0-1-sin-logx-logx-dx-

Question Number 102058 by Dwaipayan Shikari last updated on 06/Jul/20

\int_{0}^{1} \frac{s i n (l o g x)}{l o g x} d x

Answered by prakash jain last updated on 06/Jul/20

x = e^{u}

d x = e^{u} d u

I = \int_{- \infty}^{0} \frac{\sin u}{u} e^{u} d u = - \int_{0}^{\infty} \frac{e^{- u} \sin u}{u} d u

J (t) = - \int_{0}^{\infty} \frac{e^{- t u} \sin u}{u} d u

J^{'} (t) = - \int_{0}^{\infty} e^{- t u} \sin u d u = - \frac{1}{1 + t^{2}}

J (t) = - \tan^{- 1} t + C

I = J (1)

Will continue .