Let's begin with some theory on the renewal process.
In a renewal process $N(t)$, let $t$ denote the interarrival time, and $f(t)$ and $F(t)$ denote the PDF and CDF respectively. Let $M(t)=E[N(t)]$, and $M(t)$ is known to be the renewal function, which satisfies the following renewal equation:
\begin{equation} M(t)=F(t)+\int_0^tM(t-x)dF(x) ~~~~~~~~~~~~~~~~~~~(1) \end{equation}
We know that $M(t)=\sum_{n=1}^\infty F^{*n}(t)$, where $F^{*n}(t)$ denotes the $n$-fold convolution. But the $n$-fold is generally difficult to calcualte, so we resort to another way. Taking Laplace transform on both sides of equation (1) we have (we use a tilde to indicate Laplace transform)
$$\tilde{M}(s)=\tilde{F}(s)+\tilde{M}(s)\tilde{F}(s)$$
Note that the $\int_0^tM(t-x)dF(x)$ part is actually the convolution of $M(t)$ and $F(t)$, so the Laplace transform of this term is $\tilde{M}(s)\tilde{F}(s)$. Then $$\tilde{M}(s) = \frac{\tilde{F}(s)}{1-\tilde{F}(s)}$$ At last we can calculate $M(t)$ be taking the inverse Laplace transform of $\tilde{M}(s)$.
The introduction ends
For the renewal function, we can have a more general form as follows:
$$g(t)=h(t)+\int_0^tg(t-x)dF(x)~~~~~~~~~~~~~~~~~~~(2)$$
Also by taking Laplace transform on both sides, we have the unique solution for $\tilde{g}(s)$
$$\tilde{g}(s)=\frac{\tilde{h}(s)}{1-\tilde{F}(s)}$$
And $g(t)$ can be derived accordingly.
Now here comes the question, in Equation (2), if we chance the range of the integral from $[0, t]$ to $[0, \tau]$, where $\tau$ is a constant, i.e.,
$$g(t)=h(t)+\int_0^\tau g(t-x)dF(x)~~~~~~~~~~~~~~~~~~~(3)$$
In what means can I solve this euqation and get the exact expression of $g(t)$?
(As far as I can see, since I change the integral, $\int_0^\tau g(t-x)dF(x)$ is not a strict Laplace transform anymore. How would this change affect the way to solve the euqation?)
Let $H:x\mapsto F(\min(x,\tau))$, then $\mathrm dH(x)=\mathbf 1_{x\leqslant\tau}\,\mathrm dF(x)$ hence (3) reads $$g(t)=h(t)+\int_0^\infty g(t-x)\,\mathrm dH(x).$$ It follows that $$\tilde{g}(s)=\frac{\tilde{h}(s)}{1-\tilde{H}(s)},$$ where $$\tilde{H}(s)=\int_0^\infty\mathrm e^{-sx}\,\mathrm dH(x)=\int_0^\tau\mathrm e^{-sx}\,\mathrm dF(x).$$