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

16.4 Integral properties of the delta function 489
EXERCISES16.3
1 Evaluate, ifpossible,
∫ ∞
(a) e −t dt
0
∫ ∞
(b) e −kt dt k is aconstant,k > 0
0
∫ ∞ 1
(c)
1 x dx
∫ ∞ 1
(d)
1 x 2 dx
∫ 3 1
(e)
1 x −2 dx
2 Evaluate the following integrals wherepossible:
∫ 4
∫
3 3
(a)
0 x −2 dx (b) 1
0 x −1 + 1
x −2 dx
∫ 2
∫
1
∞
(c)
0 x 2 −1 dx (d) sin3tdt
0
∫ 3
(e) xe x dx
−∞
3 Find
∫ ∞
e −st costdt s>0
0
Solutions
1
1 (a) 1 (b) (c) does not exist
k
(d) 1 (e) does not exist
2 (a) does notexist (b) does notexist
3
(c) does notexist
(e) 2e 3
s
s 2 +1
(d) doesnot exist
16.4 INTEGRALPROPERTIESOFTHEDELTAFUNCTION
Thedeltafunction, δ(t −d),wasintroducedinChapter2.Thefunctionisdefinedtobea
rectanglewhoseareais1inthelimitasthebaselengthtendsto0andastheheighttends
toinfinity.Sometimesweneedtointegratethedeltafunction.Inparticular,weconsider
the improper integral
∫ ∞
−∞
δ(t −d)dt
Theintegral gives the area under the functionand this isdefinedtobe1.Hence,
∫ ∞
−∞
δ(t−d)dt =1
InChapter21 weneedtoconsiderthe improper integral
∫ ∞
−∞
f(t)δ(t −d)dt
where f (t) is some known function of time. The delta function δ(t −d) is zero everywhere
except att =d. Whent =d, then f (t)has a value f (d).Hence,
∫ ∞
−∞
f(t)δ(t −d)dt =
∫ ∞
−∞
f(d)δ(t −d)dt = f(d)
∫ ∞
−∞
δ(t −d)dt