I saw some similar questions in the website, namely here, here and here. What bothers me is that there is no general answer to this question:
When is it possible to transform a real integral to a contour integral? Always?
Or are there some requirements to satisfy, the shape of the function to integrate for example? And then, how do you decide the path to use?
I'm really confused about this. Can you point to any resource or keyword to search which may explain this in detail?
Most often when computing the real integral as a contour integral in the complex plane, one invokes Jordan's lemma.
$$C_R = \{R e^{i \theta} \mid \theta \in [0, \pi]\}$$ If the function $f$ is of the form $$f(z) = e^{i a z} g(z) , \quad z \in C_R $$ then Jordan's lemma states the following upper bound for the contour integral:
$$\left| \int_{C_R} f(z) \, dz \right| \le \frac{\pi}{a} M_R \quad \text{where} \quad M_R := \max_{\theta \in [0,\pi]} \left| g \left(R e^{i \theta}\right) \right| .$$
If $f$ is continuous on the semicircular contour $C_R$ for all large R and
$$\lim_{R \to \infty} M_R = 0$$ then by Jordan's lemma $$\lim_{R \to \infty} \oint_{C_R} f(z)\, dz = 0$$
Let $C=C_1\cup C_R$
where $C_1$ is the real line, and $C_R$ the half-circle.
We need
$$\oint_{C_R}f=0$$ for $$\oint_{C}f=\oint_{C_1}f=\int_{\mathbb{R}}f$$
to hold, resulting in
$$\int_{\mathbb{R}} f=2\pi i\sum\limits_{k=1}^nRes(f,z_k)$$
by residue theorem.
This is provided by Jordan's lemma in the limit as $R\to\infty$, assuming $\lim_{R \to \infty} M_R = 0$.
You can always make contour integrals, but the real question is whether or not you can calculate anything out of it. Often times, we follow the following path (pun intended):
$$P.V.\int_{\mathbb R}f(z)dz=\lim_{R\to\infty}\oint_Cf(z)dz-\int_{C_R}f(z)dz$$
where both integrals are taken counter clockwise, $C_R$ is the semi circle from $R$ to $-R$, and $C$ is the semi circle from $R$ to $-R$ plus a straight line from $-R$ to $R$, making $C$ a closed contour integral so that we may apply Cauchy's residue theorem.
Now, the goal is to have the scenario where
$$\lim_{R\to\infty}\int_{C_R}f(z)dz=0$$
And we can do so as follows:
$$\left|\int_{C_R}f(z)dz\right|\le\int_{C_R}|g(z)|dz\le\int_{C_R}\max_{t\in C_R}|g(t)|dz=\pi R\max_{t\in C_R}|g(t)|$$
which is simply saying that on the first step, we remove any possible chances for cancellations to occur. To account for the direction of the integral, we let $f(z)=e^{iz}g(z)$. Then, noting that the second integral is purely real, we can imagine that it must be less than or equal to it's maximum value integrated along the path. Lastly, we realize that this simplifies down to the arc length multiplied by the maximum value.
Often times, it is easy to see that
$$\lim_{R\to\infty}\pi R\max_{t\in C_R}|g(t)|=0$$
which squeezes $\int_{C_R}f(z)dz\to0$ as $R\to\infty$.
Lastly, according to the residue theorem,
$$\oint_Cf(z)dz=2\pi i\sum_{k=1}^n\operatorname{Res}(f,z_k)$$
where we evaluate over the singularities of $f$ where $\Im(z)\ge0$. This gives us the principal value for our said integral.
Note that this does not always make much sense, since we could have
$$P.V.\int_{-\infty}^\infty\sin(x)dx=0$$
when the integral doesn't exist. So do be cautious of when to use.
