Beta & Gamma Functions

Beta and gamma are the two most popular functions in mathematics. Gamma is a single variable function, whereas Beta is a two-variable function. The relation between beta and gamma function will help to solve many problems in physics and mathematics.

1. 1


Beta & Gamma Functions

By
Dr. Deepa Chauhan
Associate Professor,
Applied Science & Humanities Department,
Axis Institute of Technology, Kanpur

2. 2 Dr. Deepa Chauhan

Gamma function
If n is positive, then the definite integral 



0
1
0,ndxxe
nx , which is function of n , is called
the Gamma function (or Eulerian integral of second kind) and is denoted by n .
Thus 



0
1
0,ndxxen
nx
Put n=1

















Reduction Formula for n


































Thus




 If n is positive integer, then by repeated application of above formula, we ge









………………………………………………..

3. 3 Dr. Deepa Chauhan

Value of 






2
1
If 



0
1
0,ndxxen
nx dtte
t










0
2/1
2
1

Putting 2
xt so that xdxdt2 )1(..........................................................................................22.
1
.
2
1
00
22
dxexdx
x
e
xx













Writing y for x, we have











…………………………………………………………………………………………………………..(2)
If )(xf and )(xg are functions of x and y only, and the limits of integration are constants then
double integration can be represented as a product of two integrals. Thus  

d
c
b
a
b
a
d
c
dyygdxxfdydxygxf )()()()(

From (1) and (2) we have: 




















00
)(
00
2
2222
44
2
1
dydxedyedxe
yxyx

Changing to polar coordinates with  rdrddxdyryrx  ,sin,cos ; therefore r varies
from 0 to  and  from 0 to 2

 
























2/
00
2/
0
2/
00
2
2
2
1
44
2
1 22
 
 dderdrde
rr

Hence 






2
1

4. 4 Dr. Deepa Chauhan

5. 5 Dr. Deepa Chauhan

1. Prove that








Sol. 



0
1
0,ndxxen
nx
Let




















































2. Evaluate








Sol. Let

6. 6 Dr. Deepa Chauhan

3. Prove that















Sol.
Let







and









Let

7. 7 Dr. Deepa Chauhan

Transformations of Gamma Functions

1. dxxekn
nkxn




0
1
Sol. Let kx=t, kdx=dt

Then































Or
Proof: we have 









0
1
0
11
0
1
dyyekdykyke
dxxen
nkynnnky
nx
kdydx
kyxPut

 dxxekn
nkxn




0
1

2. Show that 0;
1
log
1
0
1
















ndx
x
n
n
Proof. Let












When






When





























3. Show that






Sol. Let





When

8. 8 Dr. Deepa Chauhan

When


































Beta Function
The definite integral





is called the Beta function and is
denoted by .
Thus






Beta function is also called the Eulerian Integral of the first kind.
Symmetry of Beta function







Since





Therefore














Hence

9. 9 Dr. Deepa Chauhan

Relation between Beta and Gamma Functions
 0,0;, 


 nm
nm
nm
nm


Proof:

We know that  dttem
mt




0
1
Putting 2
xt so that xdxdt2





















 dxxem
mx 12
0
2
2





Similarly  dyyen
ny 12
0
2
2





Now,  dyyedxxenm
nymx 12
0
12
0
22
22





 
 dydxyxenm
nmyx 12
0
12
0
)(
22
4







Changing to polar coordinates with  rdrddxdyryrx  ,sin,cos ;
therefore r varies from 0 to  and  from 0 to 2


10. 10 Dr. Deepa Chauhan
 dydxyxenm
nmyx 12
0
12
0
)(
22
4











2/
0
12121)(2
0
sincos4
2

 drdrenm
nmnmr

 




2/
0
1)(2
0
1212
2
sincos4

 drred
nmrnm





 







2
)(
2
),(
4
nmnm
nmnm  (),(

nm
nm
nm


 ,











Let









Therefore

11. 11 Dr. Deepa Chauhan

Transformations of Beta Functions

1.







Proof. We have







Put


When
When



















2.








