Is there a difference between the fourier transform on a distribution and on a function? I have to calculate the Fourier transform of the following distribution:
$$\langle F_a\mid\phi\rangle = \int_{-\infty}^\infty \cos(ax) \phi(x) \, dx, \phi \in \mathcal S (\mathbb{R})$$
Would it be wrong to calculate instead the Fourier transform on:
$$F_a=\cos(ax) \text{?}$$
In both cases, can you show me a complete solution?
 A: The fact is that $F_{a}\notin L^{1}({\bf{R}})$, so the integral $\displaystyle\int_{\bf{R}}\cos(ax)e^{-2\pi i\xi\cdot x}dx$ does not exist. Besides that, $F_{a}\notin L^{2}({\bf{R}})$, so the general Fourier transform also does not exist. So we can only speak about the distributional Fourier transform.
For the calculation, note that $\cos(ax)=\dfrac{1}{2}(e^{iax}+e^{-iax})$ and hence for a test function $\varphi\in S({\bf{R}})$, we have
\begin{align*}
\left<\widehat{\cos(ax)},\varphi\right>&=\left<\cos(ax),\widehat{\varphi}\right>\\
&=\dfrac{1}{2}\left<e^{iax},\widehat{\varphi}\right>+\dfrac{1}{2}\left<e^{-iax},\widehat{\varphi}\right>\\
&=\dfrac{1}{2}\int_{\bf{R}}e^{iax}\widehat{\varphi}(x)dx+\dfrac{1}{2}\int_{\bf{R}}e^{-iax}\widehat{\varphi}(x)dx\\
&=\dfrac{1}{2}(\widehat{\varphi})^{\vee}(a/2\pi)+\dfrac{1}{2}(\widehat{\varphi})^{\vee}(-a/2\pi)\\
&=\dfrac{1}{2}(\varphi(a/2\pi)+\varphi(-a/2\pi))\\
&=\dfrac{1}{2}\left<\delta_{a/2\pi},\varphi\right>+\left<\delta_{-a/2\pi},\varphi\right>\\
&=\left<\dfrac{1}{2}(\delta_{a/2\pi}+\delta_{-a/2\pi}),\varphi\right>,
\end{align*}
so
\begin{align*}
\widehat{\cos(ax)}=\dfrac{1}{2}(\delta_{a/2\pi}+\delta_{-a/2\pi})
\end{align*}
in the distributional sense.
