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I am asked to compute the fourier transform of the distribution $f(x) = \sin (x) \ x \in \mathbb{R}$

Let $L \in S'(\mathbb{R}^n)$. The fourier transform of $L$ is the map $\hat{L}: S(\mathbb{R}^n) \to \mathbb{C}$ definied by $\hat{L} \psi := L \hat{\psi}$ $ \forall \psi \in S(\mathbb{R}^n)$.

But I dont understand how I should "solve" for $\hat{L}$

$$ S(\mathbb{R}^n) := \{\psi: \mathbb{R}^n \to \mathbb{C} : \psi \text{ smooth }, \|\psi \|_{a,b} := \sup_{x \in \mathbb{R}^n} |x^b D^a \psi(x)| < \infty \ \forall a,b \in \mathbb{N} \cup {0} \} \\ a = (a_1, a_2, \dots, a_n) \\ b = (b_1, b_2, \dots, b_n) \\ x^b = (x^{b_1}, x^{b_2}, \dots , x^{b_n}) \\ D^a \psi(x) = \frac{\partial^{a_1}}{\partial_{x_1}^{a_1}} \dots \frac{\partial^{a_1}}{\partial_{x_n}^{a_n}} \psi(x) \\ S'(\mathbb{R^n}) := \{L :S(\mathbb{R}^n) \to \mathbb{C} : L \text{ Linear, Continuous } \} $$

My attempt.

$$ FT(\delta) = \int_{\mathbb{R}} \delta (x) e^{-ikx} dx = e^0 \int_{\mathbb{R}} \delta (x) dx = 1. $$

Thus $$ FT(\delta (x-a)) = e^{-ika} $$

Then

$$ \delta (x-a) = FT^{-1} (e^{-ika}) = \int_{\mathbb{R}} e^{ik(x-a)} dx. $$

Thus

$$ \int_{\mathbb{R}} \sin x e^{-ikx} dx = \frac{1}{2i} ( \delta (k-1) - \delta (k+1)). $$

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  • $\begingroup$ The Fourier transform is continuous and unitary $S(\mathbb{R}^n) \to S(\mathbb{R}^n)$ so it is continuous and unitary $S'(\mathbb{R}^n) \to S'(\mathbb{R}^n)$ too, i.e. $\delta$ is the FT of $1$ $\endgroup$ – reuns Dec 13 '16 at 19:48
  • $\begingroup$ What is unitary? @user1952009 $\endgroup$ – Olba12 Dec 13 '16 at 19:54
  • $\begingroup$ The Fourier transform $FT$ is a continuous linear operator $S(\mathbb{R}^n) \to S(\mathbb{R}^n)$ and it is unitary : $FT^{-1} = FT^*$ (the adjoint). And as you know $(FT^* \varphi)(x) = (FT \varphi)(-x)$ $\endgroup$ – reuns Dec 13 '16 at 19:59
  • $\begingroup$ Yeah, I'm going to need more help. @user1952009 $\endgroup$ – Olba12 Dec 13 '16 at 20:07
  • $\begingroup$ Is it clear to you that $FT(\delta) = 1$ ? Then what is $FT[\delta_a]$ ? $\endgroup$ – reuns Dec 13 '16 at 20:28
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The distribution $\sin(x)$ is $$ S(f) = \int_{\mathbb{R}}f(x)\sin(x)dx, \;\;\; f\in\mathscr{S}. $$ The Fourier transform of $S$ is defined by $$ \hat{S}(f) = S(\hat{f})= \int_{\mathbb{R}}\hat{f}(s)\sin(s)dx,\;\;\; f\in\mathscr{S}. $$ The above is simplified by using the Fourier transform inversion: \begin{align} \hat{S}(f) & = \left.\int_{\mathbb{R}}\hat{f}(s)\frac{e^{isx}-e^{-isx}}{2i}ds\right|_{x=1} \\ & = \frac{\sqrt{2\pi}}{2i}(f(1)-f(-1)) \\ & = -i\sqrt{\frac{\pi}{2}}(\delta_{1}(f)-\delta_{-1}(f)) \end{align} Therefore, $$ \hat{S} = -i\sqrt{\frac{\pi}{2}}(\delta_1 -\delta_{-1}) $$

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