# Calculating a Fourier Transform using another one

If we consider the function $f(t)=e^{-at}\cos(t) v(t)$ (where $v(t)=1\; \forall t\geq 0$, and $0$ otherwise), then I have calculated its Fourier transform, which is

$$F(f)(\omega)=\frac{a+\omega i}{1+(a+\omega i)^2}$$

Can I take limits $a\to 0$ here to deduce the Fourier transform of $\cos(t)v(t)$?

• I've deleted my answer. I've missed the fact that $v$ is not a square-integrable function, so you can't really use the $L^2$ theory for it. As for your question about whether or not the Fourier transform of $\cos(t)v(t)$ exists, it certainly doesn't exist in the usual sense (as an improper integral), but it might exist in the distribution sense. – tomasz May 16 '17 at 18:27
• Yes but the limit as $a \to 0$ of $\displaystyle\frac{a+i\omega}{1+(a+\omega i)^2}$ in the sense of distributions is $\displaystyle PV.(\frac{i\omega }{1-\omega^2 })+ \frac{\pi}{2} (\delta(\omega+1)+\delta(\omega-1))$ – reuns Jun 3 '17 at 1:39

Yes, this is a legitimate approximation argument. The Fourier transform is a continuous operator on the space of tempered distributions. The functions $f_a(t)=e^{-at}\cos(t) v(t)$ converge to $f(t)=\cos(t) v(t)$ in the sense of tempered distributions, meaning $\int f_a \varphi \to \int f \varphi$ for every Schwarz test function $\varphi$ (this holds by the dominated convergence, $|f\varphi|$ being the dominating function).
Therefore, the Fourier transforms of $f_a$ converge to the transform of $f$ in the sense of distribution. Specifically, $\hat f$ is a distribution represented by the function $\omega i/(1-\omega^2)$ in the sense of principal value, plus the delta functions pointed out in the comments below.
• You made a mistake for the limit in the sense of distributions in the Fourier domain. The FT of $1_{t > 0}$ is $\hat{H}(\omega) = PV. (\frac{1}{i \omega}) + \pi \delta(\omega)$ thus the FT of $1_{t > 0} \cos(t)$ is $$\frac{\hat{H}(\omega-1)+\hat{H}(\omega+1)}{2} = PV. (\frac{1}{2 i (\omega-1)}+\frac{1}{2 i (\omega+1)}) + \frac{\pi}{2} \delta(\omega-1)+\frac{\pi}{2} \delta(\omega+1)\\ =PV. (\frac{i\omega}{ 1-\omega^2})+ \frac{\pi}{2} \delta(\omega-1)+\frac{\pi}{2} \delta(\omega+1)$$ – reuns Jun 3 '17 at 1:32
• And $PV. (\frac{i\omega}{ 1-\omega^2})$ is the FT of $\frac12\text{sign}(t) \cos(t)$ – reuns Jun 3 '17 at 1:42