Answer

$sint \ sinht= (\dfrac{e^{t}-e^{-t}}{2})\ sint\\ L sint=\dfrac{1}{s^2+1}\\ \therefore L(e^tsint)=\dfrac{1}{(s-1)^2+1},L(e^{-t}sint)=\dfrac{1}{(s+1)^2+1},$

$Lsint sinht=\dfrac1{2}(\dfrac{1}{(s-1)^2+1} -\dfrac{1}{(s+1)^2+1} ]\\ L(\dfrac{sint sinht}{t})=\dfrac1{2}\int_s^\infty[\dfrac{1}{(s-1)^2+1} -\dfrac{1}{(s+1)^2+1} ]ds\\ =\dfrac1{2} [tan^{-1}(s-1)-tan^{-1}(s-1)]_s^\infty\\ =\dfrac{1}{2}[\dfrac{\pi}{2}-tan^{-1}(s-1)]-[\dfrac{\pi}{2}-tan^{-1}(s+1)]\\ \dfrac{1}{2}[tan^{-1}(s+1)]-tan^{-1}(s-1)]\\ $

$Now\ tan^{-1}(s+1)=\alpha,\ tan^{-1}(s+1)=\beta\\ \therefore tan\alpha =s+1,tan\beta=s-1 \\tan(\alpha-\beta)=\dfrac{tan\alpha-tan\beta}{1+tan\alpha.tan\beta}\\ =\dfrac{(s+1)-(s-1)}{1+(s+1)(s-1)}\\ =\dfrac{2}{s^2}$

$\alpha-\beta=tan^{-1}(\dfrac{2}{s^2})\\ [tan^{-1}(s-1)-tan^{-1}(s-1) ]=tan^{-1}(\dfrac{2}{s^2}) \\ L(\dfrac{sint sinht}{t}) =\dfrac{1}{2}.tan^{-1}(\dfrac{2}{s^2})\\ This\ means \\ \int _0^\infty e^{-st}.\dfrac{sint sinht}{t}.dt=\dfrac{1}{2}.tan^{-1}(\dfrac{2}{s^2})\\ put \ s=2\\ \int _0^\infty e^{-2t}.\dfrac{sint sinht}{t}.dt=\dfrac{1}{2}.tan^{-1}( \dfrac{1}{2})\\ $