How to do contour integral on a REAL function?

Suppose we are given the problem:


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

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.


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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 $R\to\infty$.