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