i have the following exercice:
Let for all $x \in \mathbb{R},$ $f(x)= \cos x$ and $g(x)= \sin x$.
Calculate $T=f \delta' + g \delta''$ for this question, i find $T=3 \delta$.
Calculate the Fourier transormates $F(f \delta')$ and $F(g \delta'')$.
How i can caculate the transformates Fourier in this question 2?. Please.
$\delta$ is Dirac distribution, and we define the Fourier transformate of an function $g$, as $$F(g)(\xi)= \displaystyle\int g(x) e^{-i x \xi} dx$$
The distributional derivative of the Dirac delta distribution is the distribution $\delta′$ defined on compactly supported smooth test functions $\varphi$ by $$ \delta'[\varphi] = -\delta[\varphi']=-\varphi'(0). $$ The first equality here is a kind of integration by parts, for if $\delta$ were a true function then $$ \int_{-\infty}^\infty \delta'(x)\varphi(x)\,dx = -\int_{-\infty}^\infty \delta(x)\varphi'(x)\,dx. $$ The $k$-th derivative of $\delta$ is defined similarly as the distribution given on test functions by $$ \delta^{(k)}[\varphi] = (-1)^k \varphi^{(k)}(0). $$
If $f$ is $n$ times differentiable around $x_0\in\Bbb R$ we have $$ f(x)\delta^{(n)}(x-x_0)=(-1)^n\sum_{k=0}^n (-1)^k \binom{n}{k}f^{(n-k)}(x_0)\delta^{(k)}(x-x_0) $$ So you have for $n=1$ and $n=2$ and $x_0=0$ \begin{align} T &=f(x) \delta'(x)+g(x) \delta''(x)\\ &=- \delta(x) f'(0)+ \delta(x) g''(0)+f(0) \delta'(x)-2 g'(0) \delta'(x)+g(0) \delta''(x)\\ &=\delta'(x)-2 \delta'(x)\\&=-\delta'(x) \end{align} using $f(0)=1,\,f'(0)=0, g(0)=0,\,g'(0)=1,\,g''(0)=0$.
For the Fourier transform use $$\mathcal F\{\delta(x)\}=\displaystyle \int_{-\infty}^{\infty}\delta(x) e^{- ix\xi}\,dx=1$$ and $$ \mathcal F\{f^{(n)}(x)\}(\xi)= (i\xi)^n \mathcal F\{f(x)\}(\xi) $$ that is $$\mathcal F\{\delta^{(n)}(x)\}= (i\xi)^n \cdot 1 $$ So we have \begin{align} \mathcal F\left\{f(x) \delta'(x)\right\}&=\mathcal F\left\{- \delta(x) f'(0)+f(0) \delta'(x)\right\}\\ &=- f'(0)+f(0) i\xi\\ &= i\xi \end{align} and \begin{align} \mathcal F\left\{g(x) \delta''(x)\right\}&=\mathcal F\left\{\delta(x) g''(0)-2 g'(0) \delta'(x)+g(0) \delta''(x)\right\}\\ &=g''(0)-2 g'(0) i\xi+g(0) (i\xi)^2\\ &=-2 i\xi \end{align} and then
$$ \mathcal F\left\{T(x)\right\}=\mathcal F\left\{f(x) \delta'(x)\right\}+\mathcal F\left\{g(x) \delta''(x)\right\}=\mathcal F\left\{- \delta'(x)\right\}=- i \xi $$