32. Final Review Day 2 - Improper Integrals

7/28/2019 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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1

0



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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1

0



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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Solution

Conclusion

1

0


( ) 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

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32. Final Review Day 2 - Improper Integrals