Inverse Laplace transform of $L^{-1}\Bigl(\frac{a}{(s^2+a^2)^2}\Bigr)$. Hence show that $\int_0^{\infty}t^3e^{-t}\sin t\,dt =0 $

Question

Any other way to find $\int_0^{\infty} te^{-2t}\cos t\, dt$. Hence show that $ \int_0^{\infty}t^3e^{-t}\sin t \, dt =0$.

My attempt:

$L(t \cos t) = \int_0^{\infty} e^{-pt}(t\cos t) \,dt$

Now evaluating this and putting p = 2, i get answer as 3/25

Now can you tell me is there any alternative way to evaluate this?

Problem says hence show $\int_0^{\infty}t^3e^{-t}\sin t\,dt = 0$

The integral = $L(t^3 \sin t) = \int_0^{\infty}t^3e^{-pt}\sin t\, dt = (-1)\frac{d^3}{dp^3} \frac{1}{p^2+1} = \frac{-8p}{(p^2-1)^2}+\frac{48p^3}{(p^2-1)^4} $

Now putting p =1 does not make sense in above equation. How to deal with this?

Answer 1

Note that we are differentiating $\frac1{1+p^2}$, so how do you get a $-1$ in the denominator?

The correct value of: $$(-1)\frac{d^3}{dp^3}\left(\frac1{1+p^2} \right) = \frac{24p\left(p^2-1\right)}{\left(p^2+1\right)^4}$$ which at $p=1$ is zero.