Page 287 - DMTH401_REAL ANALYSIS

Unit 23: Differentiation of Integrals

Notes
æ 1 + a æ x ö ö
As x varies from 0 to p, ç × tan ç ÷÷ varies through positive values from 0 to ¥ when –1 < a
è 1 - a è 2 øø
æ 1 + a æ x ö ö
< 1 and ç × tan ç ÷÷ and varies through negative values from 0 to –¥ when a< –1 or a > 1.
è 1 - a è 2 øø
Hence,

p
æ 1 + a æ x ö ö p
1
arctan ç × tan ç ÷÷ = - when - < a < 1
è 1 - a è 2 øø 2
0
and
p
æ 1 + a æ x ö ö p
arctan ç × tan ç ÷÷ = - when a < - 1 or a > 1.
è 1 - a è 2 øø 2
0
Therefore,
d
1
a
f ( ) = 0 when - < a < 1
da
d 2p
f ( ) = when a < - 1 or a > .
1
a
da a
Upon integrating both sides with respect to a, we get f(a) = C when –1 < a < 1 and f(a) = 2p In
1
|a| + C when a< –1 or a> 1.
2
C may be determined by setting a= 0 in
1
p
2
a
-
a
f ( ) = 0 ò ln(1 2 cos(x) a )dx
+
p
f (0) = 0 ò ln(1)dx
p
= 0 ò 0 dx
= 0
Thus, C = 0. Hence, f(a) = 0 when –1 < a< 1.
1
a
To determine C in the same manner, we should need to substitute in f ( ) =
2
p 2
a
0 ò ln(1 2 cos(x) + a )dx a value of a greater numerically than 1. This is somewhat
-
1
-
inconvenient. Instead, we substitute, a = , where 1 < b < 1. Then ,
b
p
-
f ( ) = 0 ò ln (1 2 cos(x)- b + b 2 ) 2 ln|b ) | dx
a
= 0 – 2p ln |b|
= 2p ln |a|
Therefore, C = 0 and f(a) = 2p ln |a| when a< –1 or a> 1.)
2
The definition of f(a) is now complete:
a
f ( ) 0 when 1 < a < 1 and
-
=
f ( ) 2 ln| |when a < - 1 or a > 1
a
p
a
=
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