Proof laplace trasform of $xf(x)$

Question

I want to prove that $\mathcal L[xf(x)] = -\frac{d \tilde{f}(s)}{d t}$

my try

By definition $$\mathcal L[x(f(x)] = \int_0^{\infty} xf(x) e^{-sx}dx = \left[x\int f(x)e^{-sx}dx\right]_0^\infty-\int_0^\infty f(x)e^{-sx}dx$$

But now how to I proceed?

Answer 1

$$ \frac{\mathrm{d}}{\mathrm{d}s}\mathcal{L}\left[f(x)\right]=\frac{\mathrm{d}}{\mathrm{d}s}\int_{0}^{+\infty}f(x)e^{-sx}\mathrm{d}x $$

$$ \frac{\mathrm{d}}{\mathrm{d}s}\mathcal{L}\left[f(x)\right]=\int_{0}^{+\infty}f(x)\frac{\mathrm{d}}{\mathrm{d}s}(e^{-sx})\mathrm{d}x $$

and

$$ \frac{\mathrm{d}}{\mathrm{d}s}(e^{-sx})=-xe^{-sx} $$

$$ \frac{\mathrm{d}}{\mathrm{d}s}\mathcal{L}\left[f(x)\right]=-\int_{0}^{+\infty}f(x)xe^{-sx}\mathrm{d}x $$

$$ \frac{\mathrm{d}}{\mathrm{d}s}\mathcal{L}\left[f(x)\right](s)=-\mathcal{L}\left[xf(x)\right](s) $$