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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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  • $\begingroup$ For some reason, this relation is not taught in many brief introductions to Laplace transforms (even though it appears early in every table of the transforms). It really makes short work of frequently-assigned problems such as finding the transform of $t \cos (kt)$, which requires two passes of integration by parts if students are limited to real-variable integration... $\endgroup$
    – colormegone
    Apr 19, 2013 at 20:33

2 Answers 2

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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.

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What is

$\displaystyle \frac{d e^{-st}}{ds}$

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  • $\begingroup$ @Amzoti: Thanks for the tip. $\endgroup$
    – bob.sacamento
    Apr 19, 2013 at 20:04

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