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Question Number 131673 by liberty last updated on 07/Feb/21

 Nice integral    ∫_0 ^( 1)  ((sin (ln x))/(ln x)) dx =?

Niceintegral01sin(lnx)lnxdx=?

Answered by mindispower last updated on 07/Feb/21

ln(x)=−t  ⇒∫_0 ^∞ ((sin(t))/t)e^(−t) dt=A  g(a)=∫_0 ^∞ ((sin(t)e^(−at) )/t)dt  g′(a)=Im−∫_0 ^∞ e^((i−a)t) dt  =−Im[(1/((a−i)))]=(1/(a^2 +1))  g(a)=tan^− (a)+c   g(0)=(π/2)⇒g(a)=tan^− (a)+(π/2)  A=g(1)=(π/4)+(π/2)=((3π)/4)

ln(x)=t0sin(t)tetdt=Ag(a)=0sin(t)eattdtg(a)=Im0e(ia)tdt=Im[1(ai)]=1a2+1g(a)=tan(a)+cg(0)=π2g(a)=tan(a)+π2A=g(1)=π4+π2=3π4

Commented by liberty last updated on 07/Feb/21

i think is (π/4) sir

ithinkisπ4sir

Commented by mathmax by abdo last updated on 07/Feb/21

sir small error  −Im((1/(a−i)))=−Im(((a+i)/(1+a^2 )))=−(1/(1+a^2 )) ⇒....

sirsmallerrorIm(1ai)=Im(a+i1+a2)=11+a2....

Answered by Dwaipayan Shikari last updated on 07/Feb/21

∫_0 ^1 ((sin(log(x)))/(log(x)))dx    log(x)=u⇒1=x(du/dx)  =∫_(−∞) ^0 ((sin(u))/u)e^u  du            u=−j  =∫_0 ^∞ ((sin(i))/j)e^(−j) du       I(α)=∫_0 ^∞ ((sinj)/j)e^(−αj) dj  I′(α)=−∫_0 ^∞ sin(j)e^(−αj) dj=−(1/(α^2 +1))  I(α)=−tan^(−1) α+C          (α→∞  C=(π/2))  I(1)=(π/2)−(π/4)=(π/4)

01sin(log(x))log(x)dxlog(x)=u1=xdudx=0sin(u)ueuduu=j=0sin(i)jejduI(α)=0sinjjeαjdjI(α)=0sin(j)eαjdj=1α2+1I(α)=tan1α+C(αC=π2)I(1)=π2π4=π4

Commented by liberty last updated on 07/Feb/21

okay

okay

Answered by liberty last updated on 07/Feb/21

I(β)=∫_0 ^1 ((sin (β ln x))/(ln x)) dx   I ′(β) = ∫_0 ^1  ln (x) ((cos (β ln x))/(ln (x))) dx   I ′(β) = ∫_0 ^1  cos (β ln x) dx   change of variables u = ln (x)  I ′(β)=∫_(−∞) ^0 cos (βu) e^u  du   I ′(β) = (1/(1+β^2 )) then I(β) = arctan (β) +C  setting β = 0 hence C = 0  I(β=1)= arctan 1 = (π/4)

I(β)=01sin(βlnx)lnxdxI(β)=01ln(x)cos(βlnx)ln(x)dxI(β)=01cos(βlnx)dxchangeofvariablesu=ln(x)I(β)=0cos(βu)euduI(β)=11+β2thenI(β)=arctan(β)+Csettingβ=0henceC=0I(β=1)=arctan1=π4

Answered by mathmax by abdo last updated on 07/Feb/21

Φ =∫_0 ^1  ((sin(lnx))/(lnx))dx  we do the changement lnx=−t ⇒x=e^(−t)   Φ =−∫_0 ^∞  ((sin(−t))/(−t))(−e^(−t) )dt =∫_0 ^∞  ((sint)/t)e^(−t)  dt let consider  f(a) =∫_0 ^∞  ((sint)/t)e^(−at)    with a>0 ⇒f^′ (a) =−∫_0 ^∞  sint e^(−at)  dt  =−Im(∫_0 ^∞  e^(it−at)  dt)  we have ∫_0 ^∞  e^((−a+i)t) dt =[(1/(−a+i))e^((−a+i)t) ]_0 ^∞   =−(1/(a−i))(−1) =(1/(a−i))=((a+i)/(a^2  +1)) ⇒f^′ (a)=−(1/(1+a^2 )) ⇒f(a)=c−arctan(a)  f(0)=c=∫_0 ^∞  ((sint)/t)dt =(π/2) ⇒f(a)=(π/2)−arctan(a)  Φ =f(1) =(π/2)−arctan(1) =(π/2)−(π/4) =(π/4)

Φ=01sin(lnx)lnxdxwedothechangementlnx=tx=etΦ=0sin(t)t(et)dt=0sinttetdtletconsiderf(a)=0sintteatwitha>0f(a)=0sinteatdt=Im(0eitatdt)wehave0e(a+i)tdt=[1a+ie(a+i)t]0=1ai(1)=1ai=a+ia2+1f(a)=11+a2f(a)=carctan(a)f(0)=c=0sinttdt=π2f(a)=π2arctan(a)Φ=f(1)=π2arctan(1)=π2π4=π4

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