What is Laplace transform of $\frac{e^{-t}}t \sin3t\sin2t$

Question

What is Laplace of :

$$\mathcal{L}\left(\frac{e^{-t}}t\sin3t\sin2t\right)$$

I m trying to use first shifting property and not able to get correct answer.

Answer 1

$$\frac{e^{-t}}{t} \sin(3t) \sin(2t) = \frac{e^{-t}}{2t}\cos(t) - \frac{e^{-t}}{2t}\cos(5t)$$ Let $f(t) = e^{-t}\cos(t)$ and $g(t) = e^{-t}\cos(5t)$, we get $$\frac{e^{-t}}{t} \sin(3t) \sin(2t) =\frac{1}{2} (\frac{f(t)}{t} -\frac{g(t)}{t})$$ Using the property that $$L(\frac{f(t)}{t}) = \int\limits_s^{\infty} F(u) \ du$$ we get $$L(\frac{e^{-t}}{t} \sin(3t) \sin(2t)) = \frac{1}{2}[\int\limits_s^{\infty} F(u) \ du - \int\limits_s^{\infty} G(u) \ du]$$ where $F(s),G(s)$ are Laplace Transforms of $f(t),g(t)$. Using $L(e^{-at}\cos( wt) ) = \frac{s+a}{(s+a)^2 + w^2}$, we can say $$F(u) = \frac{s+1}{(s+1)^2 + 1} $$ $$G(u) = \frac{s+1}{(s+1)^2 + 25} $$ Computing the integrals: $$\int\limits_s^{\infty} F(u) \ du = \int\limits_s^{\infty} \frac{u+1}{(u+1)^2 + 1} \ d u = \frac{1}{2}[\lim_{u \rightarrow \infty} \ln ((u+1)^2 + 1) - \ln ((s+1)^2 + 1)]$$ $$\int\limits_s^{\infty} G(u) \ du = \int\limits_s^{\infty} \frac{u+1}{(u+1)^2 + 25} \ d u = \frac{1}{2}[\lim_{u \rightarrow \infty} \ln ((u+1)^2 + 25) - \ln ((s+1)^2 + 25)]$$ Denote the limits as $A,B$, then: $$L(\frac{e^{-t}}{t} \sin(3t) \sin(2t)) = \frac{1}{4}[\ln ((s+1)^2 + 25)-\ln ((s+1)^2 + 1) + A-B]$$ which is $$L(\frac{e^{-t}}{t} \sin(3t) \sin(2t)) = \frac{1}{4}[\ln \frac{(s+1)^2 + 25}{(s+1)^2 + 1} + \lim_{u \rightarrow \infty} \ln \frac{(u+1)^2 + 25}{(u+1)^2 + 1}] = \frac{1}{4}\ln \frac{(s+1)^2 + 25}{(s+1)^2 + 1}$$

Answer 2

Hint

You can use $$\sin(a) \sin(b)=\frac 12( \cos(a-b)-\cos(a+b))$$ $$\sin(3t) \sin(2t)=\frac 12( \cos(t)-\cos(5t))$$ Then use the formula $$\mathcal{L} \left(\frac {f(t)}t\right )=\int_s^\infty F(u)du$$ I finally got this $$F(s)=\frac 14 \ln \left (\frac {s^2+2s+26}{s^2+2s+2} \right)$$

Answer 3

With integration of the formula ${\cal L}\left(e^{ct}\right)=\dfrac{1}{s-c}$ respect to $c$ we find $${\cal L}\left(\dfrac{e^{ct}}{t}\right)=\ln\dfrac{1}{s-c}$$ then by $\sin\alpha=\dfrac{e^{i\alpha}-e^{-i\alpha}}{2i}$ we write \begin{align} {\cal L}\left(\dfrac{e^{-t}\sin3t\sin2t}{t}\right) &= -\dfrac14{\cal L}\left(\dfrac{e^{(5i-1)t}-e^{(-i-1)t}-e^{(i-1)t}+e^{(-5i-1)t}}{t}\right)\\ &= -\dfrac14{\cal L}\left(\ln\dfrac{1}{s-(5i-1)}-\ln\dfrac{1}{s-(-i-1)}-\ln\dfrac{1}{s-(i-1)}+\ln\dfrac{1}{s-(-5i-1)}\right)\\ &= \color{blue}{-\dfrac14\ln\dfrac{(s+1)^2+1}{(s+1)^2+25}} \end{align}