Proof: We know that






(1)
Put







Also, when
And when
Thus, from (1)

12. 12 Dr. Deepa Chauhan

Or








Since beta function is symmetrical in m and n, we have

Therefore







3. Prove that



















Proof: We have







Or















Putting






We get

13. 13 Dr. Deepa Chauhan

Examples
1. Evaluate










Sol.

















































2. Prove that






















Sol. We have







Putting







Hence

14. 14 Dr. Deepa Chauhan

3. Prove that









We have







Putting



also



















Hence

15. 15 Dr. Deepa Chauhan






































































4. Evaluate







Sol. We have

16. 16 Dr. Deepa Chauhan







































































































































5. Using beta and gamma functions, evaluate 

0
4
1x
dx


Sol. We have









Putting dyydxyxyx
4/34/14
4
1



17. 17 Dr. Deepa Chauhan

4
2
4
sin
4
1
4
3
4
1
4
3
4
1
4
1
4
3
,
4
1
4
1
1
4
1
14
1
1
4
1
1 0
4
3
4
1
1
4
1
0
1
4
1
0
4/3
0
4








































 








dy
y
y
dy
y
y
dy
y
y
x
dx
I



n
nn
dx
x
x
nm
nm
m
sin
)1(
)1(
),(
0
1




























6. Prove that 


2/
0
2/
0
sin
sin
dxx
x
dx
Sol.  
2/
0
2/
0
sin
sin

dxx
x
dx
We have



















 

2/
0
02
12/
0
02
1
cossincossin

dxxxdxxx

18. 18 Dr. Deepa Chauhan








































































2
20
2
1
2
2
10
2
1
2
1
2
20
2
1
2
2
10
2
1
2
1










































4
5
2
2
1
4
3
4
3
2
2
1
4
1






















7. Evaluate  

 2/
00
4
cot
1

d
x
dx
Sol. Let 1
0
4
1
I
x
dx



 and 2
2/
0
cot Id


Putting dyydxyxyx
4/34/14
4
1



4
2
4
sin
4
1
4
3
4
1
4
3
4
1
4
1
4
3
,
4
1
4
1
1
4
1
14
1
1
4
1
1 0
4
3
4
1
1
4
1
0
1
4
1
0
4/3
0
41








































 








dy
y
y
dy
y
y
dy
y
y
x
dx
I




n
nn
dx
x
x
nm
nm
m
sin
)1(
)1(
),(
0
1







19. 19 Dr. Deepa Chauhan

2/
0
2 cot

dI



2/
0
2/12/1
sincos

 d
2
4
sin
2
1
2
4
1
1
4
1
12
4
3
4
1
2
2
2
1
2
1
2
2
1
2
1
2
1
2
1












































































424
2
cot
1
2
2/
00
4





 

d
x
dx

8. Evaluate 

1
0
3
3
1
dx
x
x


Sol. 


1
0
3
3
1
dx
x
x
I

dxxx
2/13
1
0
2/3
)1(


Putting dttdxtxtx
3/23/13
3
1


Therefore, dttttI
3/22/1
1
0
2/1
3
1
)1(


dttt
2/1
1
0
6/1
)1(
3
1


dttt
1
2
11
0
1
6
5
)1(
3
1 








2
1
,
6
5
3
1






















3
4
2
1
6
5
3
1

20. 20 Dr. Deepa Chauhan
















3
1
1
6
5
3





































































































6
1
3
2
3
6
sin
6
1
3
sin
3
2
1
3
sin
3
23
1
6
sin
6
1
3
1
1
3
1
6
1
1
3
1
3
1
6
5
3





























9. If








, then show that






Sol. We have









And























Putting m+n=1 or m=1-n in above relation

21. 21 Dr. Deepa Chauhan

10. Evaluate










Sol. We know that



















































=2





















11. Evaluate  2/3
Sol. We know that

22. 22 Dr. Deepa Chauhan

































12. Evaluate

23. 23 Dr. Deepa Chauhan

Duplication Formula
1. Prove that ,2
22
1
12
mmm
m








 where m is positive
Proof: We have )(2
cossin
12
2/
0
12
nm
nm
d
nm




 

Putting 2/1012  nn in (1), we get
)
2
1
(2
sin
2/
0
12




m
m
d
m 


.(2)
Again putting mn in (1), we get 
)2(2
cossin
2
12
2/
0
12
m
m
d
mm




 

 

)2(2
cossin2
2
1
2
2/
0
12
12
m
m
d
m
m


 





 

)2(2
2sin
2
1
2
2/
0
12
12
m
m
d
m
m


 





Putting
 dd 22

)2(22
sin
2
1
2
0
12
12
m
md
m
m


 






)2(2
sin
2
1
2
0
12
2
m
m
d
m
m


 




)2(2
sin
2
2
2
2/
0
12
2
m
m
d
m
m


 




24. 24 Dr. Deepa Chauhan

Replacing  by  , we obtain, 
)2(2
sin
2
2
2
2/
0
12
2
m
m
d
m
m







)2(2
2
sin
212
2/
0
12
m
m
d
m
m





 


From (2) and (3), we get ,2
22
1
12
mmm
m











13. Prove that 
3/1
2
3/2
6/53/1





Sol. By Duplication formula
,2
22
1
12
mmm
m













3/1
3
1
1
3
2
2
2
3
2
3
2
2
3/2
2
1
3
1
3
1
3/2
6/53/1










