Laplace transform of product of $\sinh(t)$ and $\cos(t)$

Question

If I have a function $f(t)=\sinh(t)\cos(t)$ how would I go about finding the Laplace transform? I tried putting it into the integral defining Laplace transformation: $$ F(s)= \int_0^\infty \mathrm{e}^{-st}\sinh(t)\cos(t)\,\mathrm{d}t $$ But this integral looks very hairy to me. Can $\sinh(t)\cos(t)$ be rewritten as something more manageable perhaps?

Answer 1

I would use the exponential formula for $\sinh$, namely $\sinh t = \frac{e^{t}-e^{-t}}{2}$. This turns your integral into $$\frac{1}{2}\int_0^\infty e^{-(s-1)t} \cos t \, dt - \frac{1}{2}\int_0^\infty e^{-(s+1)t} \cos t \, dt$$ Now, if you know what the Laplace transform of the cosine is, you can use that to evaluate these integrals...

Answer 2

HINT: $$\sinh t\cos t=\frac{e^t-e^{-t}}2\cdot\frac{e^{it}+e^{-it}}2=\frac{e^{t(1+i)}+e^{t(1-i)}-e^{t(i-1)}-e^{-t(1+i)}}4$$

Now, $$L\{e^{at}\}=\frac1{s-a}$$