Question Number 103198 by Dwaipayan Shikari last updated on 13/Jul/20

∫_0 ^1 sin(logx)dx

Commented byDwaipayan Shikari last updated on 13/Jul/20

sinx=((e^(ix) −e^(−ix) )/(2i))  sin(logx)=((x^i −x^(−i) )/(2i))  ∫_0 ^1 ((x^i −x^(−i) )/(2i))dx=(1/(2i))∫x^i −x^(−i) dx=(1/(2i)).[((x^(i+1) /(i+1))−(x^(−i+1) /(1−i)))]_0 ^1   =(1/(2i))((1/(i+1))−(1/(1−i)))=−(1/2)  Is it true???????

Answered by OlafThorendsen last updated on 13/Jul/20

I = ∫_0 ^1 sin(lnx)dx  x = e^u  ⇒ dx = e^u du  I = ∫_(−∞) ^0 sinu.e^u du  I = ∫_(−∞) ^0 ((e^(iu) −e^(−iu) )/(2i))e^u du  I = (1/(2i))∫_(−∞) ^0 (e^((1+i)u) −e^((1−i)u) )du  I = (1/(2i))[(e^((1+i)u) /(1+i))−(e^((1−i)u) /(1−i))]_(−∞) ^0   I = (1/(2i))[(1/(1+i))−(1/(1−i))]  I = (1/(2i))[((1−i)/(1−i^2 ))−((1+i)/(1−i^2 ))]  I = (1/(2i))(((−2i)/2)) = −(1/2)

Answered by Smail last updated on 13/Jul/20

By parts  u=sin(lnx)⇒u′=(1/x)cos(lnx)  v′=1⇒v=x  ∫_0 ^1 sin(lnx)dx=[xsin(lnx)]_0 ^1 −∫_0 ^1 cos(lnx)dx  u=cos(lnx)⇒u′=−(1/x)sin(lnx)  v′=1⇒v=x  ∫_0 ^1 sin(lnx)dx=−[xcos(lnx)]_0 ^1 −∫_0 ^1 sin(lnx)dx  2∫_0 ^1 sin(lnx)dx=−cos(ln1)  ∫_0 ^1 sin(lnx)dx=((−1)/2)

Answered by mathmax by abdo last updated on 13/Jul/20

A =∫_0 ^1  sin(lnx)dx  changement lnx =−t give  A =−∫_0 ^∞  sin(−t)(−e^(−t) )dt =−∫_0 ^∞  e^(−t)  sint dt  =−Im(∫_0 ^∞  e^(−t+it) dt) and ∫_0 ^∞  e^((−1+i)t )  dt =[(1/(−1+i))e^((−1+i)t) ]_0 ^∞   =−(1/(−1+i)) =(1/(1−i)) =((1+i)/2) ⇒ A =−(1/2)

Answered by Aziztisffola last updated on 13/Jul/20

 let x=e^t  ⇒dx=e^t dt   ∫_0 ^1 sin(logx)dx=∫_(−∞) ^( 0) e^t sin(t)dt=∫_0 ^(+∞) −e^t sin(t)dt   With Laplace transform:   L{−e^t sin(t)}= L{−sin(t)}(s−1)=((−1)/((s−1)^2 +1))   then ∫_0 ^( +∞) −e^(−st) e^t sin(t)dt=((−1)/((s−1)^2 +1))    let s=0 ⇒ ∫_0 ^( +∞) −e^t sin(t)dt=((−1)/((−1)^2 +1))=((−1)/2)   Hence  ∫_(−∞) ^( 0) e^t sin(t)dt=((−1)/2)  ∴ ∫_0 ^1 sin(logx)dx= ((−1)/2)