B.Sc. Sem V Lecture Notes

1. 1
MATHEMATICS
E – Content
ON ANALYSIS
Vinay M. Raut
Assistant Professor
Shri Shivaji Science College,
Amravati

2. 2
IMPROPER INTEGRAL
Content :
• Definition of proper and improper integral
• Converges of improper integral
• Types and Test T1 and T2 and comparison test
• Beta function and result
• Gamma function and result
Definition :
1. Proper Integral :
If
b
a
I f (x)dx= ∫
Here,
a) F:[a b] R,→ f(x) is bounded on a x b≤ ≤
b) f R[a b]∈ i.e.
b
a
f (x)dx∫ exists
c) f (x) may be discontinuous at any C [a b]∈ but f(c) ≠ ∞
then
b
a
f (x)dx∫ is called proper integral.
2. Improper Integral : An integral
b
a
f (x)dx∫ becomes an improper integral when
1. a = −∞ i.e. (lower limit)
Or 2. b = ∞ i.e. (upper limit)
Or 3. f (x) = ∞ for some x [a b]∈
First and second condition means b-a is not finite but third condition means of is not bounded on
[a b]
For ex.
1.
o
tanx x dx
−∞∫
2. 21
dx
x
∞
∫
3. sin x
e dx
∞
−∞∫
4.
1
1
dx
x−∫
Types of Improper Integral :
Improper integral is classify four types such that any improper integral can be either one of
them or sum of these integral.
Type I :
a
f (x)dx,f (x) c,a x ,a R
∞
∈ ≤ < ∞ ∈∫
Type II :
b
f(x)dx,f(x) c, x b,b R
−∞
∈ −∞ < ≤ ∈∫

3. 3
Type III :
b b
a a
f (x)dx, f (x)dx,f (x) c,a x b
+
= ∈ ≤ ≤∫ ∫ where lim
a,b R andx a f (x)+
∈ → does not
exist or lim
x a f(x)+
→ = ∞
Type IV :
b b
a a
f(x)dx, f (x)dx,f (x) c,a x b, a,b R
−
= ∈ ≤ < ∈∫ ∫ and lim
x b f(x)−
→ does not exists
or lim
x b | f (x) |−
→ = ∞
For ex :
1 0 1
2 2 2 2 21 0 1
dx dx dx dx dx
x x x x x
∞ − − ∞
−∞ −∞ − +
= + + +∫ ∫ ∫ ∫ ∫ Here, R.H.S. is sum of improper integral type II,
IV, III and I
Convergent Integral
1. If
b
lim
a
b f (x)dx→∞ = =∫ l finite then we say that
a
f (x)dx
∞
∫ converges to l and we write
a
f (x)dx
∞
=∫ l
2. If the limit does not exists then the integral is divergent i.e.
b
lim
a
b f (x)dx or→∞ = ∞ − ∞ =∫
infinite then we say that
a
f (x)dx
∞
∫ diverges
For example -
Problem 1 : 3a
dx
x
∞
∫
Solution :
b2
b
lim lim3
3 1
1
dx x
b x dx b
x 2
−
∞
−
φ
 
= →∞ = →∞  
− 
∫ ∫
lim
2
1 1 1 1
b 0
2 2b 2
   
= →∞ − = =   ∞   
∵
finite=
31
dx
x
∞
∴∫ is convergent
Problem 2 :
1
dx
x
∞
∫
Solution :
b1/2
b
lim lim1/2
1 1
1
dx x
b x dx b
1/ 2x
∞
−  
= →∞ = →∞  
 
∫ ∫
lim
b 2 b 2 = →∞ − = ∞ 
1
dx
x
∞
∴∫ is divergent
Problem 3 :
0 0
lim
a
sindx a sin xdx
−∞
= → − ∞∫ ∫
[ ]lim
a 1 cos a= →−∞ − + = does not exists
0
sin xdx
−∞
∴∫ is divergent
Problem 4 : Solve on show that 2
dx
1 x
∞
−∞ +∫ is cgt. or dgt.

