\frac{1}{u}$ and $u$, respectively). So there's some constant term contributing to the number you get from the residue theorem coming from the integral around the semi-circular arc, which explains the imaginary part of your answer. I'd recommend integration by partial fractions instead.

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Thank you, Cam! – Herbert G Mar 6 '12 at 0:49
@HerbertG: Sure thing! – Cam McLeman Mar 6 '12 at 0:53

You can use the usual integration:

$$\eqalign{ & \int\limits_{ - a}^a {\frac{u}{{{u^2} + u + 1}}du} = \frac{1}{2}\int\limits_{ - a}^a {\frac{{2u}}{{{u^2} + u + 1}}du} \cr & = \frac{1}{2}\left\{ {\int\limits_{ - a}^a {\frac{{2u + 1}}{{{u^2} + u + 1}}du} - \int\limits_{ - a}^a {\frac{{du}}{{{u^2} + u + 1}}} } \right\} \cr & = \frac{1}{2}\left\{ {\int\limits_{ - a}^a {\frac{{2u + 1}}{{{u^2} + u + 1}}du} - \int\limits_{ - a}^a {\frac{{du}}{{{{\left( {u + \frac{1}{2}} \right)}^2} + \frac{3}{4}}}} } \right\} \cr & = \frac{1}{2}\left\{ {\log \left| {{u^2} + u + 1} \right| - \frac{2}{{\sqrt 3 }}\arctan \left( {\frac{2}{{\sqrt 3 }}\left( {x + \frac{1}{2}} \right)} \right)} \right\}_{ - a}^a \cr & = \frac{1}{2}\log \left| {\frac{{{a^2} + a + 1}}{{{a^2} - a + 1}}} \right| - \frac{1}{{\sqrt 3 }}\arctan \left( {\frac{2}{{\sqrt 3 }}\left( {a + \frac{1}{2}} \right)} \right) + \frac{1}{{\sqrt 3 }}\arctan \left( {\frac{2}{{\sqrt 3 }}\left( { - a + \frac{1}{2}} \right)} \right) \cr} $$

This gives as a result $-\dfrac{\pi}{\sqrt 3}$

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(+1) Nice Answer – Amir Alizadeh Dec 29 '12 at 23:10

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