Problem with Heat Equation and Laplace Transform So, my problem is this: I have to solve the heat equation with this boundary condition:
$$
\begin{cases}
\dfrac{\partial}{\partial t} u(x, t) - D\ \dfrac{\partial^2}{\partial x^2} u(x, t) = S_0\delta(x)\delta(t) \\\\
u(x, 0) = U_0\delta(x)
\end{cases}
$$
Where $D, U_0, S_0$ are coefficients.
So the first step was to perform a Fourier transform on the whole equation (on $x \to k$) to get 
$$\hat u (k, t) + D k^2 \hat u(k, t) = S_0 \delta(t)$$
And it's ok. 
But then in the paper it's written that a Laplace transform (on $t\to s$) is performed, getting:
$$s\ \hat{\tilde u}(k, s) - U_0 + D k^2 \hat{\tilde u} (k, s)= S_0$$
Where of course $s$ (and before: $k$) are the new variables obtained because of the transformations.
My question is: Where does the $U_0$ term comes from, in the Laplace transform?
I cannot find a way to get it, I don't understand how it comes out.
EDIT
Ok, it comes out by LT. But I got stuck in computing it. Shalln't it be also an anti transform? Anybody can help me with the calculation?
 A: Laplace transforming your equation 
$$\hat{u}_{t}(k,t) + D k^{2} \hat{u}(k,t) = S_{0} \delta(t)$$
gives
\begin{align}
\int_{0}^{\infty} \hat{u}_{t} e^{-st} dt + D k^{2} \underbrace{\int_{0}^{\infty} \hat{u} e^{-st} dt}_{D k^{2} \tilde{\hat{u}}} &= S_{0} \underbrace{\int_{0}^{\infty} \delta(t) e^{-st} dt}_{1} \\ \\
\implies \hat{u} e^{-st} \bigg|_{0}^{\infty} + s \underbrace{\int_{0}^{\infty} \hat{u} e^{-st} dt}_{\tilde{\hat{u}}} + D k^{2} \tilde{\hat{u}} &= S_{0}
\end{align}
where we used integration by parts on the first integral with $v = e^{-st}$, $w' = \hat{u}_{t}$. Then note that 
\begin{align}
\lim_{t \to \infty} \hat{u}(k,t) e^{-st} &\to 0 \\
\hat{u}(k,t) e^{-st} \bigg|_{t = 0} &= \hat{u}(k,0) \\
\end{align}
But $\hat{u}(k,0)$ is just the Fourier transform of our initial condition. The Fourier transform of $\delta(x)$ is $1$ (depending on the convention, I think the transform you are using might have a scaling in front in which case it should actually be $1/\sqrt{2 \pi}$ or $1/2 \pi$). So we get
\begin{align}
\hat{u}(k,0) &= U_{0} \\
\implies -U_{0} + s \tilde{\hat{u}} + D k^{2} \tilde{\hat{u}} &= S_{0}
\end{align}
