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For this problem $\int^\infty_0 dx/{(x^3+1)}$, the answer divides the contour into three sections, which I found is really hard to understand. The question I have is why do we need to do that? Is there any simpler way to solve this problem? Like if we could just find the poles for it and we can then simply apply residue theorem to it? Thanks.

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    $\begingroup$ math.stackexchange.com/questions/367772/… $\endgroup$ – tattwamasi amrutam Nov 9 '16 at 0:44
  • $\begingroup$ The poles for $1/(x^3+1)$ are easy to find. The hard part is to "simply apply" the residue theorem, given that this is an integral from $0$ to $\infty$. $\endgroup$ – Robert Israel Nov 9 '16 at 0:48
  • $\begingroup$ Aren't them the simple poles? And if that so, we can use the formula to calculate the residue? $\endgroup$ – J.doe Nov 9 '16 at 1:26
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The problem is that you are looking for a simple formula and one doesn't apply here. Evaluating integrals via the residue theorem requires two things: an integrand and a contour. In this case, we want a contour that produces the integral in question. The contour is not a semicircle, which is what I assume is what you seem to use all the time. The reason for this is that the integrand is not symmetric, i.e., odd or even.

There are at least two well-known ways to work in this way. One is, as you say, to define a three-piece contour. The contour here is a wedge of angle $\alpha$ and radius $R$. What we do is define

$$\oint_C \frac{dz}{1+z^3} $$

where $C$ is the wedge. Then the integral is equal to

$$\int_0^R \frac{dx}{1+x^3} + i R \int_0^{\alpha} d\theta \, \frac{e^{i \theta}}{1+R^3 e^{i 3 \theta}} + e^{i \alpha} \int_R^0 \frac{dt}{1+e^{i 3 \alpha} t^3}$$

OK, note that if we are going to recover our original integral without doing any others, it should be clear that we want $e^{i 3 \alpha} = 1$, or $\alpha = 2 \pi/3$.

Note that the second integral has a magnitude that is bounded by $(2 \pi/3) R/(R^3-1)$, so that the integral vanishes as $R \to \infty$.

Now, by the residue theorem, the contour integral is $i 2 \pi$ times the residue of the poles of the integrand inside the contour. In this case, the only pole is at $z=e^{i \pi/3}$. Thus,

$$\left ( 1-e^{i 2 \pi/3} \right ) \int_0^{\infty} \frac{dx}{1+x^3} = \frac{i 2 \pi}{3 e^{i 2 \pi/3}} $$

Thus,

$$\int_0^{\infty} \frac{dx}{1+x^3} = \frac{\pi/3}{\sin{(2 \pi/3)}} = \frac{2 \pi}{3 \sqrt{3}}$$

A second way is more general but also more work. In this case, form the integral

$$\oint_C dz \frac{\log{z}}{1+z^3} $$

where $C$ is a keyhole contour of inner radius $\epsilon$ and outer radius $R$. Here, we use the multivaluedness of the log to return a value of the integral we want. In this case, we find that the contour integral is equal to, in the limit as $\epsilon \to 0$ and $R \to \infty$,

$$\int_0^{\infty} dx \frac{\log{x}}{1+x^3} + \int_{\infty}^0 dx \frac{\log{x}+i 2 \pi}{1+x^3} = -i 2 \pi \int_0^{\infty} \frac{dx}{1+x^3} $$

Now set this equal to $i 2 \pi$ times the sum of the residues of the integrand at the three poles of the integrand. I leave it to the OP to work out that the answer is identical to that I derived above.

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