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)