Proving that the Fourier transform of the identity is the Dirac delta using test functions. For every finite Schwartz function $f$ and every $t \in \mathbb{R}$, define the Dirac delta as
$$\int_{-\infty}^\infty dt f(t) \delta(t-t') := f(t').$$
How do I proof with this definition that
$$\int_{-\infty}^\infty d\omega e^{i\omega(t-t')} =2\pi \delta(t-t'). $$
When I use complex analysis, this gives me zero for all $t, t'$ since there are no poles and the contour, which clearly is wrong.
I would like to know this way of a proof to understand what the following integral is in distribution sense:
$$\int_{-\infty}^\infty d\omega e^{i\omega(t-t')} |\omega|^{a}, $$
with real $a > 0$.
 A: Given $t'$ for each $A$ let
$$g_A(t)=\int_{-A}^A  e^{i\omega(t-t')}d\omega = 2 \frac{\sin(A(t-t'))}{t-t'}$$
Then $\lim_{A\to \infty} g_A$ converges to $2\pi \delta(.-t')$ in the sense of tempered distributions.
To see this let $$h(t)=\int_{-\infty}^t \frac{2\sin(u)}{u}du$$
For any $\phi \in S(\Bbb{R})$
$$\int_{-\infty}^\infty \phi(t) g_A(t)dt = -\int_{-\infty}^\infty \phi'(t+t') h(At)dt$$
$h(At)$ converges to $2\pi 1_{t >0}$ in $L^1_{loc}$ and uniformly away from $(-\epsilon,\epsilon)$ so $-\int_{-\infty}^\infty \phi'(t) h(At)dt$ converges to
$$-\int_{t'}^\infty 2\pi\phi'(t)= \phi(t')$$
Note that this is the proof of the Fourier inversion theorem at least for the Schwartz functions.
A: Let $f(t)=e^{-i\omega t'}$ and $\phi\in \mathbb{S}$.  Then we have
$$\begin{align}
\langle \mathscr{F}^{-1}\{f\},\phi\rangle&=\langle f,\mathscr{F}^{-1}\{\phi\}\rangle\\\\
&=\int_{-\infty}^\infty e^{-i\omega t'} \frac1{2\pi}\int_{-\infty}^\infty \phi(t)e^{i\omega t}\,dt\,d\omega\\\\
&=\phi(t')
\end{align}$$
where the last equality is a consequence of the Fourier Inversion Theorem (See Appendix).  Therefore, in distribution we have
$$\mathscr{F}^{-1}\{f\}(t)=\delta(t-t')$$

APPENDIX:  PROOF OF THE INVERSION THEOREM FOR SCHWARTZ FUNCTIONS
In THIS ANSWER, THIS ONE and in the Alternative Development section of THIS ANSWER, I proved that $\frac{\sin(Lt)}{\pi t}$ is a nascent Dirac Delta.  Using this result, we see that
$$\begin{align}
\int_{-\infty}^\infty e^{-i\omega t'} \frac1{2\pi}\int_{-\infty}^\infty \phi(t)e^{i\omega t}\,dt\,d\omega&=\frac1{2\pi}\int_{-\infty}^\infty  \int_{-\infty}^\infty \phi(t+t')e^{i\omega t}\,dt\,d\omega\\\\
&=\lim_{L\to \infty}\frac1{2\pi} \int_{-\infty}^\infty \phi(t+t') \int_{-L}^L e^{i\omega t}\,d\omega\,dt\\\\
&=\lim_{L\to \infty} \int_{-\infty}^\infty \phi(t+t') \frac{2\sin(Lt)}{\pi t}\,dt\\\\
&= \phi(t')
\end{align}$$
