in doing the integral they play no role in the actual integration (other than getting in the
way!), which is basically an integration with respect to the variablet. If it helps, think ofa
andssimply as particular constants, say 3 and 4 respectively, while doing the integration.
Then you can concentrate on the actual job of integration and effectively the problem is to
integrate something likete−^4 t, for example. This is a classic case of integration by parts
(273
➤
) which you really should revise if you have the slightest doubts. We obtain, on
integrating the exponential first,
∫
te−^4 tdt=te−^4 t
− 4−(
−1
4)∫
e−^4 tdt=−1
4te−^4 t−1
16e−^4 tCount the signs carefully here! If we hadsinstead of 4 then you should now recognise
more easily that
∫
te−stdt=te−st
−s−(
−1
s)∫
e−stdt=−1
ste−st−1
s^2e−stNow we must put the limits in (279
➤
)toget
∫a0te−stdt=[
te−st
−s]a0+1
s∫a0e−stdt=[
te−st
−s]a0−1
s^2[
e−st]a
0=−1
sae−sa−1
s^2e−sa+1
s^2Agreed, this is somewhat messy. For the next step, taking the limit ofaas it tends to
infinity, it again helps to keep thinking ofsas some constant, but it is important to
remember that it is apositiveconstant. Then we need the limit result (418
➤
)xne−x→0asx→∞for any non-negative numbern. Then, applying this, sincesis positive, we get
e−sa→0andae−sa→0asa→∞and so asa→∞
∫a0te−stdt→1
s^2.
We write this as
∫∞0te−stdt=1
s^2