Laplace transform the expression $\int_0^t(t-u)y(u)du$

Question

I laplace transformed the expression $\int_0^t(t-u)y(u)du$ in Wolfram and it seems like the answer is just $\frac{Y(s)}{s^2}$.

If I change the expression to this:

$$ \int_0^t(t-u)y(u)du = t\int_0^t y(u)du-\int_0^t uy(u)du$$ then I would calculate the laplace transform to:

$$\frac{1}{s^2}Y(s)+\frac{1}{s}Y'(s)$$

But it seems that my calculation is wrong. Why does the $\frac{1}{s}Y'(s)$ just dissapear? What is wrong with my calculation?

Answer 1

The problem is in the first integral. You have

$$\mathcal{L}\left\{\int_0^ty(u)du\right\}=\frac{Y(s)}{s}$$

and

$$\mathcal{L}\left\{t\int_0^ty(u)du\right\}=-\left(\frac{Y(s)}{s}\right)'= -\frac{Y'(s)s-Y(s)}{s^2}=\frac{Y(s)}{s^2}-\frac{Y'(s)}{s}$$

If you add the term $Y'(s)/s$ from the second integral, you're left with the correct result $Y(s)/s^2$.