calculus - how to integrate $int_{-infty}^{+infty} frac{sin(x)}{x} ,dx$?

How can I do this integration using only calculus?
(not laplace transforms or complex analysis)

$$\int_{-\infty}^{+\infty} \frac{\sin(x)}{x} \,dx$$

I searched for solutions not involving laplace transforms or complex analysis but I could not find.

Answer

Putting rigor aside, we may do like this:

$$\begin{align*} \int_{-\infty}^{\infty} \frac{\sin x}{x} \; dx &= 2 \int_{0}^{\infty} \frac{\sin x}{x} \; dx \\ &= 2 \int_{0}^{\infty} \sin x \left( \int_{0}^{\infty} e^{-xt} \; dt \right) \; dx \\ &= 2 \int_{0}^{\infty} \int_{0}^{\infty} \sin x \, e^{-tx} \; dx dt \\ &= 2 \int_{0}^{\infty} \frac{dt}{t^2 + 1} \\ &= \vphantom{\int}2 \cdot \frac{\pi}{2} = \pi. \end{align*}$$
The defects of this approach are as follows:

1. Interchanging the order of two integral needs justification, and in fact this is the hardest step in this proof. (There are several ways to resolve this problem, though not easy.)


2. It is nothing but a disguise of Laplace transform method. So this calculation contains no new information on the integral.