Improper Integrals
Mathematics 54–Elementary Analysis 2Introduction
From Math 53, we have defined Z b
a
f(x)dx, with the
assumption that
[a,b] is finite and [a,b]⊂Domain off, and
Introduction
We extend the concept of definite integrals to include cases where
1 the interval of integration is infinite or
2 f(x) has an infinite discontinuity on the interval [a,b].
Some examples are Z ∞
1
dx x ,
Z 1
−∞ dx
p
x−4, Z ∞
−∞ dx x2+1 Z 4
0
dx x−4,
Z 5
1
dx
xln3x,
Z 10
1
dx
3x−7 Z ∞
0 2
x2dx, Z +∞
0
dx
Integrals with Infinite Limits of Integration
Consider the shaded region below.
Its area isAb:=
Z b 1
1 x2dx=
−1 b +1.
Supposeb→ +∞:
lim
b→+∞Ab=b→+∞lim
Z b 1
1
x2 dx=b→+∞lim µ
−1 b +1
¶ =1
Thus, we define the area of the infinite region to the right ofx=1
and bounded byy= 1
x2 and thex-axis to be
Z ∞ 1
x2dx= lim Z b 1
Infinite Interval of Integration
Definitions.
Iff(x) is continuous on [a,+∞), then
Z +∞
a f(x)dx=k→+∞lim
Z k
a f(x)dx.
Iff(x) is continuous on (−∞,b], then
Z b
−∞
f(x)dx= lim t→−∞
Z b
t f(x)dx
The above equalities hold provided the limits exist.
In which case, the improper integrals are said to beconvergent.
Infinite Interval of Integration
Definition.
Iff(x) is continuous onRandc∈R, then
Z +∞
−∞
f(x)dx= Z c
−∞
f(x)dx+ Z +∞
c
f(x)dx
provided that both improper integrals on the right-hand side areconvergent.
Example 1.
Z +∞ 0
e−xdx.
Z +∞ 0
e−xdx = lim b→+∞
Z b 0
e−xdx
= lim b→+∞
¡ −e−x¢
¯ ¯ ¯
b 0
= lim b→+∞
µ −1
eb +1
¶
Example 2.
Z −3
−∞
x dx (x2−4)2.
Z −3
−∞
x dx
(x2−4)2 =a→−∞lim Z −3
a
x dx (x2−4)2
= lim a→−∞
−1 2(x2−4)
¯ ¯ ¯
−3 a
= lim a→−∞
· −1 2·5+
1 2(a2−4)
¸ =−1
Example 3.
Z +∞ 1
lnx x dx.
Z +∞ 1
lnx
x dx =b→+∞lim
Z b 1
lnx x dx
= lim b→+∞
ln2x 2
¯ ¯ ¯
b 1
= lim b→+∞
µln2b 2 −0
¶
Example 4.
Z +∞
−∞
dx x2+2x+5
Note that: Z+∞
−∞ dx x2+2x+5=
Z1
−∞ dx x2+2x+5+
Z+∞ 1
dx
Continuation...
Z1
−∞ dx
x2+2x+5 = t→−∞lim
Z1
t
dx
(x+1)2+4=t→−∞lim
1 2tan −1 µx +1 2 ¶ ¯ ¯ ¯ 1 t = lim t→−∞ 1 2 ·
tan−1(1)−tan−1 µt
+1 2
¶¸
=12
³π 4−
³
−π2
´´
=38π
Z+∞ 1
dx
x2+2x+5 = klim→∞ Z k
1
dx
(x+1)2+4=klim→∞ 1 2tan −1 µx +1 2 ¶ ¯ ¯ ¯ k 1 = lim k→∞ 1 2 ·
tan−1 µk
+1 2
¶
−tan−1(1) ¸ =1 2 ³π 2− π 4 ´ =π 8 Therefore, Z+∞ −∞ dx x2+2x+5=
Remark.
Considerf(x)=x. Evaluating Z +∞
−∞
f(x)dx, we have
Z +∞
−∞
x dx = Z 0
−∞
x dx+ Z +∞ 0 x dx = lim a→−∞ Z 0
a x dx+b→+∞lim
Z b 0 x dx
= lim a→−∞ x2 2 ¯ ¯ ¯ 0
a+b→+∞lim
x2 2 ¯ ¯ ¯ b 0 = lim a→−∞ µ −a2
2 ¶
+ lim b→+∞
µb2 2
¶ .