4. 4
SPECIAL IMPROPER INTEGRAL OR TEST INTEGRAL
Theorem (1) (T1 - Test)
rx
1
0
T e .dx
∞
−
∫ converge if r 0> and divergent if r 0≤
Proof : Now
brx
b
rx
0
0
e
e .dx
r
−
−  
=  
− 
∫
rb rb
e 1 1 e
,r 0
r r r r
− −
 
= + = − ≠ 
− − 
Taking lim b → ∞ on both sides
b
lim limrx rb
0
1 1 1
b e .dx b e if r 0
r r r
− −
→∞ = − →∞ = >∫
if r 0= ∞ <
For case r 0=
b b
rx
0 0
e dx dx b−
= =∫ ∫
When b → ∞ then
b
rx
0
e dx−
→ ∞∫
Thus rx
0
1
e dx if r 0
r
∞
−
= >∫
if r 0= ∞ ≤
There rx
1
0
T e dx
∞
−
= ∫ converge to
1
r
if r 0> and diverge to ∞ if r 0≤
Theorem (2) (Test T2) :
2 Pa
dx
T
x
∞
= ∫ converges if P 1> and diverge if P 1≤ and a 0>
Proof : For a 0>
Now,
bP 1 1 p 1 p
b
pa
a
dx x b a
if P 1
x P 1 1 p 1 p
− + − −
 
= = − ≠ 
− + − − 
∫
1 P 1 P 1 P
b
lim lim
Pa
dx b a a
b b
x 1 P 1 P 1 P
− − −
→∞ = →∞ − = −
− − −∫
(if 1 P 0 i.e. P 1)− < >
if 1 P 0 i.e.P 1= ∞ − > <
For ( )
b b
pa a
dx dx
P 1 asx 1
x x
= = =∫ ∫
( )
b
a
log x log b log a= = − = ∞ (asb )→ ∞
Thus
1 p
Pa
dx a
if P 1
x P 1
−
∞
= − >
−∫
if P 1= ∞ ≤
2 Pa
dx
T
x
∞
∴ = ∫ converges to
1 P
a
1 P
−
−
−
if P 1> and diverges to if P 1+∞ ≤

5. 5
Theorem (3) : (Comparison test 1)
Let f (x), g(x) C, a x∈ ≤ < ∞ and 0 f (x) g(x) X a≤ ≤ ∨ ≥
1
a a
CT : g(x)dx convergen f (x)dx converges i.e.
∞ ∞
⇒ <∞∫ ∫
2
a a
CT : f (x)dx g(x)dx
∞ ∞
= ∞⇒ =∞∫ ∫
[Comparison test T2] (or) If one converges then other and if one diverges then other
Proof : Given
1. f (x),g(x) C,a x∈ ≤ < ∞
2. f (x) g(x) x a≤ ∨ ≥
3.
a
g(x)dx
∞
∫ Convergent
To prove :
a
f (x)dx
∞
∫ converges i.e. < ∞
From (3)
a
g(x)dx
∞
∫ is cgt.
b
lim
a
b g(x)dx FiniteLimit =A(Says)∴ →∞ =∫ …………(1)
From (1) f (x)andg(x) C∈
∴ they are continuous on [a b]
∴ they are integrate on [a b] (by thm)
From (2) we have
b b
a a
0 f (x)dx g(x)dx≤ ≤∫ ∫ …………(A)
Let
b
a
(b) f (x)φ = ∫ …………(2)
For 1 2b b<
2 1b b
2 1
a a
(b ) (b ) f (x)dx t(x)dxφ − φ = −∫ ∫
2
1
b a
a b
f (x)dx f (x)dx= +∫ ∫
2
1
b
b
f(x)dx= ∫
0 (asf (x) 0given)≥ ≥
2 1 1 2(b ) (b ) an b b⇒ φ ≥ φ <
∴ By definition of m.I. (b)φ is m.I. (also from equation [A])
b b
lim lim
a a
0 b f (x)dx b g(x)dx≤ →∞ ≤ →∞∫ ∫
lim lim
b f (x)dx b g(x)dx A⇒ →∞ ≤ →∞ = (by equation [A])
(b)is bounded∴φ
Thus (b)ism.I.andboundedφ
∴ by thm it is convergent
a
f (x)dx
∞
⇒ ∫ is convergent

