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

∫_0 ^1 ((sin(logx))/(logx))dx

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

x=e^u   dx=e^u du  I=∫_(−∞) ^0 ((sin u)/u)e^u du=−∫_0 ^∞ ((e^(−u) sin u)/u)du  J(t)=−∫_0 ^∞  ((e^(−tu) sin u)/u)du  J′(t)=−∫_0 ^∞ e^(−tu) sin udu=−(1/(1+t^2 ))  J(t)=−tan^(−1) t+C  I=J(1)  Will continue.