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Please ignore this thread.
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Sorry for wasting your time, but I already posted:It would have helped if you stated Theorem 4.3.3. That said, we could try it without any special theorem. We have
##Lf = \int_0^{\infty} e^{-st}f(t)dt ##
which you can split up into the sum of integrals from (0,a) (a,2a) ... ([n-1]a,na) (na,t), where na < t < (n+1)a.
Each of these is easy enough to evaluate, but be careful to separate out the case where n is even and n is odd (why?). I'll bet you can factor some stuff out of the resulting sum.
Why don't you try this out? And why don't you quote theorem 4.3.3 -- maybe it's an easier way to do things.
Actually, I think the solutions manual is wrong. Please ignore this thread.