6. 6
To prove second result (CT2)
Lei 0
a
f (x) g(x) V x a and f (x)dx
∞
≤ ≤ ≥ = ∞∫ (given)
Assume the contrary that
a
g(x)dx
∞
∫ is convergent
Then by CT1
a
f (x)dx
∞
∫ is convergent
But this is contradiction to hypothesis
a
f (x)dx
∞
= ∞∫
Hence the assumption
a
g(x)dx
∞
∫ is cgt i.e. wrong.
a
g(x)dx
∞
∴∫ is dgt.
i.e.
a
g(x)dx
∞
= ∞∫
Corollary (1) let f (x) C, a x∈ ≤ < ∞ and
a
f (x) dx
∞
< ∞∫ then
a
f (x)dx
∞
∫ converges
Proof : Given
1. Lef f (x) C, a x∈ ≤ <∞
2.
a
f (x) dx
∞
< ∞∫
To prove :
a
f (x)dx
∞
∫ is cgt.
We know that O f(x) f(x) 2 F(x)≤ − ≤ and
a
2 f (x) dx
∞
∫ is cgt. (by given 2) ∴ By comparison test
a
f (x) f (x) dx
∞
 − ∫ is cgt. …………(3)
From (2) and (3)
( )a a
f (x) dz f (x) f (x) dx
∞ ∞
 − −
  ∫ ∫ i.e. cgt.
a a a
f (x) dx f (x) dx f (x)dx
∞ ∞ ∞
− +∫ ∫ ∫ is cgt.
a
f (x)dx
∞
∫ is cgt.
EXAMPLES
(By using test T1 – T2 and comparison test)
Example 1 : Test the integral for convergence
2
8
2 7
x
dx
x 1+
∫
Solution : Let
2
2 7
x
I dx
x 1
∞
=
+
∫
Here 2 x≤ <∞
Now 7 7
x 1 x+ >

7. 7
7 7
1 1
x 1 x
<
+
(taking reciprocal)
7/27
1 1
xx 1
<
+
(taking square root)
Multiply by x2
on both sides
2
3/27
x 1
xx 1
<
+
( )7 32
2 2
− =∵
2
3/27
x 1
0
xx 1
< <
+
( )x 2≥∵ …………(1)
Here 3/22
1
dx
x
∞
∫ is cgt. by T2-test (asP 3/ 2 1)= >
∴ By comparison test (by equation 1)
2
2 7
x
dx
x 1
∞
+
∫ is cgt.
Example 2 : Test the convergence of
2
27
x 1
dx
x 1
∞
−
−
+∫
Solution : Lef
2
27
x 1
I .dx
x 1
∞
−
−
=
+∫
2 2
1
2 27 1
x 1 x 1
dx dx
x 1 x 1
∞
−
− −
= +
+ +∫ ∫ (break limit) ............(1)
In equation (1) the first integral is proper integral and has finite value i.e. cgt. Now consider.
( )22
2 21 1
x 1 2x 1
dx dx
x 1 x 1
∞ ∞ + −−
=
+ +∫ ∫
21 1
1
1dx 2 dx
x 1
∞ ∞
= −
+∫ ∫
( )1
1 1
(x) 2 tan x
∞∞ −
= −
1 1
( 1) 2 tan tan 1− −
 = ∞ − − ∞ − 
1 2( / 2 / 4)= ∞ − − π −π
2
21
x 1
dx
x 1
∞ −
= ∞
+∫
2
21
x 1
dx
x 1
∞ −
+∫ is dgt.
Example 3 : Test convergence of 2
1
sin x dx
∞
−
∫
Solution : Let 2
1
I sinx dx
∞
−
= ∫
We know that
sin for any 0ϑ θ≥
2 2
sin x x− −
≤ ( x 1)≥∵

