$\displaystyle Consider \int_0^{\infty{}}e^{-st}t^3 sin t\ dt = L (t^3sin t)\\[2ex] \displaystyle If L[f(t)] = f(s), then L[ t^nf(t)] = {(-1)}^n\ \frac{d^n}{{ds}^n} f(s)\\[2ex] \displaystyle \therefore{}\ L (t^3 \sin t) = {(-1)}^3\ \frac{d^3}{{ds}^3} L (sin t) \\[2ex] \displaystyle = -\frac{d^3}{{ds}^3}\left[\frac{1}{s^2+1}\right] \frac{d}{ds}\left[\frac{1}{s^2+1}\right]=-\ \frac{2s}{{\left(s^2+1\right)}^2} \\[2ex] \displaystyle \frac{d}{ds}\left[-\ \frac{2s}{{\left(s^2+1\right)}^2}\right] = -2\left[\frac{{\left(s^2+1\right)}^2.1-\ \ s\ .\ \ 2\left(s^2+1\right).2s}{{\left(s^2+1\right)}^4}\right] \\[3ex] \displaystyle = -2\left[\frac{{\left(s^2+1\right)}^2.1-\ \ s\ .\ \ 2\left(s^2+1\right).2s}{{\left(s^2+1\right)}^4}\right]\\[2ex] \displaystyle = -2\left[\frac{\left(s^2+1\right)\ -\ 4s^2}{{\left(s^2+1\right)}^3}\right] \\[3ex] \displaystyle = 2 \frac{\left({3s}^2-1\right)\ }{{\left(s^2+1\right)}^3} \\[2ex] \displaystyle \frac{d}{ds}\left[2\ \frac{\left({3s}^2-1\right)\ }{{\left(s^2+1\right)}^3}\right] = 2\left[\ \frac{{\left(s^2+1\right)}^3.6s\ -\ \left({3s}^2-1\right).3{\left(s^2+1\right)}^2.2s\ }{{\left(s^2+1\right)}^6}\right] \\[2ex] \displaystyle = 2\left[\ \frac{6s\ \left(s^2+1\right)\ -\ 6s\ \left({3s}^2-1\right)\ }{{\left(s^2+1\right)}^4}\right] \\[2ex] \displaystyle = -24s \frac{\left(s^2-1\right)\ }{{\left(s^2+1\right)}^4} \\[2ex] \displaystyle \therefore{} \int_0^{\infty{}}e^{-st}t^3 sin t\ dt = -24s \frac{\left(s^2-1\right)\ }{{\left(s^2+1\right)}^4} \\[2ex] \displaystyle Putting s = 1 \\[2ex] \displaystyle \therefore{}\ \int_0^{\infty{}}e^{-t}t^3 \sin t\ dt = -24s \frac{\left(1^2-1\right)\ }{{\left(1^2+1\right)}^4} = 0 \frac{d}{ds}\left[-\ \frac{2s}{{\left(s^2+1\right)}^2}\right] $