Definite Integrals 247
Example 1
Evaluate
∫∞
1
1
x
dx.
∫∞
1
1
x
dx=limk→∞
∫k
1
1
x
dx=limk→∞[lnx]k 1 =limk→∞(lnk)=∞so the integral diverges.
Example 2
Evaluate
∫∞
−∞
xe−x^2 dx.
Since both limits of integration are infinite, consider the sum of the improper inte-
grals
∫∞
−∞
xe−x
2
dx=
∫ 0
−∞
xe−x
2
dx+
∫∞
0
xe−x
2
dx. This sum is the sum of the limits
limk→−∞
∫ 0
k
xe−x^2 dx+limc→∞
∫c
0
xe−x^2 dx=klim→−∞
[
−
1
2
e−x^2
] 0
k
+limc→∞
[
−
1
2
e−x^2
]c
0
=
limk→−∞
[
−
1
2
+
1
2
e−k
2
]
+limc→∞
[
−
1
2
e−c
2
+
1
2
]
=−
1
2
+
1
2
= 0 .Since the limit exists, the integral
converges and
∫∞
−∞
xe−x^2 dx=0.
Infinite Discontinuities
If the function has an infinite discontinuity at one of the limits of integration, then
∫b
a
f(x)dx=liml→b−
∫l
a
f(x)dxor
∫b
a
f(x)dx=liml→a+
∫b
l
f(x)dx. If an infinite discontinu-
ity occurs atx=cwithin the interval of integration (a,b), then the integral can be broken
into sections at the discontinuity and the sum of the two improper integrals can be found.
∫b
a
f(x)dx=
∫c
a
f(x)dx+
∫b
c
f(x)dx=limk→c−
∫k
a
f(x)dx+liml→c+
∫b
l
f(x)dx
Example
Evaluate
∫π/ 2
0
√cosx
1 −sinx
dx.
Sincef(x)=
√cosx
1 −sinx
has an infinite discontinuity atx=
π
2
, the integral is improper.
Evaluate
∫π/ 2
0
cosx
√
1 −sinx
dx = klim→π/ 2 −
∫k
0
cosx
√
1 −sinx
dx = klim→π/ 2 −
[
− 2
√
1 −sinx
]k
0
=
klim→π/ 2 −
[
− 2
√
1 −sink+ 2
]
= 2 .Since the limit exists,
∫π/ 2
0
cosx
√
1 −sinx
dx=2.