Is the sinc function both absolutely summable (L1 norm for Continuous time signals and l1 norm for Discrete time signals) and square summable (L2 norm for Continuous time signals and l2 norm for Discrete time signals) ?
Can anyone show the integration and summation calculations that calculates above norms ?
Neither the integral
$$I_1=\int_{0}^{\infty}\left|\frac{\sin(x)}{x}\right|dx\tag{1}$$
nor the sum
$$S_1=\sum_{n=1}^{\infty}\left|\frac{\sin(n)}{n}\right|\tag{2}$$
converge. So the sinc function (sequence) is not in $L_1$ ($l_1$). A proof of the fact that $I_1$ diverges can be found here. For the sum $S_1$, note that since $|\sin(n)|\ge \sin^2(n)$ (for $n\in\mathbb{R}$) we have
$$S_1\ge \sum_{n=1}^{\infty}\frac{\sin^2(n)}{n}\tag{3}$$
Here it is shown that the series on the right-hand side of $(3)$ diverges, so $S_1$ must also diverge.
The integral
$$I_2=\int_{0}^{\infty}\left(\frac{\sin(x)}{x}\right)^2dx=\frac{\pi}{2}\tag{4}$$
can be computed in many different ways. Two possibilities are shown in the answers to this question.
Also note that
$$\int_{0}^{\infty}\left(\frac{\sin(x)}{x}\right)^2dx=\int_{0}^{\infty}\frac{\sin(x)}{x}dx\tag{5}$$
The computation of the sum
$$S_2=\sum_{n=1}^{\infty}\left(\frac{\sin(n)}{n}\right)^2=\frac{\pi-1}{2}\tag{6}$$
is discussed in this question and its answers. Note that a relation similar to $(5)$ holds:
$$\sum_{n=1}^{\infty}\left(\frac{\sin(n)}{n}\right)^2=\sum_{n=1}^{\infty}\frac{\sin(n)}{n}\tag{7}$$
So clearly, the sinc function (sequence) is in $L_2$ ($l_2$).
