Xosmas integrallar

teorema. Aytaylik f(x) va

 ( x )

funksiyalar [a,b] kesmada uzluksiz bo‟lib,

x=c nuqtada chegaralanmagan bo‟lsin. Agar chekli va nolga teng bo‟lmagan

f ( x )
lim
^{x  ñ}  ( x )


 q  0 .
limit mavjud bo‟lsa, u holda
b
_ f ( x ) d x
a
b
va _  ( x ) d x
a
integrallar

bir vaqtda yaqinlashadi yoki uzoqlashadi.

8-misol.

1
dx

1
^ ln x
2



1
integralning yaqinlashishi tekshirilsin. Integral ostidagi ^{f ( x ) } funksiya

x  1


da cheksizga intiladi.  ( x ) 

1
1  x

deb , ushbu limitni hisoblaymiz

lim
f ( x ) 1  x
 lim  1

^{x  1}  ( x )
x  1

1
integral uzoqlashadi. Demak, berilgan integral ham uzoqlashadi.

2

Agar
b
_ f ( x ) d x
a
yaqinlashuvchi bo‟lsa, u holda
b
_ f ( x ) d x
a
-absalyut

yaqinlashadi deyiladi.

Agar
b
_ f ( x ) d x -yaqinlashuvchi bo‟lsa, u holda
a
b
_ f ( x ) d x
a
integral ham

yaqinlashadi. Bu tasdiqning teskarisi o‟rinli emas. 9-misol. Ushbu

1
_ ño sx (
0
1 1
 1)  d x

integralning absalyut yaqinlashuvchiligini ko‟rsating. Aniqki,

1
co s x (
1
 1)  

bo‟lib,



1
dx

0

xosmas integral yaqinlashadi. U holda

1
1
_ ño sx (
0

 1)  d x

integralning ham yaqinlashuvchi bo‟lishi kelib chiqadi. Demak, berilgan integral absalyut yaqinlashadi.

II-BOB. XOSMAS INTEGRALLARNING BA’ZI TADBIQLARI.

2.1-§ Beta funksiya va uning xossalari.
Biz

1
_ x ^{a 1} (1 
0


x ) ^{b 1} d x

(1)

xosmas untegralni qaradik. Integral ostidagi funksiya uchun

1)a<1,


b  1
^{bo‟lganda x=0 maxsus nuqta 2) a}  ^{1, b<1 bo‟lganda x=1 maxsus}

nuqta. 3) a<1 b<1 bo‟lganda x=1 va x=0 nuqtalar maxsus nuqtalar bo‟ladi. Binobarin (1) chegaralanmgan funksiyaning xosmas integralidir. Demak, (1) integral- parametrga bog‟liq xosmas integraldir. (1) xosmas integralning a>0, b>0 da ya‟ni

M  _( a , b )  R ² : a  ( 0 ,   ),
b  ( 0 ,   )_

to‟plamda yaqinlashuvchi bo‟lishi ko‟rsatildi.
1- ta’rif: (1) integral Beta funksiya yoki birinchi tur Eyler integrali deb ataladi va B(a;b) kabi belgilanadi, demak

^B  ^{a , b}  ^
1
_ x ^{ 1} (1  x ) ^{b 1} d x ( a
0
 0 , b  0 ) _.

Shunday qilib


B ( a , b )
funksiya R ²
fazodagi

M  _( a , b )  R ² : a  ( 0 ;   ),
b  ( 0 ;   )

to‟plamda berilgandir. Endi
1⁰ (1) integral


B ( a , b )
funksiyaning xossalarini o‟rganaylik.

ixtiyoriy

1
B ( a , b )  _ x ^{a 1} (1  x ) ^{b 1} dx
0

0 0
M  _( x , b )  R ² : a  _ a ;   ),
b  _ b₀ ;   )_



0
a  0 ,



0
b  0 )

to‟plamda tekis yaqinlashuvchi bo‟ladi.

Isbot: Berilgan integralni tekis yaqinlashuvchilikka tekshirish uchun uni quyidagicha

_ 1 x ^{a 1} (1  x ) ^{b 1} d x 
1


_ ² x ^{a 1} (1  x ) ^{b 1} d x  _ 1 x ^{a 1} (1  x ) ^{b 1} d x _.

1
0 0
2

yozib olamiz. Ravshanki, a>0 bo‟lganda

1
2
_{ x} a  1 _{d x}
0

integral yaqinlashuvchi, b>0 bo‟lganda

1
_ (1  x ) ^{b 1} d x

1
2

integral yaqinlashuvchi. Parametr ^a ning



0 0
a  a ( a  0 )
qiymatlari va
 b  0 ,

1
 x  ( 0 ; )
2
uchun


x ^{a 1} (1  x ) ^{b 1} 
x ^{a 8  1} (1  x ) ^{b 1} 
_{2 x} a ₀  1

bo‟ladi. Veyrshtrass alomatidan foydalanib

1

2
_ x ^{a 1} (1 

0


x ) ^{b 1} d x

integralning tekis yaqinlashuvchi ekanligini topamiz. Shuningdek, parametr b

 a  0

 x  ^{ 1} , 1 ^

ning
b  b ( b  0 )

qiymatlari va
_{ 2} 
uchun

 

x ^{a 1} (1 
_{x )} b  1 _
x ^{a 1}  (1 
_{x )} b₀  1 _
2 (1 
_{x )} b₀  1

bo‟ladi va yana Veyrshtrass alomatiga ko‟ra

1
_ 1 x ^{a 1}  (1  x ) ^{b 1} d x
2

integralning tekis yaqinlashuvchiligi kelib chiqadi. Demak,

integral

0
a  a  0

0
va b  b  0
1
_ x ^{a 1} (1  x ) ^{b 1} d x
0

bo‟lganda, ya‟ni

0 0 0
M  _( a , b )  R ² : a  [ a ,   ), b  [b :   )_
to‟plamda tekis yaqinlashuvchi bo‟ladi.

