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Question Number 88288 by Chi Mes Try last updated on 09/Apr/20

Commented by abdomathmax last updated on 10/Apr/20

I =∫_0 ^1 ((sin(ln∣x∣))/(ln∣x∣))dx changement ln(x)=−u ⇒ I =∫_0 ^1 ((sin(lnx))/(lnx))dx =−∫_0 ^(+∞) ((−sinu)/(−u))(−e^(−u) )du = ∫_0 ^∞ ((sinu)/u) e^(−u) du let f(t) =∫_0 ^∞ ((sinu)/u)e^(−tu) du with t>0 f^′ (t) =−∫_0 ^∞ e^(−tu) sinu du =−Im(∫_0 ^∞ e^(−tu+iu) du) =−Im(∫_0 ^∞ e^((−t+i)u) du) ∫_0 ^∞ e^((−t+i)u) du =[(1/(−t+i))e^((−t+i)u) ]_0 ^(+∞) =((−1)/(−t+i)) =(1/(t−i)) =((t+i)/(t^2 +1)) ⇒f^′ (t)=−(1/(1+t^2 )) ⇒ f(t) =k−arctant lim_(t→0) f(t) =∫_0 ^∞ ((sinu)/u)du =(π/2)=k ⇒ f(t) =(π/2) −arcrant (t>0) I =f(1) =(π/2)−(π/4) =(π/4)

I=∫01sin(ln∣x∣)ln∣x∣dxchangementln(x)=−u⇒I=∫01sin(lnx)lnxdx=−∫0+∞−sinu−u(−e−u)du=∫0∞sinuue−uduletf(t)=∫0∞sinuue−tuduwitht>0f′(t)=−∫0∞e−tusinudu=−Im(∫0∞e−tu+iudu)=−Im(∫0∞e(−t+i)udu)∫0∞e(−t+i)udu=[1−t+ie(−t+i)u]0+∞=−1−t+i=1t−i=t+it2+1⇒f′(t)=−11+t2⇒f(t)=k−arctantlimt→0f(t)=∫0∞sinuudu=π2=k⇒f(t)=π2−arcrant(t>0)I=f(1)=π2−π4=π4

Answered by mind is power last updated on 09/Apr/20

u=−ln(x)⇒x=e^(−u) dx=−e^(−u) du =∫_0 ^(+∞) ((sin(u))/u)e^(−u) du f(t)=∫_0 ^(×∞) ((sin(u))/u)e^(−ut) du f′(t)=∫_0 ^(+∞) −sin(u)e^(−ut) du =−Im∫_0 ^(+∞) e^(u(i−t)) du =Im{(1/(i−t))}=((−1)/(1+t^2 )) f(t)=−arctan(t)+(π/2) f(1)=(π/2)−(π/4)=(π/4)=∫_0 ^1 ((sin(ln(x)))/(ln(x)))dx=(π/4)

u=−ln(x)⇒x=e−udx=−e−udu=∫0+∞sin(u)ue−uduf(t)=∫0×∞sin(u)ue−utduf′(t)=∫0+∞−sin(u)e−utdu=−Im∫0+∞eu(i−t)du=Im{1i−t}=−11+t2f(t)=−arctan(t)+π2f(1)=π2−π4=π4=∫01sin(ln(x))ln(x)dx=π4

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