let A be closed-interval Subset of REAL ; for f being PartFunc of REAL ,REAL
for Z being open Subset of REAL st A c= Z & Z = dom f & f = exp_R (#) (cos * exp_R ) holds
integral f,A = ((sin * exp_R ) . (upper_bound A)) - ((sin * exp_R ) . (lower_bound A))
let f be PartFunc of REAL ,REAL ; for Z being open Subset of REAL st A c= Z & Z = dom f & f = exp_R (#) (cos * exp_R ) holds
integral f,A = ((sin * exp_R ) . (upper_bound A)) - ((sin * exp_R ) . (lower_bound A))
let Z be open Subset of REAL ; ( A c= Z & Z = dom f & f = exp_R (#) (cos * exp_R ) implies integral f,A = ((sin * exp_R ) . (upper_bound A)) - ((sin * exp_R ) . (lower_bound A)) )
assume A1:
( A c= Z & Z = dom f & f = exp_R (#) (cos * exp_R ) )
; integral f,A = ((sin * exp_R ) . (upper_bound A)) - ((sin * exp_R ) . (lower_bound A))
then
Z = (dom exp_R ) /\ (dom (cos * exp_R ))
by VALUED_1:def 4;
then A4:
( Z c= dom exp_R & Z c= dom (cos * exp_R ) )
by XBOOLE_1:18;
for y being set st y in Z holds
y in dom (sin * exp_R )
then B6:
Z c= dom (sin * exp_R )
by TARSKI:def 3;
A5:
cos * exp_R is_differentiable_on Z
by A4, FDIFF_7:35;
exp_R is_differentiable_on Z
by FDIFF_1:34, TAYLOR_1:16;
then
f | Z is continuous
by FDIFF_1:33, A1, A5, FDIFF_1:29;
then
f | A is continuous
by A1, FCONT_1:17;
then A9:
( f is_integrable_on A & f | A is bounded )
by A1, INTEGRA5:10, INTEGRA5:11;
A10:
sin * exp_R is_differentiable_on Z
by B6, FDIFF_7:34;
B7:
for x being Real st x in Z holds
f . x = (exp_R . x) * (cos . (exp_R . x))
A11:
for x being Real st x in dom ((sin * exp_R ) `| Z) holds
((sin * exp_R ) `| Z) . x = f . x
dom ((sin * exp_R ) `| Z) = dom f
by A1, A10, FDIFF_1:def 8;
then
(sin * exp_R ) `| Z = f
by A11, PARTFUN1:34;
hence
integral f,A = ((sin * exp_R ) . (upper_bound A)) - ((sin * exp_R ) . (lower_bound A))
by A1, A9, B6, FDIFF_7:34, INTEGRA5:13; verum