Show that \[\mathcal{L} \{{1 \over x}f(x)\} =\int _s^∞F(s)ds\]

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Show that \[\mathcal{L} \{{1 \over x}f(x)\} =\int _s^∞F(s)ds\]

Mathematics
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The RHS is equal to \[ \int_s^\infty \int_0^\infty f(x) e^{-sx} dx \ ds \] Now swap the order of integration and you'll see it's not hard to show that this must be equal to \[ \int_0^\infty \frac{1}{x}f(x) e^{-sx} dx \]
how do we swap order of integration again
As ever, draw a diagram first and see what the region is. Then figure out how the limits change. In this case, it's pretty straight forward. Try it first and tell if you're stuck.

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figure out how the limits change when you change the order of integration that is.
They may not change at all; you'll need to convince yourself one way or another.
|dw:1328840642309:dw|
Look at the region in x,s-space.
what's a bit confusing here is that the s of limit of integration wrt to s is actually not the same variable. Which is to say, it would have been better if the RHS had been written as \[ \int_s^\infty F(s') \ ds' \]
So look at the region in x,s'-space. It's very regular.
so x,s' space yeah
Got it?
is this the right region s' from s to ∞ x form 0 to ∞|dw:1328841005870:dw|
Yes, hence when you change the order of integration, do the limits change?
well now x gos from 0 to ∞ and s' goes from s to ∞. the limits have not changed
ok i think i have it
I got there thank you James
It's a nice result, and what you might hypothesize, given the Laplace transform of xf(x)

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