Q. prove that $\mathfrak{L}\{tf(x)\}=-\frac{d\mathfrak{L}\{f(x)\}}{ds}$ where the notation used is standard one.
Attempt I tried what would seem obvious way to start: $$\mathfrak{L}\{tf(x)\}=\int_0^\infty e^{-st}tf(t)dt$$ I think integration by parts is the way but I it seems too complicated. If there are brief ways to prove this, I would really appreciate your help.
Thank you.
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$\begingroup$The quickest way, write down the definition of the Laplace transform $$\mathfrak{L}\{f(x)\}=\int_0^\infty e^{-st}f(t)dt$$
Differentiate WRT $s$, and swap differential and integral operations (assuming everything exists):
$$\frac{d}{ds}\mathfrak{L}\{f(x)\}=\frac{d}{ds}\int_0^\infty e^{-st}f(t)dt =\int_0^\infty \frac{d}{ds}\left(e^{-st}\right)f(t)dt =\int_0^\infty -t e^{-st} f(t)dt $$
Which is that $$-\frac{d\mathfrak{L}\{f(x)\}}{ds} = \int_0^\infty t f(t) e^{-st} dt $$
Lump the $t f(t)$ together and that integral is a Laplace transform, and you're done.
$\endgroup$ $\begingroup$What is
$\displaystyle \frac{d e^{-st}}{ds}$
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