emmostatwf

2022-09-06

Prove that $\mathcal{L}\left[xf\left(x\right)\right]=-\frac{d\stackrel{~}{f}\left(s\right)}{dt}$
my try
By definition
$\mathcal{L}\left[x\left(f\left(x\right)\right]={\int }_{0}^{\mathrm{\infty }}xf\left(x\right){e}^{-sx}dx={\left[x\int f\left(x\right){e}^{-sx}dx\right]}_{0}^{\mathrm{\infty }}-{\int }_{0}^{\mathrm{\infty }}f\left(x\right){e}^{-sx}dx$
But now how to I proceed?

Quinn Alvarez

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

Do you have a similar question?