Exam study haha
2. Considerf(t) = e^at for t > 0, where 'a' is a constant, then
F(s) = Integral(from 0 to infinity) e^(−st)e^(at)dt
and this integral exists fors > a, so that
L(eat) = 1/ (s-a)
for s > a.
I would have said 1/ (a-s)
Why is it not so?
Let u= (a-s)t so that du= (a-s)dt and du/(a-s)= dt. Now the integral is but we have to be careful about the limits of integration. At t= 0, u= (a-s)t= (a-s)0= 0 but if s> a, a-s< 0 so the upper limit will be .
That is, the integral is now . Integrating that gives, again, an exponential which will be 0 at the upper limit. The integral is .