The document discusses the evaluation of improper integrals using Laplace transforms, providing various examples and solutions for specific integrals. It elaborates on the relationship between integral evaluations and the properties of Laplace transforms, alongside worked-out problems demonstrating these concepts. Additionally, it includes exercises for further practice on Laplace transforms and their applications.

1. 2.18 EVALUATION OFINTEGRALSBYLAPLACETRANSFORMS]
Sometimes. evaluation ofimproper integrals, i.e. integrals having lower limit
upper linnit oo can be done easily by using Laplace transform technique.
er limit 0
SOLVEDPROBLEMS
Ex. 1:Evauate(i) (red i) fe"sin 3/dt
0
0
3
Sol. () We note that the given integral is same as
re"dtwherep= 3
But
frePd =L{}
Putting p =3, we
get
fred=32
(ii) Given integral is same as feP sin 3t dt i.e. L{sin 3r} where p
=
4.
3
But L{sin 3t = 3
9 I.e. eP sin 31 dt =
p +9
+9

2. 179
ACE TRANSFORMSs
Phatting 4,we get
sin d 16+9 25s
E.2: Evahuate (Sin2d
Sol. Given integral is the Laplace transform of with p 0
e N Sin 2 2 J/sin2414- S
tan tan'- tan - tan
Putting p0,we get
sin2 d
Ex.3: Show that (i)te"sintdt =
i) re 'sintdt=0 (0. U. O 10, M2012,S. KU. M 12,)
Sol. () We note that the integral is same as ePrsint di wherep
=
3.
But Prsint dt =
L{tsint}
2P
LUsint--D 1 +
2p
ie. eP"tsint dt =-
(p +1)
Taking p =3, we get
6 3
te sintdt= 100 50
(ii) We note that the given integral is
sameas e"rsintdt where p=1
But e"r' sintdt=L{rsinr

3. FOURIER SERIES &INTEGRAL T
TRANSFO
180
Now LI sin) =
(-1{Llsin/]=(-l+1
,d1-3p-24p(-p?
=(-) 2
dpp+1) dp(1+p) +p
Jesint dt =-24p(1-p)
(1+p
or
Putting p =1, we get
"rsintd= 24(1-1) 0
(1+1)
re@ -br
-dt S.VU. Sep, 20)
Ex.4:Evaluate(i)
2t
dt.
-dt =
log 2
(s. V. U. M. 2010)
0
0
eae-bl
Sol. (i) We have L
d =log 2+6
p+a)
i.e.,
0
Putting p = 0, we get
-d-d= loe
(ii) Taking a =
3,b=6 in the above solution, we get
d =log=log2
0
(iii) Putting a
=
1,b =2 in (i), we get the required result.
COS at- cos bt
dt.
K.
6:
UsingLaplacetransform, evaluateJ
d
Sol. Given integral is same as lePt COsal-cosbr

4. 181
APLACE TRANSFORMS
jcosaf cosr
i.. where p 0
Since L cos at cos hr
p+
cOsar-cosbr 2P
P dp
pa p
+
log+a)- log(pb) p
log
(1+b1p?
log1-log
hap
+bp
0-log
p+a)
(p+a?
ie. fpt COsar-cosbr
p+a
Take p = 0. Then
cosat cosbr
log
cos5-coS dt=|
Note : Putting a = 5 and b = 1 in the above problem, we obtain
0
a sindt
Ex.7: Uising Laplace transform, evaluate J d
Sol. The given integral is the Laplace Transform of
sins
with p= a.
Now
JL0-cos 214o= dp

5. TRANSFOR
FOURIER SERIES &INTEGRAL TRA
182
log p-Iog(p*+4)
2
log
1+4/p,
log
log-log
Putting p = a, we get
2
-at.gin dt
d t = loe +4
Note : Put a
=l in the above result. Then we have
dt=log5
Ex. 8: Using Laplace transform, evaluate te sint dt.
0
Sol. We note that the given integral is same as
te sint dt where p =1.
eP sint d =
(sint) dt =
L{t sint} =
(-1)L(sin)]
But
dp
2p
2
Putting p=1,we get te " sint dt=
EXERCISE 2(D)
Find the
Laplace Transform of
(1))rsinr (i) rsin?3 (2) rsinh at
(3) " sin at
(4) ) cos (i) sin at (ii)e3 (iv) cos at
(v)te cos 2
(vi) re cosh
(vii) cos
(5) (1+

6. L A P L A C E T R A N S F O R M s
183
(7)1J (8)1Jo6ne [Hint: L{Jo) ,LJ)}=1-
(6)1plar)
Vp+1 VP+I
(1-cosad) (i)te 4 sin 3 (iii) rcos(at +b)
a
Using Laplace transtorm, show that
10)recostde ( ) in21d (12) re 1(4/)dt =
125
25 500
Find Laplace transform of
Sinat
(13) Stnhr
(14) (15) COSaf
(16)() ()
(17)() osat-cosbt (i) e-cosbt
(19)o (20)
2
(18) () Sn" (in sin3/cos (ii) cos4/Sin 21 1-cos a
Evaluate:
21) cos 6f-cos4
(22) (23) s" di
0 0
0
(24))
(25) (i) LJ"sinhbrd (i) LJe cosht
dt (i) Le sint d
(O
0
26) () cos at dt dt (i) L{[|sinat dt dt
lo0
00
ANSWERS
( 1 ) ) 2 0 p * + 4 )
p ( p + 4 ) ? ( i ) p ' + 1 2 )
s S4(p+12) (2) ) (p-2ap +2a
2ap 2a(p-a)
p(p +36) (p-a
(iv) (p+at)-6a?p?
2a(3p (iii)
(p+ay
(p+as
(4) () 1
2a(3p2-a) -6
(4) 2p-6p
(i)
(p+a (p+3
p+2p+2
p +6p+5
( +6p +13)
(vi)
(p+2p

7. FOURIER SERIES & NTEGRAL TRA
184
P (7)
(5) 3
(p+1)
6
6)
(p+2 p+3
P
(9)) 3p+a
p'(p+a (p
(p+Bp+25)
(fdzlr-d)anh- 2
p+4)
(ii)
|(13)log 14)cot (15) Does not exist (16%) ko i)ba
oe
(17)lo ng
2 L(p-a
p+a2
i)an-
2.
(18)(log 4 4
22)
(21)
p+Np- (20) acot-1og1
(19) log P
2P
(23)log (24)(i)log1- ntog12
P P)
P- 2p+1)
b
(25)pip-a?-b?])Pp-2)( p-2p-2?0) p+d)"u) Pf-i