Using the Fourier transform, solve the temperature problem in the upper half-plane described by:
$$u_t=ku_{xx}, \; \; u=u(x,t), \; \; -\infty < x < \infty, \; t \; \geq 0, \; \; \text{u is bounded}$$ $$u(x,0)=e^{-2x^2}$$
I was given a hint, which was to first formulate a general theorem about the boundary value problems of the form
$$u_t=ku_{xx}, \; \; u=u(x,t), \; \; -\infty < x < \infty, \; t \; \geq 0, \; \; \text{u is bounded}$$ $$u(x,0)=f(x)$$
It has also been given that $\mathcal{F}\left\{\exp\left(\frac{-x^2}{2\sigma^2}\right)\right\}=\sigma\sqrt{2\pi}\exp\left(\frac{-\sigma^2\xi^2}{2}\right)$.
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My attempt at the general theorem:
So far using separation of variables I have determined that $$\hat{u}(\xi,t)=\hat{u}(\xi,0)\exp\left(-\xi^2kt\right)$$ From the Fourier transform, we have $$ \hat{u}(\xi,t)=\mathcal{F}\left\{u(x,t)\right\}=\int_{-\infty}^\infty u(x,t)\exp(i\xi x)dx$$
which tells us that
$$\hat{u}(\xi,0)=\int_{\infty}^\infty f(x)\exp(i\xi x)dx$$
I believe from the Fourier inversion theorem: $$u(x,t)=\frac{1}{2\pi} \int_{-\infty}^\infty \hat{u}(\xi,t)\exp\left(-i\xi x \right)d\xi$$
$$\; =\frac{1}{2\pi} \int_{-\infty}^\infty\hat{u}(\xi,0)\exp\left(-i\xi x-\xi^2kt\right)d\xi$$
Hence
$$u(x,t)=\frac{1}{2\sqrt{2\pi}}\int_{-\infty}^\infty\exp\left( \frac{-\xi ^2}{8}\right)\exp\left(-i\xi x-\xi^2kt\right)d\xi$$
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This https://web.math.ucsb.edu/~helena/teaching/math124b/heat.pdf leads me to believe I've done something wrong. Specifically the coefficient of the integral in the final expression for $u(x,t)$. Their denominator is $\frac{1}{\sqrt{2kt}}$ times mine (for the general theorem case). Have I done something wrong?