8. 8
2
2
1
0 sin x
x
−
< ≤ ( 1 x )≤ <∞∵
Here 21
1
dx
x
∞
∫ is cgt. by T2 – test (as P 2 1)= >
∴ By comparison test 2
1
sin x dx
∞
−
∫ is cgt.
Example 4 : Test the convergence of
2
27
x 1
dx
x 1
∞
−
−
+∫
Solution : Let
2
27
x 1
I dx
x 1
∞
−
−
=
+∫
2 2
1
2 27 1
x 1 x
dx dx
x 1 x 1
∞
−
− −
= +
+ +∫ ∫ …………(1)
Here, the 1st
integral on R.H.S. of equation (1) i.e.
2
1
27
x 1
dx
x 1−
−
+∫ is proper integral and has finite value
(i.e. cgt.)
Now consider second integral
2 2
2 21 1
x 1 (x 1) 2
dx dx
x 1 x 1
∞ ∞− + −
=
+ +∫ ∫
( )2
1
1 2 / x 1 dx
∞
= − +∫
1
1 1(x) 2(tan x)∞ − ∞
= −
1 1
( 1) 2(tan tan 1)− −
= ∞ − − ∞ −
1 2( / 2 / 4)= ∞ − − π − π = ∞
then
2
21
x 1
dx
x 1
∞ −
+∫ is dgt.
Example 5 : Show that 22
cosx
dx
x(log x)
∞
∫ converges absolutely
Solution : Let 22
cox x
I dx
x(logx)
∞
= ∫
Now, 2 2
cosx 1
0
x(logx) x(logx)
≤ ≤ ( )cosx 1≤∵
2 2
cosx 1
0
x(logx) x(logx)
≤ ≤ ( x 2)≥∵
Consider
R
lim
2 22 2
1 1
dx b dx
x(logx) x(logx)
∞
= →∞∫ ∫
R
lim 2
2
1
b (logx) dx
x
−  
= →∞  
 
∫
b1
lim
2
(log x)
b
1
−
 
= →∞ 
− 
(Put log x t 1/ xdx dt)= ⇒ =

9. 9
b
lim
2
1
b
log x
 
= − →∞  
 
lim 1 1
b
logb log2
 
= − →∞ − 
 
1
0
log 2
 
= − − 
 
1
0
 
= 
∞ 
1
finite
log2
= = …………(2)
∴ from equation (1) and (2) and by comparison test we get 22
cosx
dx
x(logx)
∞
∫ is cgt.
22
cox
dx
x(logr)
∞
∫ is absolutely cgt.
Example 6 : Test the convergence of x
1
log x.e dx
∞
−
∫
Solution :
R
limx x
1 1
log x.e dx b logx.e dx
∞
− −
= →∞∫ ∫ (Integrate by parts)
( )
x
bblim x
1 1
1 e
b log x.e / 1 . .dx
x 1
−
− 
= →∞ − − 
− 
∫
x
bblim limx
1 1
e
b logx,e b dx
x
−
−
 = − →∞ + →∞  ∫
x
b
lim limb
1
e
b logb.e 0 b dx
x
−
−
 = − →∞ − + →∞  ∫
x
b
lim
1
e
(0) b dx
x
−
= − + →∞∫
x
b
limx
1
e
logxe dx b dx
x
−
−
= →∞∫
x
1
e
dx
x
−
∞
= ∫ …………(1)
Here 1 x≤ < ∞
1/x 1≤
-x -x
e /x e≤
x
xe
0 e
x
−
−
≤ ≤ but x
1
e
∞
−
∫ is cgt.
by test T1 integral (as r 1 0)= >
∴ by comparison test we get
x
1
e
dx
x
−
∞
∫ is cgt. …………(2)
from equation (1) and (2) we get
x
1
logx . e dx
∞
−
∫ is cgt.

10. 10
Theorem : (limit test for convergences)
Let i) f (x) C, a x∈ ≤ <∞
ii) lim
x f(x) A, P 1, A R→∞ = > ∈
Them prove that
a
f (x) dx
∞
< ∞∫
(i.e.
a
f(x)dx
∞
∫ is absolutely cgt.)
Proof : we have
i) f (x) C, a x∈ ≤ <∞
ii) lim P
x x .f (x) P, P 1, A R→∞ = > ∈
since lim P
x x . f(x) A→∞ =
lim P
x x f (x) A∴ →∞ = (taking mod on both sides)
lim P
x x .f (x) A⇒ →∞ = (by definition of limit)
⇒ For given 1⇒∈= (say) there exist a +ve integer b such that
P
x b x .f(x) A 1≥ ⇒ − <
P
x f (x) 1 A⇒ < +
P
1 A
f (x)
x
+
⇒ <
P
1 A
0 f (x)
x
+
∴ ≤ < and P1
1
dx
x
∞
∫ is cgt. has P 1> (given) by T2 – test integral
∴ By comparison test we get
a
f (x) dx
∞
∫ is cgt.
a
f(n) dx
∞
⇒∫ absolutely cgt.
Statement :
Let i) f (x) C, a x∈ ≤ <∞
ii) lim
x f (x) A 0→∞ = ≠
= ±∞
Then prove that
a
f (x)dx
∞
∫ diverges and the test fails if A 0=
Example 7 : Show that
0 3
cosx
dx
1 x
∞
+
∫ converges
Solution :
0 3
cosx
I dx
1 x
∞
=
+
∫
0 3/2 3
cosx
x 1/ x 1
∞
=
+
∫

