Consider the function $f(t)$ defined such that $f(t)=0$ for $t<0$ or $t>c$ and $f(0)=1$, $f(c)=-1,$ for some $c>0$.
I want to compute the Laplace transform of $f(c-t)$.
This is given as $$\mathcal{L_{f(c-t)}(p)}=\int_0^{\infty}f(c-t)e^{-pt} \, dt.$$
I find that \begin{align*} \mathcal{L}_{f(c-t)}(p)&=\int_0^\infty f(c-t) e^{-pt} \,dt \\ & = -\int_{c}^\infty f(z)e^{-p(c-z)} \, dz\end{align*}
after making the substitution $z=c-t.$
This can be rewritten as \begin{align*} \mathcal{L}_{f(c-t)}(p) &= -e^{cp}\int_c^\infty f(z)e^{pz} \, dz \\ & = -e^{-cp} \int_0^\infty f(z)e^{pz} \, dz -e^{-cp} \int_c^0 f(z)e^{pz} \, dz \\ & = -e^{-cp}\mathcal{L}_{f(t)}(-p)-e^{-cp} \int_c^0 f(z)e^{pz} \, dz\end{align*}
I'm told that the answer should be $$-e^{-cp}\mathcal{L}_f(-p)-e^{-cp}$$
I'm not sure how to get to that from this point though.
If $z=c-t$, then $z$ should range from $c$ to $-\infty$, not $+\infty$, and there's no way to fix the argument after that.
The trick to making this work is that the integrand (call it $g(t)$) is zero outside a finite interval. Outside that interval, it doesn't matter whether you integrate out to $+\infty$ or $-\infty$ or both, it's still the same number. So the argument will look something like
$$ \int_0^\infty g(t)\,dt = \int_0^c g(t)\,dt = \int_{-\infty}^c g(t)\,dt = \int_0^\infty g(c-z)\,dz $$
except with the actual integrand thrown in.