Lower bound of Fourier transform We know the Fourier transform of the Gauss-function:
$\displaystyle\int_{\xi\in\mathbb{R}^d}e^{-\pi\, C\,|\xi|^2}e^{2\pi i \xi\cdot X}d\xi=C^{-d/2}e^{-\, \pi\, |X|^2/2}$
for any $C>0$.
Then according to my textbook:
$\displaystyle\left| \int_{\xi\in\mathbb{R}^d:|\xi|<\sqrt{d}/2}e^{-\pi\, C\,|\xi|^2}e^{2\pi i \xi\cdot X}d\xi \,\right| =\Omega(1)^d$
whenever $|X|\leq c\sqrt{d}$, if $c$ is chosen small and $C$ sufficiently large.
I don't understand this lower bound. 
While it is clear that $\displaystyle\left| \int_{\xi\in\mathbb{R}^d}e^{-\pi\, C\,|\xi|^2}e^{2\pi i \xi\cdot X}d\xi \,\right| =\Omega(1)^d$
whenever $|X|\leq c\sqrt{d}$ for some fixed $c$, why can we reduce to the set $\{\xi\in\mathbb{R}^d:|\xi|<\sqrt{d}/2\}$? I have tried with the inverse triangle inequality but cannot conclude.
 A: Since for any complex number $z \in \mathbb{C}$, obviously the inequality $$|z| \geq |\text{Re} \, z| $$ holds where $\text{Re} \, z$ denotes the real part of $z$, we have
$$\left| \int_{|\xi| < \sqrt{d}/2} e^{-\pi \cdot C \cdot |\xi|^2} \cdot e^{\imath \, 2\pi \, \xi \cdot x} \, d\xi \right| \geq \left| \int_{|\xi| < \sqrt{d}/2} e^{-\pi \cdot C \cdot |\xi|^2} \cdot \cos(2\pi \, \xi \cdot x) \, d\xi \right|$$
We can choose $c>0$ sufficiently small such that $$2\pi \cdot |\xi \cdot x| < \frac{\pi}{4}$$ for all $|\xi| < \frac{\sqrt{d}}{2}$, $|x|< c \cdot \sqrt{d}$. Consequently,
$$\cos(2\pi \cdot \xi \cdot x) \geq \frac{1}{2}$$
for all $|\xi| < \frac{\sqrt{d}}{2}$, $|x|< c \cdot \sqrt{d}$. By the positivity of the integrand, we conclude
$$\begin{align*} \left| \int_{|\xi| < \sqrt{d}/2} e^{-\pi \cdot C \cdot |\xi|^2} \cdot e^{\imath \, 2\pi \, \xi \cdot x} \, d\xi \right| &\geq \frac{1}{2} \int_{|\xi| < \sqrt{d}/2} e^{-\pi \cdot C \cdot |\xi|^2} \, d\xi \\ &\geq \frac{1}{2} e^{-\pi \cdot C \cdot d/4} \cdot \underbrace{\sqrt{d}^d}_{\geq 1} \\ &\geq \frac{1}{2} \cdot \left( e^{-\pi \cdot C/4} \right)^d \end{align*}$$
for all $|x|<c \sqrt{d}$.
