Computing partial derivative of double integral 
Let $$u(x,t) = {1\over 2c}\left(\int_{0}^{t-{x\over c}}\int_{-cs+ct-x}^{-cs+x+ct}f(y,s)\ dyds+\int_{t-{x\over c}}^t\int_{cs+x-ct}^{-cs+x+ct}f(y,s)\ dyds\right)$$
where $f$ is some $C^2$ function. Show that $u$ satisfies $u_{tt} - c^2 u_{xx} = f(x,t)$.

In the computation, I droped ${1\over 2c}$ and computed only double integral part. I tried to compute this using Leibniz rule basically but I'm not sure I applied correctly. I'll only compute $u_{xx}$ to check if my computation is right.
First, let
$$F(x,s):= \int_{-cs+ct-x}^{-cs+x+ct}f(y,s)\ dy,\ \tilde{F}(x,s): = \int_{cs+x-ct}^{-cs+x+ct}f(y,s)\ dy.$$
Then
$$u(x,t) = \int_0^{t-{x\over c}}F(x,s) ds + \int_{t-{x\over c}}^t \tilde{F}(x,s)ds.$$
Now,
\begin{align*}
u_x(x,t) = -{1\over c}F(x,t-{x\over c}) + \int_0^{t-{x\over c}}f(-cs+x+ct,s)+f(-cs+ct-x,s)ds + {1 \over c}\tilde{F}(x,t-{x\over c})+\int_{t-{x\over c}}^t f(-cs+x+ct)-f(cs+x-ct)ds
\end{align*}
\begin{align*}
u_{xx}(x,t) = -{1\over c}(f(-cs+x+ct,t-{x\over c})+f(-cs+ct-x,t-{x\over c}))+\int_{-cs+ct-x}^{-cs+x+ct} {\partial\over\partial x}f(y,t-{x\over c}) dy + f(-cs+x+ct,t-{x\over c})(-{1\over c})+f(-cs+ct-x,t-{x\over c})(-{1\over c})+\int_0^{t-{x\over c}}{\partial\over\partial x}f(-cs+x+ct,s)-{\partial\over \partial x}f(-cs+ct-x,s)ds +{1\over c}(f(-cs+x+ct,t-{x\over c})-f(cs+x-ct,t-{x\over c})-\int_{cs+x-ct}^{-cs+x+ct}{\partial\over\partial x}f(y,t-{x\over c})dy) + f(-cs+x+ct,t-{x\over c}){1\over c}-f(cs+x-ct,t-{x\over c}){1\over c}+\int_{t-{x\over c}}^t{\partial\over\partial x}f(-cs+x+ct,s)-{\partial\over\partial x}f(cs+x-ct,s)ds
\end{align*}
Am I doing correctly?
 A: Let $\left({\lambda} , {\mu}\right) = \left(x-c t , x+c t\right)$. One has
\begin{equation}{u}_{t t}-{c}^{2} {u}_{x x} =-4 {c}^{2} \frac{{\partial }^{2} u}{\partial  {\lambda} \partial  {\mu}}\end{equation}
The integral can be written
\begin{equation}2 c u = \int_{0}^{{-\frac{{\lambda}}{c}}}\left(\int_{{-c} s-{\lambda}}^{{-c} s+{\mu}}f \left(y , s\right) d y\right) d s+\int_{{-\frac{{\lambda}}{c}}}^{\frac{{\mu}-{\lambda}}{2 c}}\left(\int_{c s+{\lambda}}^{{-c} s+{\mu}}f \left(y , s\right) d y\right) d s\end{equation}
By Leibnitz' rule
\begin{equation}2 c \frac{\partial  u}{\partial  {\mu}} = \int_{0}^{{-\frac{{\lambda}}{c}}}f \left(\mu - {c} s , s\right) d s+\int_{{-\frac{{\lambda}}{c}}}^{\frac{{\mu}-{\lambda}}{2 c}}f \left({\mu}- c s , s\right) d s+0
=\int_{0}^{\frac{{\mu}-{\lambda}}{2 c}}f \left({\mu}-c s , s\right) d s\end{equation}
Hence
\begin{equation}2 c \frac{{\partial }^{2} u}{\partial  {\lambda} \partial  {\mu}} =-\frac{1}{2 c}f \left(\frac{{\lambda}+{\mu}}{2} , \frac{{\mu}-{\lambda}}{2 c}\right) =-\frac{1}{2c} f \left(x , t\right)\end{equation}
Hence
\begin{equation}
{u}_{t t}-{c}^{2} {u}_{x x} = f \left(x , t\right)
\end{equation}
