Laplace of $\sinh t\sin t$

Question

I am trying to solve a question where we have to find the Laplace of the $f(t)$ $$\int_0^\infty \frac{e^{-2t}\sinh t\sin t }{t} dt.$$ Well I approach by converting the $f(t)$ to Laplace by assuming $s=2$, then
$$\int_0^\infty \frac{e^{-st}\sinh t \sin t}{t}dt =L\left[\frac{\sinh t \sin t}{t}\right]$$
after this I easily convert the $t$ to intergral from $s$ to infinty but how to solve the Laplace of $\sinh t \sin t$, since the linearity does not apply here, I thought to use $(e^{it} - e^{-it})/ (2i) =\sin t$ and the standard definition for $\sinh t$ to solve this question. Should I use this or is there any other way to solve the question? A simple hint would be enough.

Answer 1

Expanding on @Gary's comment, famously $\int_0^\infty e^{-st}\frac{\sin t}{t}dt=\frac{\pi}{2}-\arctan s$ whenever $\Re s>0$, so$$\begin{align}\int_0^\infty\frac{e^{-st}\sinh t\sin t}{t}dt&=\frac12\left(\int_0^\infty\frac{e^{-(s-1)t}\sin t}{t}dt-\int_0^\infty\frac{e^{-(s+1)t}\sin t}{t}dt\right)\\&=\frac{\arctan(s+1)-\arctan(s-1)}{2}\end{align}$$when $\Re s>1$. For $s=2$, this is $\frac12\arctan 3-\frac{\pi}{8}$.