R
+∞−∞
x dx
is divergent
Compare to lim c→+∞
Z c
−c
f(x)dx, we have
lim c→+∞
Z c
−c
x dx= lim c→+∞
x2 2
¯ ¯ ¯
c
−c=c→+∞lim
c2 2 −
(−c)2 2 =0.
Remark.
Z +∞
−∞
f(x)dx6= lim c→+∞
Z c
Exercises
Evaluate the following integrals.
1
Z ∞
−1 4 (x+3)7dx
2
Z 1
−∞
3 x5dx
3
Z ∞ 0
1 1+x2dx
4
Z ∞
−∞
Integrals with Infinite Discontinuity
Consider the shaded region below.
Its area isAt:=
Z t 1
1 p
5−xdx= −2( p
5−t−2).
Lettingt→5−,
lim
t→5−At=limt→5−2( p
5−t−2)=4.
We define the area of the infinite region bounded byy=p1 5−x, the x−axis, and linesx=1 andx=5 to be
Z 5
1 1 p
5−xdx=tlim→5−
Z t 1
dx p
Integrals with Infinite Discontinuity
Definitions.
Iff(x) is continuous on (a,b] and lim
x→a+f(x)= ±∞, then
Z b
a
f(x)dx= lim t→a+
Z b
t
f(x)dx.
Iff(x) is continuous on [a,b) and lim
x→b−f(x)= ±∞, then
Z b
a f(x)dx=tlim→b−
Z t
a f(x)dx.
The above equalities hold if the limits exist.
In which case, we say the improper integrals areconvergent.
Integrals with Infinite Discontinuity
Definition.
Iff(x) has an infinite discontinuity atc, wherea<c<b, then we define
Z b
a
f(x)dx= Z c
a
f(x)dx+ Z b
c
f(x)dx.
The above equality holds provided that both Z c
a
f(x)dxand
Z b
c
f(x)dxare convergent. In which case, Z b
a
f(x)dxis said to
beconvergent.
If either integral on the right is divergent, then Z b
a
f(x)dxis
Example 1.
Z 2
−4 dx p
16−x2.
Note:f(x)=p 1
16−x2 has an infinite discontinuity atx= ±4.
Z 2
−4 dx p
16−x2 =t→−lim4+
Z 2
t
dx p
16−x2
= lim t→−4+
³ sin−1x
4 ´ ¯
¯ ¯ 2
t = lim
t→−4+
· sin−1
µ1
2 ¶
−sin−1 µt
4 ¶¸
=sin−1¡1 2 ¢
−sin−1(−1)=2π
Example 2.
Z π
2
0
tanx dx.
Note: The graph ofy=tanxhas a vertical asymptote atx=π2.
Z π
2
0
tanx dx = lim t→π2−
Z t 0
tanx dx
= lim t→π2−
ln|secx| ¯ ¯ ¯
t 0
= lim t→π2−
(ln|sect| −ln|sec 0|)
Example 3.
Z +∞ 0
e−px p
x dx.
Note thatf(x)=e
−px p
x has an infinite discontinuity atx=0.
Z +∞ 0
e−px p
x dx = Z 1
0 e−px
p x dx+
Z +∞ 1
e−px p
x dx
Letu= −px, du= −2dxp
x, the form is
R
(−2eu)du= −2eu+C
= lim t→0+
³ −2e−
p x´ ¯¯
¯ 1
t+k→+∞lim
³ −2e−
p x´ ¯¯
¯
k 1
= lim t→0+
µ −2
e+2e
−pt¶ + lim
k→+∞
µ
−2e− p
k +2
Exercises
1 Letf(x)=
(
x2 ifx<0
x−2 ifx>0. Evaluate Z 1
−1
f(x)dx.
2 Evaluate the following integrals.
1
Z 8
6 4 (x−6)3dx
2
Z 2
0
x2lnx dx
3
Z 2
0
dx
(2x−1)2/3
4
Z 1
0
