Is $\hat S$ a function?

101 Views Asked by At

Let $S(x)=\sum_{n=-\infty}^\infty (-1)^n\chi_{(n,n+1)}(x)$. Find the Fourier transformation of $S(x)$. Is $\hat{S}$ a function?

$$\hat{S}=\int_R \sum_{n=-\infty}^\infty (-1)^n\chi_{(n,n+1)}(x) dx=\sum_{n=-\infty}^\infty \frac{(-1)^n i e^{-iny}(e^{-iy}-1)}{y} $$

How do I show that the fourier transformation is or isn't a function?

1

There are 1 best solutions below

0
On BEST ANSWER

This is not a function in the usual sense but generalized function or distribution.

$$\sum_{n=-\infty}^\infty (-1)^ne^{-iny}=\sum_{n=-\infty}^\infty e^{in(-y+(2m+1)\pi)}=\delta(-y+(2m+1)\pi),\ \forall m\in{\mathbf Z},$$ for different branch cuts. So $$\hat S = \frac{i}{y}(e^{-iy}-1)\delta(-y+(2m+1)\pi),\ \forall m\in{\mathbf Z}$$ for different branch cuts, a Dirac delta function distribution, or generalized function.