Show that $\widehat f(\xi)=f(\xi)~,$ where $\widehat f$is the Fourier transform of $f.$ 
Let $f(x)=e^{-\pi|x|^{2}}~$ be defined on ${\bf R}^{n}$. 
Show that $\widehat f(\xi)=f(\xi)~,$ where $\widehat f$is the Fourier transform of $f.$ 

Here's my working :
First we observe that the function $s\longmapsto \displaystyle\int_{\bf R}e^{-\pi(x+is)^{2}}~dx~$ is a constant ,where $s\in{\bf R}$ and $i^{2}=-1.$Indeed , the derivative of this function is $0$ on ${\bf R}$ i.e.
\begin{align}
\frac{d}{ds}\bigg(\int_{\bf R}e^{-\pi(x+is)^{2}}~dx\bigg)&=\int_{\bf R}\frac{d}{ds}\bigg(e^{-\pi(x+is)^{2}}\bigg)~dx\\
&=\int_{\bf R}-2\pi i(x+is)e^{-\pi(x+is)^{2}}~dx\\
&=i\int_{\bf R}-2\pi(x+is)e^{-\pi(x+is)^{2}}~dx\\
&=i\int_{\bf R}\frac{d}{dx}\bigg(e^{-\pi(x+is)^{2}}~\bigg)dx\\
&=0
\end{align}
,where the first equality can be deduced by Lebesgue's Dominated Convergence Theorem and the fifth equality applies the Fundamental Theorem of Calculus to obtain the value $0~.$
Therefore , we have  $\displaystyle\int_{\bf R}e^{-\pi(x+is)^{2}}~dx=\int_{\bf R}e^{-\pi x^{2}}~dx.$
Now we use this consequence to deduce that 
\begin{align}
\widehat f(\xi)&=\int_{{\bf R}^{n}}e^{-\pi|x|^{2}}e^{-2\pi i\langle~ x,~\xi~\rangle}~dx\\
&=\int_{{\bf R}^{n}} e^{-\pi\sum_{j=1}^{n}~~x_{j}^{2}~-2\pi i\sum_{j=1}^{n}~~x_{j}~\xi_{j}}~~dx\\
&=\int_{{\bf R}^{n}} e^{-\pi\sum_{j=1}^{n}~~x_{j}^{2}~-2\pi i\sum_{j=1}^{n}~~x_{j}~\xi_{j}~-\pi\sum_{j=1}^{n}~~(i\xi_{j})^{2}~+\pi\sum_{j=1}^{n}~~(i\xi_{j})^{2}}~~dx\\
&=\int_{{\bf R}^{n}}e^{-\pi\sum_{j=1}^{n}~~(x_{j}+i\xi_{j}~)^{2}-\pi\sum_{j=1}^{n}~~\xi_{j}^{2}}~~dx\\
&=\int_{{\bf R}^{n}}e^{-\pi\sum_{j=1}^{n}~~x_{j}^{2}~-\pi\sum_{j=1}^{n}~~\xi_{j}^{2}}~dx\\
&=\int_{{\bf R}^{n}}e^{-\pi|x|^{2}}e^{-\pi|\xi|^2}~dx\\
&=e^{-\pi|\xi|^{2}}\prod_{j=1}^{n}\bigg(\int_{{\bf R}}e^{-\pi x_{j}^{2}}~~dx_{j}\bigg)\\
&=e^{-\pi|\xi|^{2}}=f(\xi)
\end{align}  
, for $x=(x_{1},...,x_{n})~,\xi=(\xi_{1},...,\xi_{n})\in{\bf R}^{n}$and the seventh equality holds by Tonelli's theorem and deal with that integral by the well-known fact $\displaystyle \int_{\bf R} e^{-x^{2}}~dx=\sqrt{\pi}$
If you have the time please check my proof for validity . Otherwise , just ignore it , that is okay . Any valuable suggestion or comment will be appreciated .
 A: The line that
\begin{align*}
\int_{{\bf{R}}}\dfrac{d}{dx}(e^{-\pi(x+is)^{2}})dx=0
\end{align*}
follows by Fundamental Theorem of Calculus needs a little more explanation. Loosely speaking, one may perform like 
\begin{align*}
\int_{{\bf{R}}}\dfrac{d}{dx}(e^{-\pi(x+is)^{2}})dx&=\lim_{N\rightarrow\infty}\int_{-N}^{N}\dfrac{d}{dx}(e^{-\pi(x+is)^{2}})dx\\
&=\lim_{N\rightarrow\infty}(e^{-(N+is)^{2}}-e^{-(-N+is)^{2}})\\
&=0,
\end{align*}
but there are several issues about this reasoning. 
First, in what sense the so called the Fundamental Theorem of Calculus is? I meant, in classical mathematical analysis we are mostly dealing with real-valued function and I believe that the theorem is written for real-valued function either, so is it really the theorem we ought to use in the case of complex-valued?
The trick is the following. Write $f(x)=u(x)+iv(x)$ for real-valued functions $u,v$. A moment of thought leads you the fact that $f'(x)=u'(x)+iv'(x)$. Since we define that
\begin{align*}
\int f'(x)dx=\int u'(x)dx+i\int v'(x)dx,
\end{align*}
then we are ready to apply the Fundamental Theorem of Calculus to the real-valued functions $u,v$ to get
\begin{align*}
\int f'(x)dx=u(x)+iv(x)+\text{some boundary terms},
\end{align*}
so the alike looking Fundamental Theorem of Calculus for complex-valued functions
\begin{align*}
\int f'(x)dx=f(x)+\text{some boundary terms}
\end{align*}
is then established.
Now turn the issue about the limit of $e^{-(N+is)^{2}}-e^{-(-N+is)^{2}}$ as $N\rightarrow\infty$. This is a limit about complex numbers. To prove that, say, $e^{-(N+is)^{2}}\rightarrow 0$, one could do 
\begin{align*}
\left|e^{-(N+is)^{2}}\right|&=\left|e^{-N^{2}+s^{2}+-2Nsi}\right|\\
&=\left|e^{-2Nsi}\right|e^{-N^{2}+s^{2}}\\
&=e^{-N^{2}+s^{2}}\\
&\rightarrow 0.
\end{align*} 
This is by far very different to say that $N+is\rightarrow\infty$ so $e^{-(N+is)^{2}}\rightarrow 0$ because saying that $N+is\rightarrow\infty$ is completely outrageous. The point is that, in what sense we mean by $N+is\rightarrow\infty$? We do have the one point compactification in complex plane, but that $\infty$ is not the infinity that we mean in real analysis. We do have $N+is\rightarrow\infty$ in the complex compactification sense, still, this does not justify the consequence that $e^{-(N+is)^{2}}\rightarrow 0$.
