The D'Alembert solution of the wave equation $$u_{tt}-c^2u_{xx}=0, x \in \mathbb{R}, t>0$$ $$u(x,0)=\phi(x), x \in \mathbb{R}$$ $$u_t(x,0)=\psi(x), x \in \mathbb{R}$$
is the following:
$$u(x,t)=\frac{1}{2}\begin{bmatrix} \phi(x-ct)+\phi(x+ct) \end{bmatrix}+\frac{1}{2c} \int_{x-ct}^{x+ct} \psi(\tau)d \tau$$
I am looking at the proof of this solution, which is the following:
The general solution of the wave equation is given from: $\displaystyle{u(x,t)=f(x+ct)+g(x-ct)} \ \ \ \ \ (1)$
$\displaystyle{u(x,0)=f(x)+g(x)=\phi(x)} \ \ \ \ \ \ (2)$
$(1): \displaystyle{u_t(x,t)=c(f'(x+ct)-g'(x-ct))} \ \ \ \ \ (3)$
$\displaystyle{u_t(x,0)=c(f'(x)-g'(x))=\psi(x)} \ \ \ \ \ \ (4)$
$(2): \displaystyle{f'(x)+g'(x)=\phi'(x)} \ \ \ \ \ (5)$
From the relations $(4)$ and $(5)$ we get:
$$f'(x)=\frac{1}{2}\begin{bmatrix} \phi'(x)+\frac{\psi(x)}{c} \end{bmatrix}$$ $$g'(x)=\frac{1}{2}\begin{bmatrix} \phi'(x)-\frac{\psi(x)}{c} \end{bmatrix}$$
Integrating these equations we get:
$$f(s)=\frac{1}{2}\begin{bmatrix} \phi(s)+\frac{1}{c} \int_0^s \psi(\tau)d \tau \end{bmatrix}+a$$ $$g(s)=\frac{1}{2}\begin{bmatrix} \phi(s)-\frac{1}{c} \int_0^s \psi(\tau)d \tau \end{bmatrix}+b$$
Why are the limits of the integral $0$ and $s$, and not $-\infty$ and $+\infty$??
This is the fundamental theorem of calculus.
Under reasonable conditions, ${d\over dx}\int_a^xf(x')dx' = f(x)$.
Think about what happens as you vary $x$ to $x+\delta x$. The area under the curve gets an extra little sliver with a width of $\delta x$ and a height of $f(x)$. So the rate of change of this integral with respect to the right endpoint is the height of the integrand at the endpoint.
It would be bizarre if the integrand were between fixed limits. You'd be saying that the derivative of something that didn't depend on $s$ depends on $s$.