What is the Fourier transformation of H(-t)? I want to start by the definition of Fourier transformation. It is
$$\mathcal{F}[H(-t)]=\int_{-\infty}^0 e^{-j{\omega}t}\,dt =\left.\frac{1}{-j\omega}e^{-j{\omega}t}\right|_{-\infty}^0 $$
But when $t \rightarrow{-\infty}$, the result goes to the infinity, right?


EDIT:  ALTERNATIVE AND RIGOROUS DEVELOPMENT
In THIS ANSWER, I evaluated the Fourier Transform of the Heaviside (i.e., unit step) function using a rigorous distributional approach.



Finding the Fourier Transform of the unit step function, $H(-t)$, is as easy as $1,2,3$.

STEP $1$:
The Fourier Transform of $f(t)=1$ is$\int_{-\infty}^\infty (1)e^{i\omega t}\,dt =2\pi \delta(\omega)$, since the inverse Fourier Transform of $2\pi \delta(\omega)$ is $\frac{1}{2\pi}\int_{-\infty}^\infty(2\pi \delta(\omega))\,e^{-i\omega t}\,d\omega=1$.

STEP $2$:
The Fourier Transform of the signum function can be evaluated as
$$\begin{align}
\mathscr{F}\{\text{sgn}(t)\}(\omega)&=\lim_{a\to 0}\int_{-\infty}^\infty \text{sgn}(t)e^{-a|t|}e^{i\omega t}\,dt\\\\
&=\lim_{a\to 0}\left(\frac{i2\omega}{\omega^2+a^2}\right)\\\\
&=\frac{i2}{\omega}
\end{align}$$

STEP $3$:
We note the $H(-t)=\frac12-\frac12\text{sgn}(t)$ and hence
$$\mathscr{F}\{H(-t)\}(\omega)=\pi \delta(\omega)-\frac{i}{\omega}$$
And we are done!
When applying $\mathscr{F}\{H(-t)\}(\omega)$ to a test function $\phi(\omega)$, the integral of the term $\frac{i}{\omega}$ must interpreted as a Cauchy Principal Value.


