Continuous set of eigenvalues-the solution is $u(x,t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty}{\widetilde{f}e^{ikx} e^{-k^2t}}dk$ $$u_t=u_{ xx}, x \in \mathbb{R},t>0  $$
$$u(x,0)=f(x)$$
The solution is of the form $u=X(x)T(t)$.
Firstly we have to solve the problem $X''+ \lambda X=0$:
$\lambda=-k^2$ The solutions from this case go to infinity when $x \rightarrow \infty$
$\lambda=0$ The only solution that I want from this case is $X=c_0$
$\lambda=k^2$ $X(x)=c_1e^{ikx}+c_2e^{-ikx},k>0$. So $X_k=e^{ikx},k \in \mathbb{R} $
The solution of the problem $T'+k^2T=0$ is $T_k(t)=e^{-k^2t}$
So $$u_k(x,t)=X_k T_k=e^{ikx}e^{-k^2t}, k \in \mathbb{R}$$
$$u(x,t)=\int_{-\infty}^{\infty}{\widetilde{f}(k)X_k T_k}dk$$
So $$u(x,t)=\frac{1}{2 \pi} \int_{-\infty}^{\infty}{\widetilde{f}(k)e^{ikx} e^{-k^2t}}dk$$
where do we have $\frac{1}{2 \pi}$ from?? From the eigenfunction $X=c_0$??     

$\ddot \smile$
Both forms are equivalent since you can choose $\tilde f$ to contain an extra constant.
It will have been rewritten like that so that it has the form of an inverse fourier transform. For $t=0$ the inverse fourier transform maps exactly to $f$, which is the boundary condition.
