Maths4

Országok listájaHungaryBudapesti Corvinus EgyetemGazdálkodástudományi KarNemzetközi gazdálkodás (angol nyelven)Mathematics2JegyzetekMaths4

Extention of Riemann-integral

4.1

Improper Integral

The Riemann-integral of a function

f

is not dened if the interval is not nite, or if

the integrand is not bounded in the interval of integration. The improper integral is the extension of Riemann-integral for these cases.

Innite interval of integration
b Denition 4.1 Let us assume that the Riemann-integral a f (x) dx exists for any b b a and limb a f (x) dx exists (nite number). Then the improper integral b

f (x) dx = lim
a

b

f (x) dx
a

is said to be convergent, otherwise it is divergent.
Consider the following example:

Example 4.1 Find the integral (of the density function of the exponential distribution):


e-x dx
0

where > 0 is a given constant.
Solution: For any constant

b>0

we have:

b

e-x dx = -e-x
0
Thus

b 0

= 1 - e-b



e-x dx = lim (1 - e-b ) = 1
0
for any constant

b

> 0.

b Denition 4.2 Let us assume that the Riemann-integral a f (x) dx exists for any b a b and lima- a f (x) dx exists (nite number). Then the improper integral b b

f (x) dx = lim
-

a-

f (x) dx
a

is said to be convergent, otherwise it is divergent.

1

4.1.

IMPROPER INTEGRAL

Denition 4.3 The improper integral
a

f (x) dx =
- -

f (x) dx +
a

f (x) dx,

where a is an arbitrary real number. That is - f (x) dx is said to be convergent a if both - f (x) dx and a f (x) dx converge.
For instance, the integral

-

2x dx 1 + x2 b>0

does not converge (does not exist), though for any

b -b

2x dx = 0 , 1 + x2

because the integrand is an odd function. But

b 0
and its limit is not exist).

2x dx = ln(1 + b2 ) 1 + x2

+

as

b ,

therefore the above improper integral diverges (does

Example 4.2 Evaluate the improper integral
0

xe-cx dx

2

where c > 0 is a given constant.
Solution: For any

b>0

we have

b 0
and

xe-cx dx = -

2

1 -cx2 e 2c

b 0

b

lim -

1 -cb2 1 (e - 1) = . 2c 2c

Integrals of unbounded functions
b Denition 4.4 Let limxa+ f (x) = ±, but the Riemann-integral a+ f (x) dx b exists for any > 0 and lim0+ a+ f (x) dx exists (nite number). Then the improper integral b b

f (x) dx = lim
a

0+

f (x) dx
a+

is said to be convergent, otherwise it is divergent.
b- Denition 4.5 Let limxb- f (x) = ±, but the Riemann-integral a f (x) dx exists for any > 0 and lim0+ ab- f (x) dx exists (nite number). Then the improper integral b b-

f (x) dx = lim
a

0+

f (x) dx
a

is said to be convergent, otherwise it is divergent.
2

4.1.

IMPROPER INTEGRAL

b Denition 4.6 Let limxc f (x) = ±, c (a, b), then the improper integral a f (x) dx exists, if both improper integrals ac f (x) dx and cb f (x) dx are convergent and b c b

f (x) dx =
a a

f (x) dx +
c

f (x) dx

Example 4.3 Find the integral
1


0

1 dx 1 - x2 [0; 1],
for if

Solution: The integrand is not bounded on the interval

x1-
But for any

lim

1 =. 1 - x2

>0

we have

1-


0
and

1 dx = arcsin(1 - ) 1 - x2 , 2

0

lim arcsin(1 - ) = arcsin 1 =

therefore the improper integral

1


0

1 dx = . 2 2 1-x

Exercise 4.1 Find the following improper integrals: 1 1 1) dx 2) dx 2

3) 5) 7) 9) 11) 13) 15)

1 0

x

1 dx 2+x+1 x xe-x dx
2

4) 6) 8) 10) 12) 14)

1

x

e-ax dx
0 0 0

1+x dx 1 + x2 - 1 dx x + e-x - e 0 1 dx 1 - x2 -1 1 1 -x e dx x2 0 2 1 dx x2 - 4x + 3 0

0

xn e-x dx (n

is positive integer)

e-ax cos bx dx


- 2 0 2 1

x2 dx 1 + x2 2x
3

(x2 - 1)2 1 dx x ln x

dx

Exercise 4.2 Determine the improper integral



f (x) dx,
-

where

a)

x e 2 f (x) = -x e 2

if x 0, if x > 0.

b)

1 x2 x2 f (x) = 1 x2

if x < -1, if -1 x 1, if x > 1.

