1
$\begingroup$

Let $\theta(t) = (1-|t|)^+$ for $t \in \mathbb{R}$. In order to show that \begin{align} \int_{-\infty}^\infty \theta(t) e^{-i \xi t}\ dt = \bigg(\frac{sin\big(\frac{\xi}{2}\big)}{\frac{\xi}{2}}\bigg)^2, \qquad (*) \end{align} I want to compute the Fourier transform of $\phi(t) = \mathbb{1}_{\{-1/2 \leq t \leq 1/2\}}$ and relate this to $\theta$.

It could be shown that the Fourier transform $\hat{f}$ of $\phi$ equals \begin{align} \hat{f}(\xi) = \frac{\sin(\pi \xi)}{\pi \xi}. \qquad (**) \end{align} Making the substitution $u = \frac{1}{2}t$ in $(*)$ gives \begin{align} \int_{-\infty}^\infty \theta(t) e^{-i \xi t}\ dt = 2 \int_{-1/2}^{1/2} (1-|2u|)^+ e^{-2i \xi u}\ du \qquad (***) \end{align} and could be worked out. However, I do not see how $(**)$ could be related to $(***)$. Any help is appreciated!

$\endgroup$
1
$\begingroup$

You can write $\theta$ as $1_I\ast1_I$ for an interval $I$. Say $I=[-a,a]$. Then you have to compute $$ \int_{-a}^a1_{[-a,a]}(x-t)dt. $$

$\endgroup$
  • $\begingroup$ Thanks for your reply. The statement follows indeed from there. However I am quite unfamiliar with these calculations. Could you maybe elaborate this further? $\endgroup$ – iJup Apr 6 '17 at 16:48
  • $\begingroup$ $$\mathbb{1}_{[-1/2,1/2]} * \mathbb{1}_{[-1/2,1/2]} (x) = \int_{\max(-1/2, x-1/2)}^{\min(1/2, x+1/2)} dt$$ $$ = \begin{cases} 0 \qquad \text{if $x \geq \frac{1}{2}$} \\ 1-x \qquad \text{if $0 \leq x < \frac{1}{2}$} \\ 1+x \qquad \text{if $\frac{-1}{2} < x <0$} \\ 0 \qquad \text{if $x \leq \frac{-1}{2}$}\end{cases}$$ could be plugged in $(***)$. To make ends meet, how do we have to use the Fourier transform of $\phi$? $\endgroup$ – iJup Apr 7 '17 at 18:54
  • $\begingroup$ @iJup Careful with Tex! All the text boxes in the cases -environment missed the closing dollar-sign around the formula. Also, it is NEVER a good idea to use more than 80 characters of TeX without a break in the comments. The comments are parsed differently, and extra spaces will be added by the sparser - with random side effects. $\endgroup$ – Jyrki Lahtonen Apr 7 '17 at 20:25
  • $\begingroup$ Okido, I will do! @Julián, I solved my question. Thanks! I used that $\int_\mathbb{R} (\phi * \phi)(t) e^{-i \xi t}\ dt = (\int_\mathbb{R} \phi(t)e^{-i \xi t}\ dt)^2$. $\endgroup$ – iJup Apr 8 '17 at 13:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.