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 $0$.