let A be closed-interval Subset of REAL ; :: thesis: 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 = x ) & dom (ln * (f ^ )) = Z & dom (ln * (f ^ )) = dom f2 & ( for x being Real st x in Z holds
f2 . x = - (1 / x) ) & f2 | A is continuous holds
integral f2,A = ((ln * (f ^ )) . (sup A)) - ((ln * (f ^ )) . (inf A))
let f, f2 be PartFunc of REAL , REAL ; :: thesis: for Z being open Subset of REAL st A c= Z & ( for x being Real st x in Z holds
f . x = x ) & dom (ln * (f ^ )) = Z & dom (ln * (f ^ )) = dom f2 & ( for x being Real st x in Z holds
f2 . x = - (1 / x) ) & f2 | A is continuous holds
integral f2,A = ((ln * (f ^ )) . (sup A)) - ((ln * (f ^ )) . (inf A))
let Z be open Subset of REAL ; :: thesis: ( A c= Z & ( for x being Real st x in Z holds
f . x = x ) & dom (ln * (f ^ )) = Z & dom (ln * (f ^ )) = dom f2 & ( for x being Real st x in Z holds
f2 . x = - (1 / x) ) & f2 | A is continuous implies integral f2,A = ((ln * (f ^ )) . (sup A)) - ((ln * (f ^ )) . (inf A)) )
assume A1: ( A c= Z & ( for x being Real st x in Z holds
f . x = x ) & dom (ln * (f ^ )) = Z & dom (ln * (f ^ )) = dom f2 & ( for x being Real st x in Z holds
f2 . x = - (1 / x) ) & f2 | A is continuous ) ; :: thesis: integral f2,A = ((ln * (f ^ )) . (sup A)) - ((ln * (f ^ )) . (inf A))
then A2: ( f2 is_integrable_on A & f2 | A is bounded ) by INTEGRA5:10, INTEGRA5:11;
A3: ln * (f ^ ) is_differentiable_on Z by A1, FDIFF_8:5;
A4: for x being Real st x in dom ((ln * (f ^ )) `| Z) holds
((ln * (f ^ )) `| Z) . x = f2 . x

proof

let x be Real; :: thesis: ( x in dom ((ln * (f ^ )) `| Z) implies ((ln * (f ^ )) `| Z) . x = f2 . x )
assume x in dom ((ln * (f ^ )) `| Z) ; :: thesis: ((ln * (f ^ )) `| Z) . x = f2 . x
then A5: x in Z by A3, FDIFF_1:def 8;
then ((ln * (f ^ )) `| Z) . x = - (1 / x) by A1, FDIFF_8:5
.= f2 . x by A1, A5 ;
hence ((ln * (f ^ )) `| Z) . x = f2 . x ; :: thesis: verum

end;

dom ((ln * (f ^ )) `| Z) = dom f2 by A1, A3, FDIFF_1:def 8;
then (ln * (f ^ )) `| Z = f2 by A4, PARTFUN1:34;
hence integral f2,A = ((ln * (f ^ )) . (sup A)) - ((ln * (f ^ )) . (inf A)) by A1, A2, A3, INTEGRA5:13; :: thesis: verum