Is there an elementary way to proof that $$\int_0^\infty \frac{\sin{x}}{x} \mathrm{d}{x} = \frac{\pi}{2}$$ or equivalently $$\int_{-\infty}^{+\infty} \frac{\sin{x}}{x} \mathrm{d}{x} = \pi \;\;\; \text{?}$$
Prequesites (which might be assumed):
- Basic knowledge about limits, complex numbers, differentiation, series and power series as well as indefinite Riemann-integration in one dimension may be assumed (including Riemann-sums).
- No complex or multi-dimensional calculus as well as Fourier-analysis.
Note (1):
Sorry if the question is covered somewhere already I couldn't find it. Also it is absolutely clear that the integral can be solved by a multitude of approaches if the prerequisites are dropped among them fourier analysis, polar coordinates and complex analysis.
Note (2): The closest thing I found was a combination of Robin Chapman's Approach 1 and Zarax's method used in Approach 2. This essentially reduces to first showing by integration by parts that $$\int_{-\infty}^{+\infty} \frac{\sin{x}}{x} \mathrm{d}{x} = S \Longleftrightarrow \int_{-\infty}^{+\infty} \frac{\sin^2{x}}{x^2} \mathrm{d}{x} = S$$ and then showing that $$ S = \int_{-\infty}^{+\infty} \frac{\sin^2{x}}{x^2} \mathrm{d}{x} = \pi$$ as described in Approach 1. Which is what I will post as fleshed out answer later.
Note (3): For completness sake possible solutions to the problem are already collected in Solving the dirichlet Integral but as stated in the comments I was looking for something more basic.
Nevertheless any other basic approaches (not already collected there) are highly welcome!