Weak limits and computing the Fourier transform of the Heaviside function A common problem on this site is to compute the Fourier transform of the Heaviside function that is $0$ on the negative reals and $1$ on the positive reals. A standard technique is to consider the approximations $H_\alpha(x)$ defined to be $0$ on the negative reals and $e^{-\alpha x}$ on the positive reals. We then compute the Fourier transform and obtain (ignoring a constant factor)
$$\widehat{H_\alpha}(w) = \frac{1}{\alpha + iw}.$$
One can show that taking the limit as $\alpha$ goes to zero gives the correct answer. 
I am uncomfortable with one part of this argument. We are claiming that since $H_\alpha\rightarrow H$ as distributions, $\widehat{H_\alpha}\rightarrow \widehat H$. On one hand, this seems true because $\widehat H_\alpha(\phi)=H(\hat{\phi})$, so just staring at the definitions seems to prove it. On the other hand, I am having trouble making sense of the following computation.
Let's consider approximating $H$ a different way, using the characteristic functions of the interval $[0,M]$. We should obtain
$$\widehat H(w) = \lim_{M\rightarrow \infty} \int_0^M e^{-iwt} \ dt = \lim_{M\rightarrow \infty} \frac{1}{\sqrt{2\pi}} \frac{1-e^{iwM}}{iw}.$$
But this limit does not appear to be $\hat H$. In fact, it doesn't seem to exist. 
What is going on here? If my reasoning is correct, how can one show the latter limit gives the right answer?
 A: The latter limit does not exist because you think about it pointwisely. Instead, you should check the limit in the topology of tempered distributions, where it will be seen to exist and be exactly what it should be. To this end, pick $\varphi \in \mathcal S(\Bbb R)$ and check whether (I'm omitting multiplicative constant factors) $\lim \limits _{M \to \infty} \langle \frac {1 - \Bbb e ^{-\Bbb i w m}} {\Bbb i w}, \varphi \rangle = \langle \frac 1 {\Bbb i w}, \varphi \rangle$, i.e. whether $\lim \limits _{M \to \infty} \int \limits _{-\infty} ^{\infty} \frac {\Bbb e ^{-\Bbb i wM}} w \varphi \Bbb d w = 0$. Now, I'm writing in a rush, please check whether the following is right.
Let $I(M) = \int \limits _{-\infty} ^{\infty} \frac {\Bbb e ^{-\Bbb i wM}} w \varphi \Bbb d w$, so $I'(M) = \int \limits _{-\infty} ^{\infty} \Bbb e ^{-\Bbb i wM} \varphi \Bbb d w = \hat \varphi (M)$, so $I(M) = \int \limits _0 ^M \hat \varphi (t) \Bbb d t = \int \limits _0 ^1 \hat \varphi (Mu) M \Bbb d u$. It is known that the Fourier transform of a Schwartz function is again a Schwartz function, therefore $\hat \varphi$ decreases rapidly towards infinity, so a combination of this fact and of Lebesgue's dominated convergence theorem shows that $I(M) \to 0$ as desired, saving the coherence of mathematics once more!
