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Say I have this integral $\int_{-\infty}^\infty$ $\frac{x^2}{x^6+1}$ dx . Now I know that it has six poles according to this denominator which are the six roots for -1. The question is after I split the denominator up into the 6 brackets which are the poles, is there any simpler way to evaluate the sum of the residues ? So that residue theorem can be used. Does seem like a pain to handle the 5 remaining brackets for each pole that I want to find the residue for.
Duplicate edit: The other question is about finding the poles only. That I already know how to do. I don't see how that is even relevant to my question.

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    $\begingroup$ Do you have to calculate the integral using complex analysis? Writing the function as $$\frac{1}{3}\frac{3x^2}{1+(x^3)^2}$$ it is clear that $\frac{1}{3}\arctan(x^3)$ is a primitive function. $\endgroup$
    – mickep
    Nov 1 '15 at 14:26
  • $\begingroup$ @Surb: This doesn't look like a duplicate to me. The other question asks to find the poles: this question states the poles. The other question says to use a contour integral, while this one asks if there is an easier way. $\endgroup$ Nov 2 '15 at 0:46
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As shown in many many places throughout this site, one can limit the number of residues to evaluate by using the symmetry of the integrand and making a wise choice of contour that exploits this symmetry. In this case, consider

$$\oint_C dz \frac{z^2}{z^6+1} $$

where $C$ is a wedge-shaped contour in the upper-half plane of radius $R$ and of angle $\pi/3$. The reason why the wedge has an angle of $\pi/3$ will become apparent as we carry out the calculation.

The contour integral is equal to

$$\int_0^R dx \frac{x^2}{1+x^6} + i R \int_0^{\pi/3} d\theta \, e^{i \theta} \frac{R^2 e^{i 2 \theta}}{1+R^6 e^{i 6 \theta}}+ e^{i \pi/3} \int_R^0 dt \, \frac{(e^{i \pi/3})^2 t^2}{1+(e^{i \pi/3})^6 t^6}$$

As $R \to \infty$, the magnitude of the second integral is bounded by

$$\left | i R \int_0^{\pi/3} d\theta \, e^{i \theta} \frac{R^2 e^{i 2 \theta}}{1+R^6 e^{i 6 \theta}} \right | \le \frac{\pi}{3} \frac{R^3}{R^6-1}$$

which clearly vanishes in this limit.

In this limit, the first integral becomes what we are looking for and the third integral becomes a multiple of what we are looking for. The sum of these two is equal to the contour integral which is

$$2 \int_0^{\infty} dx \frac{x^2}{1+x^6} = \int_{\infty}^{\infty} dx \frac{x^2}{1+x^6} $$

By the residue theorem, this is equal to $i 2 \pi$ times the sum of the residues of the poles inside $C$. Note that we have the advantage of having only one pole inside $C$ at $z=e^{i \pi/6}$ instead of three had we carried out the computation with a semicircle. Thus,

$$\int_{\infty}^{\infty} dx \frac{x^2}{1+x^6} = i 2 \pi \frac{(e^{i \pi/6})^2}{6 (e^{i \pi/6})^5}= \frac{\pi}{3}$$

Note that the wedge angle of $\pi/3$ made the integrands of the first and third integrals essentially the same within a constant factor.

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I think you want to use the formula to calculate residue for a fractional function $f(z)$ defined as $$f(z) = \frac{g(z)}{h(z)}.$$ The residue at a simple pole $z_0$ is given by $$Res(f(z),z_0) = \frac{g(z_0)}{h'(z_0)}.$$ Essentially it is the same formula, as mentioned by @Dustin, but with an application of L'Hopital rule. You can follow this to get better idea of how it works.

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The function $f(z) = \frac {z^2}{z^6+1}$ has six poles of order 1, which means that you can compute the residue for each pole p as follows:

$$Res (f,p) = \lim_{z\to p}f(z)(z-p) $$

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  • $\begingroup$ what if there was an additional $x^2$ on the numerator though? How would that affect the formula? $\endgroup$
    – jaybeedee
    Nov 1 '15 at 14:43
  • $\begingroup$ Sorry about that, misread! Will edit in a moment, but it doesn't make a difference, since the roots of $x^2$ do not coincide with those of $x^6-1$. $\endgroup$ Nov 1 '15 at 14:46

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