082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

488 Chapter 16 Further topics in integration
Now
E =
∫ ∞
0
= V2 RC
−2R
V 2 e −2t/RC
R
lim
t→∞ e−2t/RC = 0
[
e
−2t/RC ] ∞
0
dt = V2
R
∫ ∞
0
e −2t/RC dt
and so the energy storedinthe capacitor isgiven by
E = CV2
2
Example16.12 Find
∫ ∞
0
e −st sintdt s>0
Solution Using integration by parts, withu = e −st and dv = sint, wehave
dt
∫ ∞
0
e −st sintdt = [ −e −st cost ] ∫ ∞
∞
−s e −st costdt
0
Considerthefirsttermonther.h.s.Weneedtoevaluate [ −e −st cost ] ast → ∞andwhen
t = 0.Notethat −e −st cost → 0ast → ∞becausewearegiventhatsispositive.When
t = 0, −e −st cost evaluates to −1, and so
∫ ∞
0
e −st sintdt =1−s
∫ ∞
0
0
e −st costdt
Integrating by parts for a second time yields
∫ ∞
{ [e
e −st sintdt =1−s
−st sint ] ∫ ∞
∞
+s 0
0
∫ ∞
=1−s 2 e −st sintdt
0
0
}
e −st sintdt
because [ e −st sint ] ∞
0 evaluatestozeroatbothlimits.Atthisstagethereadermightsuspectthatwehavegonearoundinacircleandstillneedtoevaluatetheoriginalintegral.
However, somealgebraic manipulation yields the required result.We have
∫ ∞
0
∫ ∞
e −st sintdt +s 2 e −st sintdt = 1
(1+s 2 )
0
∫ ∞
0
∫ ∞
0
e −st sintdt = 1
e −st sintdt = 1
1+s 2