Question Number 101073 by Dwaipayan Shikari last updated on 30/Jun/20

∫_0 ^∞ ((sin(logx))/(logx))dx

Answered by mathmax by abdo last updated on 30/Jun/20

A =∫_0 ^∞  ((sin(lnx))/(lnx))dx   changement lnx =−t give  x =e^(−t)   A =∫_(+∞) ^(−∞)  ((sin(−t))/(−t)) (−e^(−t) )dt =∫_(−∞) ^(+∞)  ((sint)/t) e^(−t)  dt  =∫_(−∞) ^0  ((sint)/t) e^(−t)  dt(→t =−u) +∫_0 ^∞  ((sint)/t) e^(−t)  dt  =∫_(+∞) ^0  ((−sinu)/(−u)) e^u  (−du) +∫_0 ^∞  ((sint)/t) e^(−t)  dt =∫_0 ^(+∞ ) ((sint)/t) e^t  dt +∫_0 ^∞  ((sint)/t) e^(−t) [dt  =∫_0 ^∞ ((sint)/t)(e^t  +e^(−t) )dt   =∫_0 ^∞  ((sint )/t) e^t  dt (diverrgent) +∫_0 ^∞  ((sint)/t) e^(−t)  dt (convervent)  ⇒∫_0 ^∞  ((sin(lnx))/(lnx))dx is divergent