Fourier Transform Confusion According to Wikipedia, the Fourier transform $\hat f$ of an integrable function $f:\mathbb{R}\to\mathbb{C}$ is defined by
$$
\hat f\left(\xi\right)=\int_{-\infty}^{\infty}f\left(x\right)e^{-2\pi ix\xi}dx.\tag{1}
$$
Consider $f\left(x\right)=e^{-x^2}$. Then, using $\left(1\right)$, I obtain
$$\hat f\left(\xi\right)=e^{-\left(\pi\xi\right)^2}\sqrt\pi.$$
However, WolframAlpha computes
$$
\mathcal F_x\left[e^{-x^2}\right]\left(\omega\right)=\frac{e^{-\frac{\omega^2}{4}}}{\sqrt 2},
$$
the back of my PDEs book yields
$$
\mathcal F\left(e^{-x^2}\right)\left(\xi\right)=\frac{e^{-\frac{\xi^2}{4}}}{2\sqrt\pi},
$$
and my professor asserts
$$
\mathcal F\left(e^{-x^2}\right)\left(\xi\right)=\left(2\pi\right)^\frac12e^{-\frac{\xi^2}2}.
$$
What in the world is going on!?
 A: In different settings, the Fourier Transform uses different characters, $e^{-\lambda ix\xi}$, with different values of $\lambda$. Normalized so that $\mathcal{F}_\lambda(\mathcal{F}_\lambda(f))(x)=f(-x)$, the Fourier Transform is
$$
\mathcal{F}_\lambda(f)(\xi)=\sqrt{\frac{\lambda}{2\pi}}\int_{-\infty}^\infty f(x)\,e^{-\lambda ix\xi}\,\mathrm{d}x\tag{1}
$$
Applying $(1)$ to the function $e^{-\alpha x^2}$ yields
$$
\begin{align}
\mathcal{F}_\lambda\left(e^{-\alpha x^2}\right)(\xi)
&=\sqrt{\frac{\lambda}{2\pi}}\int_{-\infty}^\infty e^{-\alpha x^2}\,e^{-\lambda ix\xi}\,\mathrm{d}x\\
&=\sqrt{\frac{\lambda}{2\pi}}\,e^{-\frac{\lambda^2}{4\alpha}\xi^2}\color{#C00000}{\int_{-\infty}^\infty e^{-\alpha(x+\frac{\lambda}{2\alpha} i\xi)^2}\,\mathrm{d}x}\\
&=\sqrt{\frac{\lambda}{2\pi}}\,e^{-\frac{\lambda^2}{4\alpha}\xi^2}\color{#C00000}{\int_{-\infty}^\infty e^{-\alpha x^2}\,\mathrm{d}x}\\
&=\sqrt{\frac\lambda{2\alpha}}\,e^{-\frac{\lambda^2}{4\alpha}\xi^2}\tag{2}
\end{align}
$$
Where the equality of the red integrals is assured by contour integration over a very wide rectangle.
Usually, either $\lambda=2\pi$, so that $\sqrt{\frac{\lambda}{2\pi}}=1$ in $(1)$, or $\lambda=1$, to simplify the character. Using $(2)$ with $\alpha=1$, we can determine which value of $\lambda$ is used in each case cited in the question.
In the first case, $\lambda=2\pi$ is used:
$$
\mathcal{F}_{2\pi}\left(e^{-x^2}\right)(\xi)=\sqrt{\pi}\,e^{-\pi^2\xi^2}\tag{3}
$$
Wolfram Alpha seems to use $\lambda=1$:
$$
\mathcal{F}_1\left(e^{-x^2}\right)(\xi)=\sqrt{\tfrac12}\,e^{-\xi^2/4}\tag{4}
$$
The PDE book uses $\lambda=1$, but seems to have chosen an asymmetric Fourier Transform, where the Fourier Transform takes the factor $\frac1{2\pi}$ instead of $\frac1{\sqrt{2\pi}}$ in $(1)$, and the Inverse Fourier Transform takes the factor $1$ instead of $\frac1{\sqrt{2\pi}}$. Thus, their answer is $(4)$ divided by $\sqrt{2\pi}$.
The form given by your professor would need to use $\lambda=\sqrt{2}$ and a factor of $\frac1{\sqrt{\pi}}$ instead of $\frac1{\sqrt[4]{2\pi^2}}$. My guess is that this is not what was intended. However, it looks somewhat like the pdf for Normal Distribution with variance $1$ and mean $0$.
A: What is going on is that there are different conventions for the placement of $2\pi$. You stick it as a factor in the forward Fourier transform, or the inverse transform, or you use a factor $\sqrt{2\pi}$ in both, or you put it in the exponent, or you heroically try to hide it from view altogether by changing the measure itself, as Rudin famously does in one of his books. The bottom line is, the theory of the Fourier transform needs $2\pi$ somewhere,and there is no universal agreement on where it should go.
