I found the following in my lecture notes:
$$u_t=u_{xx}, x \in \mathbb{R}, t>0 \\ u(x,0)=f(x)$$
$$u(x,t)=X(x)T(t)$$
$$\Rightarrow \frac{T'(t)}{T(t)}=\frac{X''(x)}{X(x)}=-\lambda \in \mathbb{R}$$
$$X''(x)+\lambda X(x)=0, x \in \mathbb{R}$$ $$X \text{ bounded }$$
The characteristic equation is $\rho^2+ \lambda=0 \Rightarrow \rho^2=-\lambda$
- $\lambda<0$: $X(x)=c_1e^{ \sqrt{- \lambda}x}+c_2 e^{- \sqrt{- \lambda}x} $
So that $X$ is bounded $\Rightarrow c_1=c_2=0$.
- $\lambda=0$: $X(x)=c_1x+c_2 \overset{c_2=0}{\Rightarrow } X_0(x)=1$
$\lambda>0 \Rightarrow \rho^2= \lambda i^2 \Rightarrow \rho= \pm i \sqrt{\lambda}$
$\Rightarrow X(x)=c_1 \cos(\sqrt{\lambda}x)+ c_2 \sin (\sqrt{\lambda} x), x \in \mathbb{R}$
$T'(t)+ \lambda T(t)=0$
$\Rightarrow T(t)=c e^{- \lambda t}$
$$u_{\lambda}(x,t)=e^{- \lambda t} \cos(\sqrt{\lambda} x) \\ e^{- \lambda t} \sin(\sqrt{\lambda}x)$$
General solution:
$$u_1(x,t)= \int_0^{\infty} c_1(\lambda) e^{- \lambda t} \cos(\sqrt{\lambda}x)d \lambda$$
$$u_2(x,t)= \int_0^{\infty} c_2(\lambda) e^{- \lambda t} \sin(\sqrt{\lambda}x)d \lambda$$
$$u(x,t)= \int_0^{\infty} c(\lambda) e^{- \lambda t} e^{i \sqrt{\lambda} x} d \lambda$$
$$( c_1, c_2: [0,+\infty) \to \mathbb{R}, c:[0,+\infty) \to \mathbb{C}$$
Initial conditions:
$$u(x,0)=f(x) \Leftrightarrow f(x)= \int_0^{\infty} c(\lambda) e^{i \sqrt{\lambda}x} d \lambda$$
Could you explain me why in this case the general solution is of the form $u(x,t)= \int_0^{\infty} c(\lambda) e^{- \lambda t} e^{i \sqrt{\lambda} x} d \lambda$?
Why isn't it $u(x,t)= \frac{a_0}{2}+\sum_{k=1}^{\infty} (a_k \cos (kx)+ b_k \cos (kx))$ ?
This is because there are no boundary conditions specified. The $\lambda$ can only be determined by the boundary conditions. With the two boundaries $0,L$, you can get the discrete form of sine and cosine functions, namely, $\lambda=\frac{n\pi}{L}$.