By definition of L.T. $\int\limits_0^{\infty} e^{-st} f(t)dt= L[f(t)]\\ \int\limits_0^{\infty} e^{-st} [\dfrac {\sin ht\space \sin t}t]dt=L[\sin ht \dfrac {\sin t}t] =L[(\dfrac {e^t-e^{-t}}2\dfrac {\sin t} t] \\ =\dfrac 12\{ L [e^t\dfrac {\sin t}t ] - L [e^{-t} \dfrac {\sin t} t]\} \\ =\dfrac 12 \{[L\dfrac {\sin t}t]_{s\rightarrow s-1} - L[\dfrac {\sin t}t]_{S\rightarrow s+1}\}------ (1) \\ But \space L [\dfrac {\sin t}t] =\int\limits_S^\infty L[\sin t]ds=\int\limits_S^\infty \dfrac 1{s^2+1} ds\\ =\tan^{-1}+[S]^\infty_S= \dfrac \pi2 -\tan^{-1}S ----- \text {put in } (1) \\ =\dfrac 12[\dfrac \pi2-\tan^{-1} (S-1) -\dfrac \pi2 +\tan^{-1} (S+1)] =\dfrac 12[\tan^{-1} (S+1) -\tan^{-1}(S-1)] \\ =\dfrac 12\tan^{-1} [\dfrac {(S+1)-(S-1)}{1+(S+1)(S-1)}]=\dfrac 12\tan^{-1} [\dfrac 2{3^2}] \\ Put \space S=2 \\ \int\limits_0^\infty e^{-2t} [\dfrac {\sin ht \space \sin t}t]dt=\dfrac 12\tan^{-1} [\dfrac 24]=\dfrac 12\tan^{-1} [\dfrac 12]$