32. final review day 2 - improper integrals
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Improper Integrals, Part 3:
The Concept Reloaded (Rigorous)Definition of Improper Integrals
Singularities in the Interval of Integration
Basic Improper IntegralsComparison Theorem
The Gamma FunctionThe Gamma Function
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Improper Integrals
An integral is improper if either
the interval of integration is infinitely long or
if the function has singularities in the interval of integration.
One cannot apply numerical methods like LEFT or RIGHT sums toapproximate the value of such integrals.
1
20
1is improper because the integrand has a singularity.dxx
0The integral e is improper because the
interval of integration is infinitely long.
x dx
0
eis improper for two reasons: infinitely long
interval of integration and a singularity of the integrand.
x
dxx
Definition
Examples
1
2
3
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Improper Integrals: Visualization
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Example 1
Evaluate
Integration by parts:
dxx
x
1 2
ln
xvx
dx
du
xdxdvxu
1
,
,ln2
dxx
x
dxx
x b
b
121 2
ln
lim
ln
L I P E TL I P E T
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Example 1
Evaluatedx
x
x
1 2
ln
111
1
lim
1
11ln1lnlim
lnlim
1 2
b
b
bdx
x
x
b
b
b
b
What rule do we use to evaluate this?
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Example 2
Evaluate
This is arctangent x
dxx
21
1
dx
x
dx
x
b
baa
0
2
0
2
1
1lim
1
1lim
0tantanlimtanlim0tan 1111
baba
22tanlimtanlim
11ba
ba
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Example 4
Evaluatedx
x 1
0 1
1dx
x
b
b
01 1
1lim
01ln1lnlim11
lim 101 bdxxb
b
b
01ln1lnlim1 bb
00
x 1ln
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This was covered in lastweeks lecture on improperintegrals. Please review it.
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1121lim
2
21lim
2
1
12
1
lim
xx
x
x
xxxx
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1121lim
2
21lim
21
12
1
lim
xx
x
x
xxxx
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Example: The Gamma Function
1
0
The gamma function is defined for 0 by setting
( ) e . Show that this integral converges.x t
x
x t dt
Problem
Solution The integral defining the gamma function is improper
because the interval of integration extends to infinity.
If 0 < x < 1, the integral is also improper because then
the function to be integrated has a singularity at x = 0,
the left endpoint of the integration interval.
Case 1.
0 < t < 1. Observe first that the integral converges if p > -1.
1
0
pt dt
11 1 1 10 0 0
0
1 1lim lim lim
1 1 1 1
p pp p
a a aa a
t at dt t dt
p p a p
This computation requires the assumption that p > -1, i.e., that p + 1 > 0. This allows you toconclude thatap+1 0 as a 0.
NB: Interval here is ( 0, 1 ]
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Example: The Gamma Function
1
0
The gamma function is defined for 0 by setting
( ) e . Show that this integral converges.x t
x
x t dt
Problem
Solution(contd)
Case 1.Case 1.(0 0,
1 1e .
x t x
t t
Hence the integral converges by the ComparisonTheorem and by the fact that converges for p > -1.
1
0
pt dt
11
0e
x tt dt
To show the convergence of the integral ,
use the fact that
1
1e
x t
t dt
1 2lim e 0.
t
x
tt
This holds for all values of x. Hence there is a number bbxx
such that for t >bbxx. 1 2e 1x tt
This means that for t >bx. 1 1 2 2 2e e e ex t x t t tt t
Case 2.Case 2.
t > 1.t > 1.
NB: Interval here is [1,
)
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Example: The Gamma Function
1
0
The gamma function is defined for 0 by setting
( ) e . Show that this integral converges.x t
x
x t dt
Problem
Solution(contd)
To show the convergence of the integral ,
use the fact that
1
1
ex tt dt
1 2lim e 0.
tx
tt
This holds for all values of x. Hence there is a number bbxx
such that for t >bbxx. 1 2e 1x tt
This means that for t >bx. 1 1 2 2 2e e e ex t x t t tt t
Hence the integral converges by the Comparison Theorem,
since the integral converges (by direct computation).
1 tex
x
bt dt
2
e
x
t
bdt
Case 2.Case 2.
t > 1.t > 1.Write this out, prove it.
The main idea here was to find a convergent
funct ion that we know alwaysalways lies above t x-1 e - t
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Example: The Gamma Function
Solution
Conclusion
1
0
The gamma function is defined for 0 by setting
( ) e . Show that this integral converges.x tx
x t dt
Problem
Let x > 0 be fixed.
1
1 1 1 1
0 0 1
e e e e
x
x
b
x t x t x t x t
b
t dt t dt t dt t dt
We split the improper integral defining the Gamma functionto three integrals as follows:
converges by part 1. Here we needed to assume that x > 0.
1
1
0
ex tt dt 1
is an ordinary integral.1
1
e
xb
x tt dt 2
1e
x
x t
b
t dt
3 converges by part 2.
We conclude that the integral converges.1
0
ex tt dt