let a be Real; for A being non empty closed_interval Subset of REAL
for f, f2 being PartFunc of REAL,REAL
for Z being open Subset of REAL st A c= Z & ( for x being Real st x in Z holds
( f . x = a + x & f . x > 0 ) ) & dom (ln * f) = Z & dom (ln * f) = dom f2 & ( for x being Real st x in Z holds
f2 . x = 1 / (a + x) ) & f2 | A is continuous holds
integral (f2,A) = ((ln * f) . (upper_bound A)) - ((ln * f) . (lower_bound A))
let A be non empty closed_interval Subset of REAL; for f, f2 being PartFunc of REAL,REAL
for Z being open Subset of REAL st A c= Z & ( for x being Real st x in Z holds
( f . x = a + x & f . x > 0 ) ) & dom (ln * f) = Z & dom (ln * f) = dom f2 & ( for x being Real st x in Z holds
f2 . x = 1 / (a + x) ) & f2 | A is continuous holds
integral (f2,A) = ((ln * f) . (upper_bound A)) - ((ln * f) . (lower_bound A))
let f, f2 be PartFunc of REAL,REAL; for Z being open Subset of REAL st A c= Z & ( for x being Real st x in Z holds
( f . x = a + x & f . x > 0 ) ) & dom (ln * f) = Z & dom (ln * f) = dom f2 & ( for x being Real st x in Z holds
f2 . x = 1 / (a + x) ) & f2 | A is continuous holds
integral (f2,A) = ((ln * f) . (upper_bound A)) - ((ln * f) . (lower_bound A))
let Z be open Subset of REAL; ( A c= Z & ( for x being Real st x in Z holds
( f . x = a + x & f . x > 0 ) ) & dom (ln * f) = Z & dom (ln * f) = dom f2 & ( for x being Real st x in Z holds
f2 . x = 1 / (a + x) ) & f2 | A is continuous implies integral (f2,A) = ((ln * f) . (upper_bound A)) - ((ln * f) . (lower_bound A)) )
assume that
A1:
A c= Z
and
A2:
for x being Real st x in Z holds
( f . x = a + x & f . x > 0 )
and
A3:
dom (ln * f) = Z
and
A4:
dom (ln * f) = dom f2
and
A5:
for x being Real st x in Z holds
f2 . x = 1 / (a + x)
and
A6:
f2 | A is continuous
; integral (f2,A) = ((ln * f) . (upper_bound A)) - ((ln * f) . (lower_bound A))
A7:
f2 is_integrable_on A
by A1, A3, A4, A6, INTEGRA5:11;
A8:
ln * f is_differentiable_on Z
by A2, A3, FDIFF_4:1;
A9:
for x being Element of REAL st x in dom ((ln * f) `| Z) holds
((ln * f) `| Z) . x = f2 . x
dom ((ln * f) `| Z) = dom f2
by A3, A4, A8, FDIFF_1:def 7;
then
(ln * f) `| Z = f2
by A9, PARTFUN1:5;
hence
integral (f2,A) = ((ln * f) . (upper_bound A)) - ((ln * f) . (lower_bound A))
by A1, A3, A4, A6, A7, A8, INTEGRA5:10, INTEGRA5:13; verum