Solution of the PDE $u_{tt}-\alpha u_{xt}=0$ Consider the PDE $$u_{tt}-\alpha u_{xt}=0$$ with $u(x,0)=\phi(x)$ and $u_t(x,0)=\psi(x).$
I'm told that a solution to this PDE is $$u(x,t)=\phi(x)+\frac{1}{\alpha}\int_x^{x+\alpha t}\psi(s) \, ds$$
Now I'm told that $\phi(x)=0$ and $\psi(x)=\delta(x)$ where $\delta(x)$ is the delta function. I'm asked to express the solution in terms of the Heaviside function.
Attempt:
\begin{align*} u(x,t) & = \frac{1}{\alpha}\int_x^{x+\alpha t}\delta(s) \, ds \\ 
& = \frac{1}{\alpha}\left(\int_x^{-\infty} \delta(s) \, ds + \int_{-\infty}^{x+\alpha t}\delta(s) \, ds \right) \\ & = \frac{1}{\alpha}\left(\int_{-\infty}^{x+\alpha t}\delta(s) \, ds-\int_{-\infty}^{x} \delta(s) \, ds\right)\\ & = \frac{1}{\alpha}\left( H(x+\alpha t)-H(x) \right)\end{align*}
Is this correct?
I'm also asked to find the limits as $\alpha \to 0$ and as $\alpha \to \infty$ of the general solution 
$$u(x,t)=\phi(x)+\frac{1}{\alpha}\int_x^{x+\alpha t}\psi(s) \, ds$$
How do I do this part?
 A: I can answer the second point, the limits. I've found that simply $c=\alpha$. Then, we can manage the general solution: First, let define $\Psi(s)$ as being a primitive of $\psi(s)$: $\Psi'(s)=\psi$(s), then,
$$\lim_{\alpha\to 0}\frac{1}{\alpha}\int_x^{x+\alpha t}\psi(s) \,\mathrm ds=\lim_{\alpha\to0}\frac 1\alpha\left(\Psi(x+\alpha t)-\Psi(x)\right)=$$
$$=\lim_{u\to0}\frac tu\left(\Psi(x+u)-\Psi(x)\right)=t\Psi'(x)=t\psi(x)$$
So, in the limit for $\alpha\to 0$,
$$u(x,t)=\phi(x)+t\psi(x)$$
Not completely sure for the limit $\alpha\to\infty$, but I can try. First, it's supposed for $\psi(s)$ some kind of "nice" asymptotic behaviour. I consider that $\Psi(s)=\int_0^s\psi(s)\mathrm ds\to C$ as $s\to\infty$ ($\psi\in L^1(\mathbb R)$ or as much as necessary)
$$\lim_{\alpha\to\infty}\frac{1}{\alpha}\int_x^{x+\alpha t}\psi(s) \,\mathrm ds=\lim_{\alpha\to\infty}\frac 1\alpha\left(\Psi(x+\alpha t)-\Psi(x)\right)=$$
$$=\lim_{\alpha\to\infty}\frac{\Psi(x+\alpha t)}{\alpha}+\lim_{\alpha\to\infty}\frac{\Psi(x)}{\alpha}=0$$
In the limit,
$$u(x,t)=\phi(x)$$
