MATH 1502: CALCULUS II
26 JAN 2010: HOMEWORK 2
Let s be a fixed, strictly positive number. Show the improper integral
ˆ ∞
e−st sin(t) dt
0
converges and find its value, which depends, of course, on s.
Solution. By definition,
ˆ
ˆ
∞
−st
e
e−st sin(t) dt.
sin(t) dt = lim
0
b→∞
b
0
We begin by using integration by parts to find the antiderivative of e−st sin(t). Setting
u = sin(t) and dv = e−st dt gives us du = cos(t) dt and v = −1
e−st .
s
ˆ
ˆ
ˆ
−1
−1 sin(t)
−1
−st
−st
−st
−st
sin(t)e −
− e cos(t) dt .
e cos(t) dt =
e sin(t) dt =
s
s
s
est
We integrate by parts once further to find the antiderivative of e−st cos(t). Setting u = cos(t)
and dv = e−st dt gives us du = − sin(t) dt and v = −1
e−st .
s
ˆ
ˆ
ˆ
−1
−1 cos(t)
−1
−st
−st
−st
−st
cos(t)e −
−e sin(t) dt =
+ e sin(t) dt .
e cos(t) dt =
s
s
s
est
Plugging in the second equation to the first yields
ˆ
ˆ
−1 sin(t) −1 cos(t)
−st
−st
e sin(t) dt =
−
+ e sin(t) dt
s
est
s
est
ˆ
ˆ
− sin(t) − cos(t)
1
−st
−st
e sin(t) dt =
+
− 2
e sin(t) dt
sest
s2 est
s
ˆ
1
− sin(t) − cos(t)
−st
1+ 2
e sin(t) dt =
+
+C
s
sest
s2 est
ˆ
s2 + 1
− sin(t) − cos(t)
−st
e
sin(t)
dt
=
+
+C
s2
sest
s2 est
ˆ
−1
s sin(t) cos(t)
−st
e sin(t) dt = 2
+ st
+C
s +1
est
e
Written for Dr. Andrew’s Spring 2010 class.
This is the general antiderivative of e−st sin(t). We use it to compute
ˆ
b
−st
e
0
´b
0
e−st sin(t):
b
−1
s sin(t) cos(t)
sin(t) = 2
+C
+ st
s +1
est
e
0
s sin(b) cos(b) 0 1
−1
= 2
+ sb − −
s +1
esb
e
1 1
s sin(b) cos(b)
1
−1
+ 2
.
= 2
+ sb
sb
s +1
e
e
s +1
To find the value of the improper integral, we take the limit as b → ∞:
ˆ ∞
−1
s sin(b) cos(b)
1
−st
e sin(t) dt = lim 2
+ 2
=
+ sb
sb
b→∞
s +1
e
e
s +1
0
1
sin(b)
cos(b)
−1
s lim
+ lim
+ 2
2
sb
sb
b→∞ e
b→∞ e
s +1
s +1
We cannot evaluate these limits by “plugging in” b = ∞, as sin(b) and cos(b) have no welldefined limit as b → ∞. However, observe that sin and cos are bounded functions: no matter
what b we choose, we have
−1 ≤ sin(b) ≤ 1 and − 1 ≤ cos(b) ≤ 1.
1
Since e−1
sb → 0 and esb → 0 as b → ∞ (this is true since s is fixed and positive), by the
squeeze theorem we have that
sin(b)
cos(b)
→ 0 and
→0
sb
e
esb
as b → ∞. Thus,
ˆ
∞
e−st sin(t) dt =
0
−1
1
1
[s · 0 + 0] + 2
= 2
.
+1
s +1
s +1
s2
Since this is a finite number, the integral is indeed convergent.