How do you integrate $e^{-st}t\cos(t)$?

Question

I'm doing differential equations and specifically studying Laplace Transformations, where of course the Kernel is:

$K(s,t) = e^{-st}$

And the Laplace Transformation $\mathcal{L}$ of a function $f(t)$ is:

$\mathcal{L\{f(t)\}} = \int_0^{\infty}K(s,t)f(t)\,dt$

So if I'm given $f(t) = t\cos(t)$, which gives me:

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

How do I go about integrating this? It's not nearly as simple as doing Integration By Parts since I have three different functions here $(e^{-st}, t, \cos(t))$. No ideas for $u$-substitution either. Is there a trig sub that I'm not seeing here?

Answer 1

There's a cool way to do this integral, by differentiating under the integral sign for the following integral:

$$ \mathcal{L_1[f(t)]} = \int_0^{\infty}e^{-st}\cos t \hspace{3 pt}\text{d}t $$

$$ \frac{d}{ds}\mathcal{L_1[f(t)]} = \int_0^{\infty}\frac{\partial}{\partial s}\left(e^{-st}\cos t\right)\text{d}t = -\int_0^{\infty}te^{-st}\cos t \hspace{3 pt}\text{d}t $$

So, find the Laplace transform of $\mathcal{L_1[f(t)]}$ in terms of $s$ and differentiate the expression with respect to $s$. Change the sign to get your answer!

Answer 2

Hint

Consider $$I=\int e^{-st}\,t\,\cos(t)\,dt \qquad J=\int e^{-st}\,t \,\sin(t)\,dt$$ $$I+iJ=\int e^{-st}\,t\, e^{it}\,dt=\int e^{-(s-i)t}\,t\,dt$$ Integrate by parts, take the real part of the result and apply bounds.

I am sure that you can take it from here.

Answer 3

$$\int_0^\infty t\cos t e^{-st}dt=-\frac{d}{ds}\int_0^\infty\cos t e^{-st}dt=-\frac{d}{ds}\int_0^\infty\ \Re{e^{-(s-i)t}}dt$$

$$=-\frac{d}{ds}\Re{\int_0^\infty e^{-(s-i)t}dt}=-\frac{d}{ds}\Re{\frac{1}{s-i}}=-\frac{d}{ds}\frac{s}{s^2+1}=\frac{s^2-1}{(s^2+1)^2}$$