Improper Integral
Since the concept of Laplace transform involves an integral from zero to infinity, the
knowledge of improper integral is needed.
Definition: An improper integral over an unbounded interval is defined as a limit of
integrals over finite intervals; thus
!
"a
A
"
A#! a
f(t)dt = lim
f(t)dt
where A is a positive real number.
If the integral from a to A exists for each A > a, and if the limit as A → ∞ exists, then the
improper integral is said to converge to that limiting value. Otherwise the integral is said
to diverge, or fail to exist,
Examples:
1) Let f(t) = ect, t ≥ 0 and c is a real nonzero constant.
" ct
A ct
e ct A
1
e
dt
=
lim
e
dt
=
lim
= lim e cA $ 1
0
#0
#
A!" 0
A!" c
A!" c
cA
if c < 0 , e → 0 as A → ∞, then the improper integral converges to -1/c,
if c > 0, ecA → ∞ as A → ∞, then the improper integral diverges,
if c = 0 , ect = 1, then the improper integral diverges.
(
)
2) Let f(t) = 1/t, t ≥ 1,
" dt
#1
t
the improper integral diverges.
A dt
#
A!" 1
= lim
t
= lim ln A = "
A!"
3) Let f(t) = t –p, t ≥ 1, where p is any real number, p ≠ 1, case p =1 was example 2.
# !p
A
1
lim $ t !p dt = lim
A1!p ! 1
$1 t dt = A"#
1
A"# 1 ! p
as A → ∞, as A1-p → 0 if p > 1 and the improper integral converges to 1/(p -1) for p > 1.
But if p ≤ 1, the improper integral diverges.
(
)
Definition: A function f(t) is said to be piecewise continuous on a closed interval [a, b]
if the interval can be partitioned by a finite number of points a = t0 < t1 < …… < tn = b so
that:
1) f(t) is continuous in each open subinterval ti-1 < t < ti.
2) lim+ f(t)!exists!!and!! lim" f(t)!exists!for!all!i.
t!t i"1
t!t i
In other words, f is piecewise continuous on [a, b] if it is continuous in the interval except
for a finite number of jump discontinuities.
Definition: If f(t) is continuous on [a, b] for every b > a, then we say that f(t) is piecewise
continuous on t ≥ a
Example:
#!2!if!!2 " t < !1
% !1!if!!1 < t < 0
%
f(t) = $
% t!if!0 < t < 1
%& 3!if!1 < t " 3
If f(t) is piecewise on [a, A], then it can be shown that
piecewise continuous for t ≥ a, then
A
!a
A
!a
f(t)dt exists. Hence, if f(t) is
f(t)dt exists for every A > a.
However, piecewise continuity is not enough to ensure the convergence of the improper
integral
!
"a
f(t)dt as we have seen in the examples.
The most convenient way to test the convergence or divergence of an improper integral is
by the following comparison theorem.
Theorem: Comparison Test for Improper Integrals
If f(t) is piecewise continuous for t ≥ a, if |f(t)| ≤ g(t) when t > M for some positive
constant M, and if
!
!
"M g(t)dt converges, then "a
f(t)dt also converges.
On the other hand, if f(t) ≥ g(t) ≥ 0 for t ≥ M. and if
also diverges.
!
!
"M g(t)dt diverges, then "a
f(t)dt
Examples:
2
1) For all x ≥ 1, -x2 ≤ -x, and since ex is an increasing function, then e !x " e !x
Now,
# !x
$1
A !x
$
A"# 1
e dx = lim
e dx = lim ! e !x
A"#
A
1
= lim %& !e ! A + e '( = e
A"#
So the improper integral converges,
Then, the improper integral
" !x 2
#1
e
"
dx $ # e !x dx is also convergent.
1
1
ex
!
2x + 1 2x + 1
" dx
A dx
$1
= lim #
= lim & ln ( 2x + 1)
Now, #
1 2x + 1
A!" 0 2x + 1
A!" % 2
So the improper integral diverges.
x
! e
dx also diverges.
Therefore, "
0 2x + 1
2) For all x ≥ 0, 1 < ex, then
A'
0 )
= lim $ ln ( 2A + 1) 2 * 0 ' = "
(
( A!" %
1