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Question Number 64745 by mathmax by abdo last updated on 21/Jul/19

calculate ∫_0 ^1 ((sin(lnx))/(lnx))dx

calculate∫01sin(lnx)lnxdx

Commented by mathmax by abdo last updated on 21/Jul/19

let A =∫_0 ^1 ((sin(lnx))/(lnx))dx changement lnx =−t give x=e^(−t) A =−∫_0 ^∞ ((sin(−t))/(−t)) (−e^(−t) )dt = ∫_0 ^∞ ((sint)/t) e^(−t) dt let consider the parametric function ϕ(x) =∫_0 ^∞ ((sint)/t) e^(−xt) dt with x≥0 we have ϕ^′ (x) =−∫_0 ^∞ sint e^(−xt) dt =−Im(∫_0 ^∞ e^(it−xt) dt) ∫_0 ^∞ e^((−x+i)t) dt =[(1/(−x+i)) e^((−x+i)t) ]_0 ^(+∞) =−(1/(−x+i)) =(1/(x−i)) =((x+i)/(x^2 +1)) ⇒ϕ^′ (x)=−(1/(x^2 +1)) ⇒ϕ(x) =−arctan(x)+c ϕ(0) =(π/2) =c ⇒ϕ(x) =(π/2) −arctan(x) and ∫_0 ^∞ ((sint)/t) e^(−t) dt =ϕ(1) =(π/2) −(π/4) =(π/4) .

letA=∫01sin(lnx)lnxdxchangementlnx=−tgivex=e−tA=−∫0∞sin(−t)−t(−e−t)dt=∫0∞sintte−tdtletconsidertheparametricfunctionφ(x)=∫0∞sintte−xtdtwithx⩾0wehaveφ′(x)=−∫0∞sinte−xtdt=−Im(∫0∞eit−xtdt)∫0∞e(−x+i)tdt=[1−x+ie(−x+i)t]0+∞=−1−x+i=1x−i=x+ix2+1⇒φ′(x)=−1x2+1⇒φ(x)=−arctan(x)+cφ(0)=π2=c⇒φ(x)=π2−arctan(x)and∫0∞sintte−tdt=φ(1)=π2−π4=π4.

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