Menu Close

0-sin-logx-logx-dx-

Tinku Tara 0 Comments
Pin




Question Number 101073 by Dwaipayan Shikari last updated on 30/Jun/20
∫_0 ^∞ ((sin(logx))/(logx))dx
∫0∞sin(logx)logxdx
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
A=∫0∞sin(lnx)lnxdxchangementlnx=−tgivex=e−tA=∫+∞−∞sin(−t)−t(−e−t)dt=∫−∞+∞sintte−tdt=∫−∞0sintte−tdt(→t=−u)+∫0∞sintte−tdt=∫+∞0−sinu−ueu(−du)+∫0∞sintte−tdt=∫0+∞sinttetdt+∫0∞sintte−t[dt=∫0∞sintt(et+e−t)dt=∫0∞sinttetdt(diverrgent)+∫0∞sintte−tdt(convervent)⇒∫0∞sin(lnx)lnxdxisdivergent

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *