I am trying to prove the 3D Diffusion Equation $$\begin{cases}u_t(\vec x,t) &= c\nabla u\\ u(\vec x,0)&=g(\vec x)\end{cases}$$ From the 3D Fourier Transform, where $\vec k,\vec x\in\mathbb R^3$.
I took the Fourier Transform of both sides of the 3D Diffusion to get:
$$\widehat u_t(k,t)+c|\vec k|^2\widehat u(k,t) = 0$$
Now, this gives that the solution to the ODE is: $$\widehat u(k,t) = \widehat g(k) e^{-ct|\vec k|^2} = \widehat g(\vec k)\cdot\widehat f(\vec k)$$
Now, to find the solution, I want to find the Inverse FT of $f$, whcih leads to: $$\frac{1}{(2\pi)^3}\iiint_{\mathbb R^3}e^{-ct|\vec k|^2+ikx} d\vec k$$ So isn't $|\vec k|^2$ a scalar? If not, how would I take the integral of the integrand?
$|k|^2$ is a scalar, but $\vec k$ is a vector. Note that also, in your last equation, $x$ should be $\vec x$, so the last expression is $$\frac{1}{2\pi}\iiint_{\mathbb R^3}e^{-ct|\vec k|^2+i\vec k\vec x} d\vec k$$ You can either solve the integral in Cartesian coordinates, or change it to spherical coordinates. $|\vec k|^2$ will transform into $k^2$, $\vec k \vec x$ is $kx\cos\theta$, and $d\vec k$ is $k^2\sin\theta dk d\theta d\phi$. The integration limits for $\phi$ is from $0$ to $2\pi$, $theta$ goes from $0$ to $\pi$, and $k$ is from $0$ to $\infty$