11. 11
Here,
3/2 3
cosx
f (x) C, 0 x
x 1/ x 1
= ∈ ≤ <∞
+
(By applying limit test taking P = 5/4)
lim lim5/4 5/4
3/2 3
cosx
x x f (x) x x .
x 1/ x 1
→∞ = →∞
+
5
P 1
4
 
= > 
 
lim
1/4 3
cosx
x
x 1 1/ x
= →∞
+
0= r 1
D 0 as 0
 
→ = 
∞ 
∵
∴ Limit for
0
f (x) dx
∞
∫ is cgt.
0 3
cosx
dx
1 x
∞
⇒
+
∫ is cgt.
0 3
cosx
dx
1 x
∞
⇒
+
∫ is absolutely cgt.
Example 8 : Show by limit test
0 2
dx
1 2x
∞
= ∞
+
∫
Solution :
Let 20
dx
I
1 2x
∞
=
+∫
Here 2
1
f (x) C, 0 x
1 2x
= ∈ ≤ <∞
+
lim lim
2
x
x x.f (x) x
1 2x
→∞ = →∞
+
lim lim
2 2
x 1
x x
x 1/ x 2 1/ x 2
= →∞ = →∞
+ +
1
2
=
∴ by limit test (here P = 1)
0
f (x)dx
∞
∫ is dgt.
0
f (x)dx
∞
⇒ = ∞∫

12. 12
BETA – GAMMA FUNCTION
Main application of this function is to solve improper integral.
Beta Function :
The integral
1
m 1 n 1
0
x (1 x) dx− −
−∫ with m 0, n 0> > is called Beta function of is denoted by
(mn)β
1
m 1 n 1
0
(mn) x (1 x) .dx− −
∴β = −∫
Prove symmetry property of Beta function
i.e. (mn) (n m)β =β
Proof : By definition
1
m 1 n 1
0
(mn) x (1 x) .dx− −
β = −∫
Put 1 x t dx dt− = ⇒ = −
And limit when x 0 t 1= ⇒ =
when x 1 t 0= ⇒ =
0
m 1 n 1
1
(mn) (1 t) .t ( dt)− −
∴β = − −∫ (Change the limit)
1
n 1 m 1
0
t (1 t) .dt− −
= −∫
1
n 1 m 1
0
x (1 x) .dx,− −
= −∫ (Change t to x)
(n m)= β (by definition)
(n m)= β
(mn) (nm)∴β = β
i.e. m n are interchangeable in beta function
Some Result :
1. Prove
m 1
m n0
x
(mn) .dx
(1 x)
−
∞
+
β =
+∫
Proof : By definition
1
m 1 n 1
0
(mn) x (1 x) .dx− −
β = −∫ …………(1)
Put 2
1 1
x dx .dt
1 t (1 t)
−
= ⇒ =
+ +
Limits when x 0 t= ⇒ = ∞
When x 1 t 0= ⇒ =
n 1
m 1 20
1 1 1
(mn) 1 . dt
(1 t) 1 t (1 t)
−
∞
−
− 
∴β = − 
+ + + 
∫ (by equation 1)
( )
n 1
m n0
t
.dt
1 t
−
∞
+
=
+
∫
n 1
m n0
x
dx
(1 x)
−
∞
+
=
+∫ (Change t to x)
n 1 m 1
m n m n0 0
x x
B(mn) dx dx
(1 x) (1 x)
− −
∞ ∞
+ +
∴ = =
+ +∫ ∫

