0-1-sin-ln-x-ln-x-dx-

Question Number 85256 by john santu last updated on 20/Mar/20

\int \underset{0}{\overset{1}{}} \frac{\sin (\ln x)}{\ln (x)} dx

Commented by john santu last updated on 20/Mar/20

\sin (z) = \frac{e^{iz} - e^{- iz}}{2 i}

\sin (\ln x) = \frac{e^{i \ln (x)} - e^{- i \ln (x)}}{2 i} = \frac{x^{i} - x^{- i}}{2 i}

\int \underset{0}{\overset{1}{}} \frac{x^{i} - x^{- i}}{2 i \ln (x)} dx = I (1)

let I (b) = \int \underset{0}{\overset{1}{}} \frac{x^{bi} - x^{- i}}{2 i \ln (x)} dx

I^{'} (b) = \int \underset{0}{\overset{1}{}} \frac{\partial}{\partial b} [\frac{x^{bi} - x^{- i}}{2 i \ln (x)}] dx

I^{'} (b) = \int \underset{0}{\overset{1}{}} \frac{x^{bi} \ln (x) i}{2 i \ln (x)} dx

I^{'} (b) = \frac{1}{2 (1 + bi)} x^{bi + 1}]_{x = 0}^{x = 1}

I^{'} (b) = \frac{1}{2 (1 + bi)} \Rightarrow I (b) = \int \frac{1}{2 (1 + bi)} db

I (b) = \frac{1}{2 i} \ln (1 + bi) + C

Commented by john santu last updated on 20/Mar/20

\frac{1}{2 i} \ln (1 + bi) + C = \int \underset{0}{\overset{1}{}} \frac{x^{bi} - x^{- i}}{2 i \ln (x)} dx

b = - 1 \Rightarrow \frac{1}{2 i} \ln (1 - i) + C = 0

C = - \frac{1}{2 i} \ln (1 - i)

I (1) = \frac{1}{2 i} \ln (\frac{1 + i}{1 - i}) = \frac{1}{2 i} \frac{π i}{2} = \frac{π}{4}

∴ \int \underset{0}{\overset{1}{}} \frac{\sin (\ln x)}{\ln (x)} dx = \frac{π}{4}

Commented by mathmax by abdo last updated on 20/Mar/20

I = \int_{0}^{1} \frac{s i n (l n x)}{l n x} d x c h a n g e m e n t l n x = - t g i v e x = e^{- t} \Rightarrow

I = - \int_{0}^{+ \infty} \frac{s i n (t)}{t} (- e^{- t}) d t = \int_{0}^{\infty} \frac{s i n t}{t} e^{- t} d t

l e t f (a) = \int_{0}^{\infty} \frac{s i n t}{t} e^{- a t} d t w i t h a ⩾ 0 w e h a v e

f^{^{'}} (a) = - \int_{0}^{\infty} s i n t e^{- a t} d t = - I m (\int_{0}^{\infty} e^{- a t + i t} d t)

\int_{0}^{\infty} e^{(- a + i) t} d t = {[\frac{1}{- a + i} e^{(- a + i) t}]}_{0}^{+ \infty} = - \frac{1}{a - i} {- 1} = \frac{1}{a - i}

= \frac{a + i}{a^{2} + 1} \Rightarrow f^{^{'}} (a) = - \frac{1}{1 + a^{2}} \Rightarrow f (a) = k - a r c t a n (a)

f (0) = \frac{π}{2} = k \Rightarrow f (a) = \frac{π}{2} - a r c t a n (a)

I = f (1) = \frac{π}{2} - \frac{π}{4} = \frac{π}{4}