Prove that ∫ 0 1 1 1 + ln 2 ⁡ x d x =...

oleifere45

oleifere45

Answered

2022-06-27

Prove that 0 1 1 1 + ln 2 x d x = 1 sin ( x 1 ) x d x

Answer & Explanation

sleuteleni7

sleuteleni7

Expert

2022-06-28Added 28 answers

Hint. One may observe that
1 1 + ln 2 x = 1 i ln x = 0 e ( i ln x ) t d t , x ( 0 , 1 ) ,
gives
0 1 1 1 + ln 2 x d x = 0 1 0 e ( i ln x ) t d t d x = 0 ( 0 1 x t d x ) e i t d t = 0 1 t + 1 e i t d t = 0 sin t t + 1 d t = 1 sin ( x 1 ) x d x
as announced.
Lucille Cummings

Lucille Cummings

Expert

2022-06-29Added 5 answers

Define I ( a ) = 0 e ( x + 1 ) a sin ( x ) x + 1 d x. Then I ( a ) = 0 e ( x + 1 ) a sin ( x ) d x = e a a 2 + 1 , and since lim a I ( a ) = 0, we have   I ( 0 ) = 0 sin ( x ) x + 1 = 0 e a a 2 + 1 d a = 0 1 d x 1 + ln 2 x .

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