13. 13
2. Prove
/2
2m 1 2n 1
0
(mn) 2 sin cos .do
π
− −
β = θ θ∫
Proof : We have that
1
m 1 m 1
0
(mn) x (1 x) .dx− −
β = −∫
Put 2
x sin dx 2sin .cos .d= θ ⇒ = θ θ θ
Limit when x 0then 0= θ =
x 1then π/2= θ=
( )
/2 m 12 2n 1 2 n 1
0
(mn) sin cos [1 sin ] .2sin cos d
π − − −
∴β = θ − θ θ θ θ∫
/2
2m 1 2n 1
0
2 sin .cos .d
π
− −
= θ θ θ∫
/2
2m 1 2n 1
0
(mn) 2 sin .cos d
π
− −
∴β = θ θ θ∫
Gamma Function :
The integral x n 1
0
e .x . dx, n 0
∞
− −
>∫ is called a Gamma function.
It is denoted by n
Properties of Gamma function :
1. 1 1=
Proof : By definition x n 1
0
n e .x .dx, n 0
∞
− −
= >∫
Putting n 1=
x
0
1 e .dx
∞
−
∴ = ∫
b blim limx x
00
b e dx b e− −
 = →∞ = →∞ − ∫
lim limb 0 b
b e e b 1 e 1− −
   = →∞ − + = →∞ − =   
b
b
1 1
e 0
e
− 
= = = 
∞ 
2. Recurrence formula
n 1 n n+ =
Proof : By definition
b
limn x n x
0 0
n 1 x .e dx b x .e dx
∞
− −
+ = = →∞∫ ∫
( )
bblim n x n 1 x
0 0
b x e n.x ( e )dx− − − = →∞ − − −
  ∫
b
lim n b n 1 x
0
b b .e 0 n x .e .dx− − − = →∞ − + +
  ∫
b
lim n 1 x
0
0 b n x .e dx− −
= + →∞ ∫
n 1 x
0
n x .e dx
∞
− −
= ∫
n n= (by definition of gamma)

14. 14
3. Prove n 1 n! for n 0, 1, 2,...+ = =
Proof : n 1 n n+ = by recurrence formula
n(n 1) n 1 if n 1 0= − − − > ( )n 1 n n+ =
n(n 1)(n 2) n 2 if n 2 0= − − − − >
n(n 1)(n 2) ...... 2 1 1= − − × ×
n(n 1)(n 2) ...... 2 1= − − × × × 1 1=∵
n!=
4. Prove n 1 kx
n0
n
x e dx
k
∞
− −
=∫ where n,k 0>
Proof : Put
t
x
k
=
dt
dx
k
=
Limits for
x 0 t 0= ⇒ =
x t= ∞⇒ = ∞
n 1 kx n 1 t n 1 t
n n0 0 0
dt 1 n
x .e dx (t/k) .e . t .e dt
k k k
∞ ∞ ∞
− − − − − −
= = =∫ ∫ ∫
Relation between Beta – Gamma Function :
The relation between beta and gamma function is stated as
m n
(mn)
m n
β =
+
Proof : We know that by definition
t m 1
0
m e .t dt
∞
− −
= ∫
Put 2
t x=
dt 2xdx∴ =
2
x 2m 2
0
m e x 2x dx
∞
− −
∴ = ∫
2
x 2m 1
0
m 2 e x dx
∞
− −
⇒ = ∫ …………(1)
y2 2n 1
0
n 2 e .y dy
∞
− −
∴ = ∫ …………(2)
( )( )2 2
x 2m 1 y 2n 1
0 0
m n 2 e .x .dx 2 e .y .dy
∞ ∞
− − − −
= ∫ ∫
2 2
(x y ) 2m 1 2n 1
0 0
4 e .x .y .dx dy
∞ ∞
− + − −
= ∫ ∫
Put, 2 2 2
x rcos , y rsin x y r , dx dy rdr d= θ = θ ⇒ + = = θ
and r varies from 0 to ∞
θ varies from 0 to π/2as show below