3

4.2.

DOUBLE INTEGRALS

4.2

Double Integrals

Let us consider the following rectangle on the real plane

T = {(x, y) : a x b, c y d}
and let

f

be continuous on the region

T .
on the region

We dene the double integral of

f

T

as follows:

d c
For example,

b

d

b

f (x, y) dxdy =
a c a

f (x, y) dx

dy

1 0 0

1

1

1

1

4xy dxdy =
0 0

4xy dx

dy =
0

2x2 y

1 0

dy = 1

Theorem 4.1 If the function f is continuous on the region T, then
b a c d d b

f (x, y) dy

dx =
c a

f (x, y) dx

dy

Exercise 4.3 Determine the double integral
4 2 0 2

1 (6 - x - y) dxdy 8

in two dierent ways, rst integrate with respect to x and then w.r. to y and second integrate w.r.to y and then w.r. to x. Show that the value of the double integral is 1 in both cases.
4.3 Double improper integrals

The double improper integrals of the form

c a







f (x, y) dxdy =
c a

f (x, y) dx dy f
is a continuous function

have essential roles in probability theory. If the integrand (here we omit the details), then

and the improper integrals with respect to each variables are uniformly convergent

c a







f (x, y) dx dy =
a c

f (x, y) dy dx

that is the order of integration can be exchanged. We will not check the uniform convergence in the following examples, but it can be shown that it is satised.

integral:

Gaussian integral

As an example we determine the value of the Gaussian



G=
0

e-x dx

2

By the geometric interpretation of the integral, the Gaussina integral must be convergent, because

e-x < e-x

2

for any

x > 1.

4

4.3.

DOUBLE IMPROPER INTEGRALS

Determine the double improper integral of the function

f (x, y) = xe-x

2

(1+y 2 )

on the rst quadrant of the plane. It can be veried that the function satises the uniform convergence condition, and carry out the integration in both orders. If we integrate rst with respect to the variable

y,
2

then

0 0



xe-x

2

(1+y 2 )





dy dx =
0

xe-x

e-x
0

2 2

y

dy dx

If we make the substitution the right side is

t = xy , e-x
2 2

then the value of the integral in parenthesis on

0

y

dy =

1 x

0

e-t dt =

2

1 G x

Thus we have

0 0



xe-x

2

(1+y 2 )



dy dx =
0

xe-x · x,
then

2

1 G dx = G2 . x

Now, if we integrate rst with respect to

0

= -1 2 = = -1 2 -1 2

2 2 xe-x (1+y ) dx dy = 0 2 2 1 -2x(1 + y 2 )e-x (1+y ) dx 0 1+y 2 0 2 2 1 e-x (1+y ) dy = 0 1+y 2 0 1 1 (-1) dy = 2 [arctan y]0 = 4 0 1+y 2

dy =

Comparing with the previously obtained result, we have

G2 = /4,

that is,



e-x dx =
2

G=
0

. 2

The integrand is an even function, therefore it follows that

-

e-x dx =

2



.

Exercise 4.4 Evaluate the following integrals:

1) 3) 5) 7) 9)

3

1

(1 + 4xy) dxdy
1 3 0 1

2) 4) 6) 8) 10)

4

1

(x2 + y 2 ) dydx
2 4 -1 2



x + y dxdy

(x +
1 1 0 2



y) dxdy

0

2

0

1

(2x + y)8 dxdy
0 4 0 2

0

1

1 ln 2 0

x y + y x
ln 5

1 /2

xex dydx y
/2

dydx

sin(x + y) dydx
0 1 0 1 0

e2x-y dydx
1

xy x2 + y 2 + 1

dydx

0

5



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