Lower bounding decay of Fourier transform of a discontinuous function Consider the function: 
$$g(t) = \frac{1}{2\pi} \int_0^{\infty} \exp(itu) \exp(-u^2) du$$
This is the Fourier transform of the function $u \mapsto \chi_{[0,\infty)}(u)\cdot \exp(-u^2)$, where $\chi_{[0,\infty)}(u)$ is $1$ for $u \in [0,\infty)$ and $0$ otherwise.
By a direct argument one may check the following upper bound on the decay of $g$: There exists a constant $M >0$ such that for all $t \in \mathbb R$:
$$ |g(t)| \leq \frac{M}{|t|+1}$$
My questions is the following: Is the above inequality sharp? In other words, does there exist $m>0$ such that for all $t \in \mathbb R$:
$$ |g(t)| \geq \frac{m}{|t|+1}$$
 A: This question is interesting, and although I can't prove the bounds asserted in the OP I can provide some well-founded conjectures.
It is easy to show with various methods that 
$$g(t)=\frac{1}{4\sqrt{\pi}}e^{-t^2/4}~\Big(1+i~\text{erfi}(t/2)\Big)$$
Thus for the absolute value $$h(t)\equiv|g(t)|=\frac{e^{-t^2/4}}{4\sqrt{\pi}}\sqrt{1+\text{erfi}^2\Big(\frac{t}{2}\Big)}$$
It is not hard to show using L' Hopital's rule that 
$$\lim_{t\to\infty}(t+1)|g(t)|=\frac{1}{4\sqrt{\pi}}$$
I conjecture that the bound is sharp with $m=\frac{1}{4\sqrt{\pi}}$. We get lucky here with our conjecture since $h(0)=\frac{1}{4\sqrt{\pi}}$ and therefore the vertical distance
$$d(x)=h(x)-\frac{1}{4\sqrt{\pi}(x+1)}$$
has the property that $d(0)=d(\infty)=0$. That means there is a maximum inbetween, and a good idea for a proof would exploit this and the fact that $h(x)$ for small $x$ is concave ,which means that $d(x)$ is initially increasing. 
I also conjecture that there is a best value for $M$ as well which can be obtained by solving the equations
$$\begin{align}h(x)&=\frac{M}{x+1}\\h'(x)&=-\frac{M}{(x+1)^2}\end{align}$$
or upon some rearrangement
$$M=(x_0+1)h(x_0)$$
where $x_0$ is the unique solution of the equation
$$(x_0+1)\frac{h'(x_0)}{h(x_0)}=-1$$
where
$$\frac{h'(x)}{h(x)}=\frac{1}{4}\frac{e^{t^2/4}}{\text{erfi}^2(t/2)+1}-\frac{t}{2}$$
Here's the image that has been guiding my intuition

showing the upper and lower bounds for $m=\frac{1}{4\sqrt{\pi}}$ and $M\approx 0.3$. 