15. 15
2π/2
r 2m 1 2n 1 2m 1 2n 1
0 0
m n 4 e r r .cos sin rdrd
∞
− − − − −
∴ = θ θ θ∫ ∫
( )( )2/2
2m 1 2n 1 r 2m 2n 1
0 0
2 cos sin d 2 e r dr
π ∞
− − + −
= θ θ θ∫ ∫
m n (mn) m n= β +
m n
B(mn)
m n
∴ =
+
/2
2m 1 2n 1
0
From equati
i
on (1) and e
2 cos )
r s
s
ult
n d (mn
π
− −
 
 
 β

θ θ =

θ∫
Result : To prove
/2
2m 1 2n 1
0
2 sin sin d (mn)
π
− −
θ θ θ = β∫
Proof :
1
m 1 n 1
0
(mn) x (1 x) dx− −
β = −∫
Put 2
x sin= θ and solve it
Note : From
m n
(mn)
m n
β =
+
and
/2
2m 1 2n 1
0
(mn) 2 sin cos d
π
− −
β = θ θ θ∫
/2
2m 1 1n 1
0
1 m n
sin d cos d (mn)
2 2 m n
π
− −
∴ θ θ θ θ = β =
+
∫ …………(A)
If we put 2m 1 P and 2n 1 q− = − =
then
/2
P 2
0
p 1 q 1
2 2sin .cos d
p q 2
2
2
π
+ +
θ θ θ =
+ +
∫ …………(B)
Result : Prove 1/ 2 = π
Now by equation (A)
/2
2m 1 2n 1
0
m n
sin d .cos d
2 m n
π
− −
θ θ θ θ =
+
∫
Put m n 1/ 2= =
( )
/2 2
0
d / 2 2
π
θ = π∫ ( )1 1=∵
( )
2
/ 2 1/ 2 2π =
( )
2
1/ 2 1/ 2= π ⇒ = π

16. 16
EXAMPLE
1.
7 6!
30
2(2!)(3!)2 3 4
= = ( )n 1 n!+ =∵
,
( )n (n 1)!= −
2. 4 x x 5 1
0 0
x .e dx e x dx 5 4! 24
∞ ∞
− − −
= = = =∫ ∫
3. 3 2x 2x 4 1
40 0
4 3! 3
x e dx e x dx
2 16 8
∞ ∞
− − −
= = = =∫ ∫
n 1 kx
n0
n
by x e dn
k
∞
− −
 
=  
 
∫
Prove
2
t 2n 1
0
n 2 e t dt
∞
− −
= ∫
Solution : We know that y n 1
0
n e y dy, n o
∞
− −
= >∫
Put 2
y t= (we take any variable)
dy 2dt⇒ =
2 2 2n 1
t 2 n 1 t t
0 0
n e (t ) 2t dt 2 e dt
−∞ ∞
− − −
∴ = =∫ ∫
Example : Prove : ( )
1 n 1
0
n log1/ x dx
−
= ∫
Solution : Put y logx= − and solve it
EVALUATE
1. Prove 1/4 x
0
x e dx
∞
−
∫
Solution : Put 2
x t dx 2tdt= ⇒ =
1/4
1/4 x t 2
0 0
x e dx e (t ) 2tdt
∞ ∞
− −
=∫ ∫
t 3/2
0
2 e t dt 2 5 / 2
∞
−
= =∫ (by definition)
2 3 / 2 1= +
( )2 3/ 2 3/ 2=
( ) ( )2 3/ 2 1/ 2 1 2 3/21/2 1/ 2 3/ 2= + = = π
( )as 1/ 2π =
2. Prove 3 x
0
x.e dx
∞
−
∫
Put 3 2
x t dx 3t dt= ⇒ = then

17. 17
t 3 1/2 2 t 7/2
0 0
e (t ) .3t dt 3 e t dt 3 9 / 2
∞ ∞
− −
= = =∫ ∫
7 5 3 1 1 315
3 . . .
2 2 2 2 2 16
= × = π
3. Prove
1
0
dx
log n
= π
−
∫
4. Prove that
2 2
h x
0
e dx
24
∞
− π
=∫
5. Evaluate 1/ 2, 3 / 2, 5 / 2− − −
n 1
put n 1/ 2, 3 / 2, 5 / 2 in n
n
 +
= − − − =  
 
