# Fourier transform - Evans - help with understanding a step

I'm having trouble understanding a step in Evans' book Partial Differential Equations.

In section 4.3, page 187 (2nd edition), it says

DEFINITION. If $u \in L^1(\mathbb{R}^n)$, we define its Fourier Transform $\mathcal{F}u = \hat{u}$ by $$\hat{u}(y) := \frac{1}{(2 \pi)^{n/2}} \int_{\mathbb{R}^n} e^{-ix\cdot y}u(x)dx\qquad (y \in \mathbb{R^n})$$ and its inverse Fourier transform $\mathcal{F}^{-1}u = \check{u}$ by $$\check{u}(y) := \frac{1}{(2 \pi)^{n/2}} \int_{\mathbb{R}^n} e^{ix\cdot y}u(x)dx\qquad (y \in \mathbb{R^n})$$

He then goes on to say, on page 188:

(...) as we will explicitly compute below in Example 1, $$\int_{\mathbb{R}^n} e^{ix\cdot y - t |x|^2}dx = \left( \frac{\pi}{t} \right)^{n/2} e^{-\frac{|y|^2}{4\epsilon}} \qquad (t > 0)$$ Consequently if $\epsilon > 0$ and $v_\epsilon(x) := e^{-\epsilon |x|^2}$, we have $\hat{v}_\epsilon (y) = \frac{e^{-\frac{|y|^2}{4\epsilon}}}{(2\epsilon)^{(n/2)}}$.

What I don't get is, in this last part, isn't he calculating $\check{v}_\epsilon$, rather than $\hat{v}_\epsilon$?

I know this must be really simple but I'm stuck on it. I'd appreciate any help.

Strictly speaking, yes, but notice that the sign of $y$ turns out not to matter in this case (this definition of the Fourier transform acting on even functions with enough regularity is its own inverse, as may be seen by setting $u=−x$).