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Suppose we are given the problem:

Evaluate:

$$\int_{0}^{\infty} \frac{1}{x^6 + 1} dx$$

Where $x$ is a real variable. A real variable function (no complex variables).

I was reading Schaum's outline to this solution and it defines:

$f(z) = \displaystyle \frac{1}{z^6 +1}$

And it says consider a closed contour $C$

Consisting of the line from $-R$ to $R$ and the semi circle $\Gamma$ traversed in positive, counter-clockwise sense.

Then it finds the poles etc..

A few questions arise:

(1:) What does it mean to integrate along a contour $C$?

(2:) It defined $f(z)$ but what is $z$ equal to?

From the very first chapter, it defined: $z = x + iy$ and $w = u + iv$ where $w = f(z)$

(3:) It finds $z^6 + 1 =0$ at the $z$ values:

$Z: \{e^{\pi i/6},e^{3\pi i/6}, e^{5\pi i/6}, e^{7\pi i/6}, e^{9\pi i/6}, e^{11\pi i/6}\} $ Then it says only $e^{\pi i/6},e^{3\pi i/6}, e^{5\pi i/6}$ lie within the Contour.

How do you know which lie in the contour, when $-R, R$ has no limit yet?

Thank you, I hope this wasnt overwhelming, any answer is appreciated.

Thanks!

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To answer your questions

(1) Integrating along a contour means integrating along a parametrized path, i.e. given a parametrization $\gamma: [0,1]\to \Bbb C$ which has the property that $\gamma([0,1])=C$, the contour, integrating along the path means to compute

$$\int_C f=\int_0^1(f\circ\gamma)(t)\gamma'(t)\,dt$$

(2) $z$ is the argument of the function, it's not equal to anything specific.

(3) You don't need a limit, the contour is a curve in the complex plane, it has an inside and an outside, they are just noting that only things in the upper half-plane--i.e. poles with positive imaginary part--are inside the contour (since the inside is in the upper half-plane). They are skipping a bit where you figure out the dependence on $R$, because it won't matter, eventually every point is inside the contour for some large $R$. This is easily seen because

$$\Bbb C=\bigcup_{n=1}^\infty B_n(0)$$

is the union of balls of radius $n$ centered around $0$, of which the curves where $R=n$ are the boundary. So every point in the upper half-plane is inside the curve for a large enough $R$. The idea is that eventually you will take a limit, so it doesn't matter if you start with some large $R$ where you're sure you have all the upper half-plane zeroes, since the eventual limit will have $R\to\infty$.

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  • $\begingroup$ you are an amazing person. Thank you for answering. Can I ask a question? For (2) the book points that $z = x + iy$ explicitly in the earlier chapters. $z= x + iy$ is a requirement isnt it? Otherwise we can extend $R \to \infty$ along the real $x$ axis.?? For (1) and integral is area under a curve isnt it? Then what is the concept of integration along a path? I hope I didnt overwhelm you, and I hope you can answer these question. Thanks a lot Adam! $\endgroup$ – Amad27 Dec 22 '14 at 10:45
  • $\begingroup$ @Amad27 (2) They are changing letters from $x$ to $z$ to emphasize that you intend to plug non-real numbers into $f$. Yes, a typical $z$ will have the form $x+iy$, this is fine, don't worry about it too much. (1) No, integrals are only areas under curves when you're integrating functions on intervals, you shouldn't try too hard to think geometrically in this setting, there's no simple geometric way to interpret integrals in the complex plane. If you're further interested in the concept of integration abstractly, try checking wikipedia. :-) $\endgroup$ – Adam Hughes Dec 22 '14 at 10:50
  • $\begingroup$ So I want to get some final things in order before I do other integrals. Please say if you agree or disagree: (2) $z = x + iy$ is the form they are using. How do you know where the other solutions (the rejected ones) of $z^6 + 1 = 0$ lie? I know they lie below the x-axis because they were rejected but how do you know? $\endgroup$ – Amad27 Dec 22 '14 at 10:58
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    $\begingroup$ @AdamHughes: this is a very complete answer (+1). I might add in (2) that $z$ is a point in $\mathbb{C}$. Often it is used to represent a point on the parametrized curve $\gamma(t)$. In (3) usually a limit is taken since it is often the case that the integral goes to $0$ along the part of the curve where $|z|\to\infty$. $\endgroup$ – robjohn Dec 22 '14 at 11:29
  • $\begingroup$ @Amad27 you can solve using the fact that $e^{i\theta}=\cos\theta+i\sin\theta$ and the fact that $\sin\theta<0$ when $\pi<\theta<2\pi$. I agree with the $z=x+iy$ thing as well. $\endgroup$ – Adam Hughes Dec 22 '14 at 19:44

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