Fourier transform and approximate identity on $L^1$ I've got a question about something I don't succeed to do. We consider $s>0$ : $G_s(x) = \frac{1}{\sqrt(s)}e^{\frac{-\pi x^2}{s}}$. We can verify that it's an approximate identity on $L^1$. Now, we want to calculate the fourier transform of $G_s$.
So I have :
$\hat{G}(s)(x) = \displaystyle \int_{\mathbb{R}} G_s(t)e^{-itx} \, \mathrm{d}t = \frac{1}{\sqrt{s}}e^{-\frac{x^2s}{4\pi}} \displaystyle \int_{\mathbb{R}} e^{-\frac{\pi}{s}(t-\frac{isx}{(\pi)^2})^2} \, \mathrm{d}t$
I would like to do the change of variable $u=t-\frac{isx}{\pi^2}$, because then I can conclude, but I of course I can do with integreal on $\mathbb{R}$, and with line integral, it gave me nothing... But, what happen if I say that $[-R;R]$ and $[-R-\frac{isx}{\pi^2}; R-\frac{isx}{\pi^2}\ ]$ are two homotopic path ? Because as then the integral are the same cause the function on the integral is holomorphic, and then I have : $\hat{G_s}(t) = \frac{1}{\sqrt{s}}e^{-\frac{x^2s}{4\pi}}$ ?
Then, I've tried to do that with the theory of holomorphic function and line integral, but It didn't worked. So, I don't know how to calculate this function... 
If someone could help me, thank you very much !
 A: 
But, what happen if I say that $[-R;R]$ and $[-R-\frac{isx}{\pi^2};
 R-\frac{isx}{\pi^2}\ ]$ are two homotopic path ? Because as then the
   integral are the same cause the function on the integral is
   holomorphic, and then I have : $\hat{G_s}(t) =
 \frac{1}{\sqrt{s}}e^{-\frac{x^2s}{4\pi}}$ ?

No, you cannot draw that conclusion, because the two paths have different endpoints. Otherwise, by the same reasoning you might also conclude that $\int_0^1 zdz=\int_1^2zdz$, which is clearly not true.
However, your conclusion holds for the limit $R\to \infty$, that is, the integrals on $(-\infty-\frac{isx}{\pi^2},
 +\infty-\frac{isx}{\pi^2}\ )$ and $(-\infty,+\infty)$ agree. To see this, consider for any $R>0$ the circuit given by the edges of the rectangle
$$[-R,R]\times \left[-\frac{isx}{\pi^2},0\right]$$
The integral of $G_s(x)$ over this circuit is identically zero because it is an integral of a holomorphic function over a closed curve.
As $R\to \infty$, you will find that the integral on the vertical sides vanish, therefore the integral on the horizontal sides, which are precisely $\left(-\infty-\frac{isx}{\pi^2},
 +\infty-\frac{isx}{\pi^2}\right )$ and $(-\infty,+\infty)$, must be the same.
In general with your convention for the Fourier transform (non-unitary, angular frequency) we have
$$f(x)=e^{-\alpha x^2}\Rightarrow \hat{f}(x)=\sqrt{\frac{\pi}{\alpha}}e^{-\frac{x^2}{4\alpha}} $$
so in your case $\hat{G}_s(x)=e^{-\frac{x^2s}{4\pi}}$.
