\[I= \int^{\infty}_0 \frac{e^{-x}sinx}{x} dx\]
consider \[F( \alpha )= \int^{\infty}_0 \frac{e^{- \alpha x}sinx}{x} dx\]
.\[I=F(1)\]
.\[\frac{dF(\alpha)}{d \alpha}=\int^{\infty}_0 \frac{- x e^{- \alpha x}sinx}{x} dx= - \int^{\infty}_0 e^{- \alpha x}sinx\]
.Integration by parts returns for this integral
\[\frac{dF(\alpha )}{d \alpha} = [ \frac{e^{- \alpha x}}{2} (- \alpha sinx-cosx)]^{\infty}_0 = - \frac{1}{\alpha^2+1}\]
.Then
\[\frac{dF(\alpha )}{d \alpha }= - \frac{1}{ \alpha^2+1} \rightarrow F(\alpha )= \int - \frac{1}{\alpha^2+1} d \alpha =- tan^{-1} (\alpha )+c\]
/When
\[\alpha = \infty, \; F(\alpha )=0\]
so \[0=-tan^{-1} ( \infty )+c=- \frac{ \pi }{2} +c \rightarrow c= \frac{ \pi }{2}\]
.\[\int^{\infty}_0 \frac{e^{-x}sinx}{x} dx = F(1) = \frac{ \pi}{2}- tan^{-1} (1) = \frac{\pi}{4}\]
.