Fourier transforming the function: $$f(t) = \left\{ \begin{array}{ll} 1; & \mbox{ } |t| \leq 1 \\ 0; & \mbox{otherwise} \end{array} \right.$$ We get: $$F(y)=2 \frac{\sin y}{y}$$ And now applying Parseval's identity for the Fourier transform,
$$\int_{- \infty}^\infty {4 \frac{\sin ^2 y}{y^2}dy}=\int_{-1}^1 {f(t)^2}dt=2$$ By linearity of the integral we get the result $\frac{1}{2}$, however the result is $\pi$. Where did I go wrong? Thank you.
Your result for the Fourier transform of $f$ indicates that you use the definition
$$\mathcal{F}[f](y) = \int_{-\infty}^\infty f(t) e^{-ity}\,dt$$
of the Fourier transform (or maybe even with $e^{ity}$ instead of $e^{-ity}$).
With that definition, the Fourier transform is not an isometry of $L^2(\mathbb{R})$, and Plancherel's theorem becomes
$$\int_{-\infty}^\infty \lvert \mathcal{F}[f](y)\rvert^2\,dy = 2\pi \int_{-\infty}^\infty \lvert f(t)\rvert^2\,dt.$$
Then the result you get is indeed $\pi$.