4 8
1/ 2 2 , 3 / 2 , 5 / 2
3 15
π − π
⇒ − = − π − = − =
6.
( )
3
70
x
dx
1 x
∞
+
∫
Solution :
3 4 1
7 4 30 0
x x 4 3 3! 2! 1
dx (4 3)
(1 x) (1 x) 6! 607
−
∞ ∞
+
= = β = = =
+ +∫ ∫
7.
1 1
3 3 4 1 4 1
0 0
4 4 3! 3!
x (1 x) dx x (1 x) dx (4 4)
7!8
− −
− = − = β = =∫ ∫
1/140=
8.
1/2 1
1/2 1/20 0
dx x
dx
(1 x)x(1 x)
−
∞ ∞
+
=
++
∫ ∫
1/ 2 1/ 2
(1/ 2 1/ 2)
1
= β = = π
9.
1 1
3/2 1 3/2 1
0 0
x (1 x) dx x (1 x) dx (3/2 3/2)− −
− = − = β∫ ∫
1
1/ 2
3 / 2 23 / 2
3 2! 8
π × π
π
= = =
10. Find (3, 1/2)β
Solution :
3 1/ 2 2! 1/ 2 16
(3, 1/ 2)
157 / 2 5/2 3/2 1/21/ 2
β = = =
× ×

18. 18
Prove that
/2
0
tan .d
2
π π
θ θ =∫
Solution :
1/2/2 /2
1/2
0 0
tan .d (sin ) (cos ) .d
−π π
θ θ = θ θ θ∫ ∫
1/ 2 1 1/ 2 1 1/ 2 1 (1/ 2) 1
2
2 2 2
 + − + + − +
=  
 
3 / 4 1/ 4 1
1/ 4 1 1/4
22 1
= = −
1
2 sin / 4 2
π π
= =
π
Byusing n. 1 n
sin n
 π
− = 
π 
Ex. Prove the following result
1.
/2
2 4
0
2 1 4 1
2 2sin cos d
2 1 4 1
2
2
π
+ +
θ θ θ=
+ + +
∫
1 1 3 1 1
2 2 2 2 2
2 4
× ×
=
3
16(3!) 32
π π
= =
By using
/2
P 2
0
P 1 q 1
2 2sin d cos d
P q 2
2
2
π
 + +
 
 θ θ θ =
 + +
 
 
∫
Ex.
/2
6
0
sin d
π
θ θ∫
Solution :
/2
6 0
0
sin (cos) d
π
θ θ∫
5 3 1
1/ 2 1/ 2
6 1 6 1 10 1 2 2 22
22 2 2 4
× × ×
+ + ++
= =
15 5
16 (3!) 32
π π
= =
Ex.
/2 /2
0 0
d
sin d
sin
π πθ
θ θ
θ
∫ ∫
Solution : { }{ }/2 /2
1/2 0 1/2 0
0 0
(sin ) (cos ) d (sin ) (cos ) d
π π
−
θ θ θ θ θ θ∫ ∫

19. 19
1/ 2 1 0 1 1/ 2 1 0 1
2 2 2 2.
1/ 2 1 1 1/ 2 1 1
2 2
2 2
− + + + +
=
− + + + + +
1/ 4 . 3 / 4 / 4
42 3 / 4 . 2 5 / 4 1/4 1/4
π π π π
= =
1
5 / 4 1/ 4
4
 
= 
 
∵
= π
Objective Questions :
1.
rx
0
e dx
∞
−
∫ converges if …………
2. rx
0
e dx
∞
−
∫ diverges if …………
3. pa
dx
x
∞
∫ converges if …………
4. pa
dx
x
∞
∫ diverges if …………
5. An Improper Integral of the type n 1 x
0
x .e dx,n 0
∞
− −
>∫ is called …………
6. n is denoted by
0
............
∞
∫ where n 0>
7. Recurrence formula for gamma function …………
8. Find the value of
1
2
−
= …………
9. An Improper Integral of the type
1
m 1 n 1
0
x (1 x) dx− −
−∫ is called …………
10. In Beta function m and n are inter changeable …………
11. Relation between Beta and Gamma function …………
*****