I believe I have found the Fourier transform of $f_c(x)=\exp(-cx^2)$ (where $0<c<\infty$) by noting first that $f_c'(x)=-2cx\exp(-cx^2)$. Taking the Fourier transform of both sides of this equation leads to
$(i\cdot t)\hat{f}_{c}(t)=-2c\cdot i\cdot\hat{f}_{c}'(t)$.
Letting $\phi(t):=\hat{f}_c$, I get that the above inequality can be written as
$\phi'(t)+\frac{t}{2x}\phi(t)=0$
Using $M(t)=\exp\Big(\frac{t^2}{4c}\Big)$ as an integrating factor, I can solve this ODE to be
$\phi(t)=K\exp(-t^2/4c)$
Where
$K=\phi(0)=\hat{f}_c(0)=\int_\mathbb{R}\exp(-cx^2)\cdot\exp(0)dx=\sqrt\frac{\pi}{c}$
giving me that
$\hat{f}_c(t)=\sqrt\frac{\pi}{c}\exp(-t^2/4c)$
If this is correct, my next task is to show that there exists a unique $c$ such that $\hat{f}_c=f_c$, but I don't see how this is possible considering it would mean that $c=\pi$ and $c=\frac{1}{4c}\implies c=\frac{1}{2}$ simultaneously. Can anyone point out what I am doing incorrectly? Thanks in advance!
You cannot have $\hat{f}=f$ unless you have normalized the Fourier transform so that it is isometric., meaning that $\|\hat{f}\|=\|f\|$, where $\|\cdot\|$ is the $L^{2}(\mathbb{R})$. That means $$ \hat{f}(s) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}f(t)e^{-ist}\,dt. $$