A question about Laplace Transform: $\mathcal{L}\{yf(y)\}$ I want to find the Laplace transform of 
$\mathcal{L}\{yf(y)\}$ where
$$\mathcal{L}\{f(y)\}=\int_{0}^{\infty} e^{-sy}f(y)\text{d}y=F(s).$$
How can one help me?
 A: Hint. By differentiating
$$
F(s):=\int_{0}^{\infty} e^{-sy}f(y)\text{d}y \tag1
$$ one just gets
$$
-F'(s)=\int_{0}^{\infty} y\:e^{-sy}f(y)\text{d}y=\mathcal{L}\{yf(y)\}, \tag2
$$ assuming all expressions exist.
A: By definition:
$$\text{F}\left(\text{s}\right)=\mathcal{L}_t\left[\text{f}\left(t\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\text{f}\left(t\right)e^{-\text{s}t}\space\text{d}t$$
So, when we use the derivative:
$$\text{F}'\left(\text{s}\right)=\mathcal{L}_t\left[\text{f}\left(t\right)\right]'_{\left(\text{s}\right)}=\frac{\partial}{\partial\text{s}}\left\{\int_0^\infty\text{f}\left(t\right)e^{-\text{s}t}\space\text{d}t\right\}=-\int_0^\infty t\text{f}\left(t\right)e^{-\text{s}t}\space\text{d}t=-\mathcal{L}_t\left[t\text{f}\left(t\right)\right]_{\left(\text{s}\right)}$$
So:
$$-\text{F}'\left(\text{s}\right)=-\mathcal{L}_t\left[\text{f}\left(t\right)\right]'_{\left(\text{s}\right)}=\mathcal{L}_t\left[t\text{f}\left(t\right)\right]_{\left(\text{s}\right)}$$
But you've to pay attention to that all the functions exist, otherwise this does not hold.

In general:


*

*When $\text{n}\in\mathbb{N}$:
$$\text{F}^{\left(2\text{n}\right)}\left(\text{s}\right)=\mathcal{L}_t\left[\text{f}\left(t\right)\right]^{\left(2\text{n}\right)}_{\left(\text{s}\right)}=\mathcal{L}_t\left[t^{2\text{n}}\text{f}\left(t\right)\right]_{\left(\text{s}\right)}$$

*When $\text{n}\in\mathbb{N}$:
$$-\text{F}^{\left(1+2\text{n}\right)}\left(\text{s}\right)=-\mathcal{L}_t\left[\text{f}\left(t\right)\right]^{\left(1+2\text{n}\right)}_{\left(\text{s}\right)}=\mathcal{L}_t\left[t^{1+2\text{n}}\text{f}\left(t\right)\right]_{\left(\text{s}\right)}$$

