In calculating the integral
$$\int_{-\infty}^{\infty} \frac{\sin x}{x}dx$$
by contour integration, we use
$$\int_{-\infty}^{\infty} \frac{\sin x}{x}dx = \int_{-\infty}^{\infty} \frac{\operatorname{Im}(e^{ix})}{x}dx =\operatorname{Im}\left(\int_{-\infty}^{\infty} \frac{e^{ix}}{x}dx \right)$$
but in the process, we have gone from an integral which is well-defined with no real singularities to one with a real singularity which in fact is just undefined as an improper integral. Therefore, in the source I am reading, we take the cauchy principal value (CPV) of the integral on the RHS instead of treating it as an improper integral. This principal value is calculated by use of the Residue Theorem.
My question: There are different ways to treat singularities in integrals. How do we know that this one (the CPV) will give us the correct result for $\int_{-\infty}^{\infty} \frac{\sin x}{x}dx$? Of course, knowing the answer by other methods, we can compare and see it was correct, but I'm looking to understand why the reasoning is valid.
Response to 1st answer: Simply saying that the integral converges is not enough. We need some way to know that in particular the CPV is the correct notion of integration for the exponential integral. Clearly, not any notion of integration which converges must give the correct result.
Response to 2nd answer: The question I ask is: by what reasoning is the notion of CPV in $\int_{-\infty}^{\infty} \frac{\sin x}{x}dx =\operatorname{Im}\left(CPV\int_{-\infty}^{\infty} \frac{e^{ix}}{x}dx \right)$ justified. Of course the first integral is the same as an improper integral or CPV, this just doesn't answer the question.
Best Answer
Since this is a question about rigor we should state what integral we are using. From (1) This integral is not Lebesgue integrable as it is not absolutely convergent, it is an improper Riemann integral though so we will use the Riemann integral.
Let us have a visual of the function being integrated
When we state the purely real integral $$I_1 = \int_{-\infty}^{\infty} \frac{\sin x}{x}dx$$ this is short-hand for $$I_1 = \lim_{a \to -\infty} \lim_{b \to \infty} \int_{a}^{b} f(x) dx$$ where $$f(x) = \begin{cases} \frac{\sin(x)}{x} & x \not = 0 \\ 1 & x = 0 \\ \end{cases}$$
The convergence of this integral must be addressed. It does not converge absolutely so we should attempt to cancel parts of the integral with itself. We will show that the right side (from $2\pi$ to $\infty$) converges by chopping it up and subtracting the negative part of each sine wave from the positive part. For each segment we have
$$\begin{align} &\, \left|\int_{\pi 2 k}^{\pi (2 k + 2)} \frac{\sin(x)}{x} dx\right| \\ \le&\, \left|\int_{\pi 2 k}^{\pi (2 k + 1)} \frac{\sin(x)}{x} - \frac{\sin(x)}{x + \pi} dx\right| \\ \le&\, \left|\int_{\pi 2 k}^{\pi (2 k + 1)} \frac{\sin(x)}{\pi (2 k + 1)} - \frac{\sin(x)}{\pi (2 k + 2)} dx\right| \\ \le&\, \left|\int_{\pi 2 k}^{\pi (2 k + 1)} \frac{\sin(x)}{\pi} \left[\frac{1}{2 k + 1} - \frac{1}{2 k + 2}\right] dx\right| \\ \end{align}$$
and the alternating harmonic series converges. Note that each segment is positive and their sum converges absolutely. Note that we showed the integral from $0$ to $\infty$ converges independently of the integral from $-\infty$ to $0$.
The finite part from $0$ to $2 \pi$ easily converges as we always have $\sin(x)/x \le 1$.
When we state a contour integral like
$$I_2 = \mathrm{p.v.} \int_\gamma \frac{e^{i z} - e^{-i z}}{2 i \cdot z} dz$$
($\gamma$ denoting a line from $-\infty$ to $\infty$).
the meaning (2) is that the we delete from the contour $\gamma$ an $\epsilon$ sized ball around the singularity and take the limit of $\epsilon$ to 0.
So $$I_2 = \lim_{\epsilon \to 0} \lim_{a \to -\infty} \lim_{b \to \infty} \left(\int_a^{-\epsilon} + \int_{\epsilon}^b\right) \frac{e^{i z} - e^{-i z}}{2 i \cdot z} dz$$
Regarding convergence of $I_2$, this integral (or half of it, to be precise) drops out algebraically from an application of Cauchy's Theorem (that a contour integral around a closed curve not containing poles gives 0) to the meromorphic function $e^{iz}/z$. The details are here.
Now to show $I_1 = I_2$. The single exceptional point at $x=0$ has no bearing on the value of the integral $I_1$ so we may split $I_1$ around 0 and add a $\lim_{\epsilon \to 0}$ on it. Now $I_1 - I_2 = 0$ can be shown. Rewrite $I_2$ to use the complex $\sin$ function.
$$\begin{align} I_1 - I_2 =& \left[\lim_{a \to -\infty} \lim_{b \to \infty} \int_{a}^{b} f(x) dx \right] - \left[\lim_{\epsilon \to 0} \lim_{a \to -\infty} \lim_{b \to \infty} \left(\int_a^{-\epsilon} + \int_{\epsilon}^b\right) \frac{e^{i z} - e^{-i z}}{2 i \cdot z} dz\right] \\ =& \left[\lim_{\epsilon \to 0} \lim_{a \to -\infty} \lim_{b \to \infty} \left(\int_a^{-\epsilon} + \int_{\epsilon}^b\right) \frac{\sin(x)}{x} dx \right] - \left[\lim_{\epsilon \to 0} \lim_{a \to -\infty} \lim_{b \to \infty} \left(\int_a^{-\epsilon} + \int_{\epsilon}^b\right) \frac{\sin(z)}{z} dz\right] \\ =& \lim_{\epsilon \to 0} \lim_{a \to -\infty} \lim_{b \to \infty} \left(\int_a^{-\epsilon} + \int_{\epsilon}^b\right) \left[\frac{\sin(x)}{x} - \frac{\sin(x)}{x} \right] dx \\ =& 0 \end{align}$$