Calculate fourier transform for function $h$:
$$\hat{h}(\nu)=\int_{-\infty}^\infty e^{-i2\pi\nu t} h(t) dt $$
When $h(t)=1$ and $|t|\le\frac 12$.
And when $h(t)=0$ and $|t|\gt\frac 12$
Also does it hold, that $\int_{-\infty}^\infty |\hat{h}(\nu)|d\nu \lt \infty$? What about $\int_{-\infty}^\infty |\hat{h}(\nu)|^2d\nu \lt \infty$ or $\lim_{|\nu| \to \infty}\hat{h}(\nu)=0$?
The Sinc Function is the fourier transform of a discrete pulse: $$ \begin{align} \hat{h}(\nu) &=\int_{-\infty}^\infty h(t)e^{-2\pi i\nu t}\,\mathrm{d}t\\ &=\int_{-1/2}^{1/2}e^{-2\pi i\nu t}\,\mathrm{d}t\\ &=\frac{e^{\pi i\nu}-e^{-\pi i\nu}}{2\pi i\nu}\\ &=\frac{\sin(\pi\nu)}{\pi\nu}\\[7pt] &=\mathrm{sinc}(\pi\nu) \end{align} $$
Note that $$ \begin{align} \int_k^{k+1}\left|\frac{\sin(\pi\nu)}{\pi\nu}\right|\,\mathrm{d}\nu &\ge\int_k^{k+1}\left|\frac{\sin(\pi\nu)}{\pi(k+1)}\right|\,\mathrm{d}\nu\\ &=\frac2{\pi^2(k+1)} \end{align} $$ Since the Harmonic Series diverges, the integral $$ \int_{-\infty}^\infty|\hat{h}(\nu)|\,\mathrm{d}\nu $$ also diverges.
The Plancherel Theorem says that the $L^2$ norm of a function and its Fourier Transform are equal. Since $$ \int_{-1/2}^{1/2}1^2\,\mathrm{d}t=1 $$ we get $$ \int_{-\infty}^\infty|\hat{h}(\nu)|^2\,\mathrm{d}\nu=1 $$
Since $$ \int_{-1/2}^{1/2}|1|\,\mathrm{d}t=1 $$ The Riemann-Lebesgue Lemma says that $$ \lim_{\nu\to\infty}\hat{h}(\nu)=0 $$ Of course, since we've computed it, we can just see that $$ \lim_{\nu\to\infty}\frac{\sin(\pi\nu)}{\pi\nu}=0 $$
Isn't it pretty straightforward, or am I missing something?:
$$\hat h(v)=\int\limits_{-1/2}^{1/2}e^{-2\pi ivt}dt=\left.-\frac1{2\pi iv}e^{-2\pi ivt}\right|_{-1/2}^{1/2}=-\frac1{2\pi iv}\left(e^{-\pi iv}-e^{\pi iv}\right)=\frac1{\pi v}\sin\pi v$$