Eslatma.

B ( a , b )
ning
M  _( a , b )  R ² : a  ( 0 ;   ), b  ( 0 ;   )_
to‟plamda

notekis yaqinlashuvchiligini ko‟rish qiyin emas.

2⁰.


B ( a , b )
funksiya


M

 _( a , b )  R ² : a  ( 0 ,   ), b  ( 0 ;   )_

to‟plamda uzluksiz funksiyadir. Haqiqatan ham,

B ( a , b ) 
1 x ^{a 1} (1 


0
x ) ^{b 1} d x

0
integralning M to‟plamda tekis yaqinlashuvchi bo‟lishidan va integral ostidagi

funksiyaning
 ( a , b )  M
da uzluksizligidan teoremaga asosan


B ( a , b )
funksiya

M  _( a , b )  R ² : a  ( 0 ;   ), b  ( 0 ;   )_


to‟plamda uzluksiz bo‟ladi.

3⁰.  ( a , b )  M
uchun
B ( a , b ) = ^{B ( b , a )}
bo‟ladi. Darhaqiqat


B ( a , b )  1 x ^{a 1} (1  x ) ^{b 1} d x
0

integralda x=1-t almashtirish bajarilsa, unda

 
B ( a , b )  1 x ^{a 1} (1  x ) ^{b 1} d x  1 t ^{b 1}  (1  t ) ^{a 1} d t  B ( b , a )
0 0


bo‟lishini topamiz.

4⁰.


B ( a , b )
funksiya quyidagicha ham ifodalanadi;

B ( a , b ) 
_{ t} a  1


d t
⁰ (1  t ) ^{a  b}
(2)

Haqiqatan ham, (1) integralda
t
x 
1  t
almashtirish bajarilsa, u holda

1  
t t d t
^{ } _t a  1

B ( a , b ) 
_ x ^{a 1} (1  x ) ^{b 1} dx  _
( ) ^{a 1}  (1 
1  t

1  t

₎ b  1 _{ }
(1  t ) ^2
 (1  t ) ^{a  b} d t

0 0 0

bo‟ladi. Xususan, b  1  a ( 0  a  1) bo‟ganda

B  ( a , 1  a ) 
^ _t   1

0
_{ 1  t} d t 

s in 
(3)

bo‟ladi (3) munosabatdan quyidagini topamiz:

₅⁰_.  ( a , b )  M
uchun


'( M
1 1
B ( ; )  
2 2

'  _( a , b )  R ² : a  ( 0 ;   ), b  (1;   )_)

bo‟ladi


B ( a , b ) 
b  1


a  b  1


B ( a , b  1)
(4)

1. 
   
   
   
   
   integralni bo‟laklab integrallaymiz:
   


¹ a  1



a

0
b  1
¹ b  1 ^x
1 a b  1 1
b  1 ¹




a b  2

B ( a , b )  x
0
(1  x )
d x 
(1  x )
0
d ( )  x a a
(1  x )
x (1  x )
_a 0
d x 

b  1 ¹ _a
 _ x (1 


x ) ^{b  2} d x

_a 0

(a>0, b>1) . Agar


x ^a (1 


_{x )} b  2

 x ^{a 1} _1  (1 

x ) _ (1 


_{x )} b  2 _

 x ^{a 1}  (1 
_{x )} b  2 _
x ^{a 1}  (1 
_{x )} b  1
ekanligini e‟tiborga olsak, u holda

  
1 x ^a (1  x ) ^{b  2} d x  1 x ^{a 1} (1  x ) ^{b  2} d x  1 x ^{a 1} (1  x ) ^{b 1} d x  B ( a ; b  1)  ( a , b )
0 0 0

natijada
bo‟lib,

B ( a , b ) 

bo‟ladi. Bu tenglikdan esa

b  1


a
 ^{B ( a , b  1)  B ( a , b )} 

B ( a , b ) 
b  1


a  b  1


B ( a , b  1) ( a  0 ,


b  1)

bo‟lishini topamiz. Xuddi shunga o‟xshash  ( a , b )  M '' uchun

( M ''  _( a , b )  R ² : a  (1;   ), b  ( 0 ;   )_)

B ( a , b ) 
a  1


a  b  1


B ( a  1, b )

bo‟ladi. Xususan, b  n ( n  N ) bo‟lganda

B ( a , b )  B ( a , n ) 
n  1


a  n  1


B ( a , n  1)

bo‟lib (4) formulani takror qo‟llab, quyidagini topamiz.

Ravshanki,


B ( a , n ) 
n  1

a  n  1
n  2

a  n  1
1


n  1
B ( a , 1) .

1
Demak,


B ( a 1) 
¹ x ^{a 1}  d x  ¹ :


a _a

B ( a , n ) 
1  2 ( a  1)
a ( a  1)( a  2 ) ( a  n  1)
(5).

Agar (5) da
a  m ( m  N )
bo‟lsa, u holda

Biz
1  2 ....( n  1) ( n  1) !( m  1) !
B ( m , n )  
m ( m  1)...( m  n  1) ( m  n  1) !