Basic Fourier question 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 $\lim_{|\nu| \to \infty}\hat{h}(\nu)=0$?
 A: 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
$$
\lim_{\nu\to\infty}\frac{\sin(\pi\nu)}{\pi\nu}=0
$$
A: Isn't it pretty straightforward, or am I missing something?:
$$\hat h(v)=\int\limits_{-1/2}^{1/2}e^{-2\pi ivt}dt=\left.-\frac1{2\pi iv}e^{-2\pi ivt}\right|_{-1/2}^{1/2}=-\frac1{2\pi iv}\left(e^{-\pi iv}-e^{\pi iv}\right)=\frac1{\pi v}\sin\pi v$$
