Evaluate $\hat f(x):=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-itx}dt$ for $f(t)=\sum_{n=0}^{\infty}\frac{1}{2^n}1_{[-2,2]}(x-2n)$

Question

$\hat f(x):=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-itx}dt$, for $f \in L^1(\mathbb R)$

Let $f(t)=\sum_{n=0}^{\infty}\frac{1}{2^n}1_{[-2,2]}(t-2n)$, where $1_{[.,.]}$ is the indicator function

I want to determine $\hat f(x).$

I think we need to interchange the series and the integral which should be no problem because

$\frac{1}{2^n}1_{[-2,2]}(x-2n) \ge 0$ so we can use Fubini.

Now, using $1_{[-2,2]}(x-2n)=1_{[2n-2,2n+2]}(x)$:

$\hat f(x)=\frac{1}{\sqrt{2\pi}}\sum_{n=0}^{\infty}\frac{1}{2^n}\int_{-\infty}^{\infty}1_{[2n-2,2n+2]}(t)e^{-itx}dt=\frac{1}{\sqrt{2\pi}}\sum_{n=0}^{\infty}\frac{1}{2^n}\frac{e^{-i(2n+2)}-e^{-i(2n-2)}}{-ix}$

Is that correct so far? I thought about using $\sin(x)=\frac{1}{2i}(e^{ix}-e^{-ix})$ and $\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix})$ to simplify this term but I don't know if that will help

Answer 1

It's correct, but at end there is no need to that formulas for simplification of sum. Just observe that all terms display in positive and negative, so annihilate each other, but these two: $$\hat{f}(x) = \frac{1}{\sqrt{2\pi}}(\frac{e^{-2i}}{ix}+\frac{1}{2}.\frac{1}{ix})$$

Answer 2

If you are also required to be rigorous you have to justify the interchange of the integral and the infinite sum. You can do this by observing that your sereis converges in $L^{1}$ norm and $s_n \to f$ in $L^{1}$ norm implies $\hat {s_n} \to \hat {f}$ pointwise. [Take $s_n$ to be the n-th partial sum of the series].