Question

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

Answer 1

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
\[ \int_0^\infty \frac{1}{x}f(x) e^{-sx} dx \]

Answer 2

how do we swap order of integration again

Answer 3

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.

Answer 4

figure out how the limits change when you change the order of integration that is.

Answer 5

They may not change at all; you'll need to convince yourself one way or another.

Answer 6

Look at the region in x,s-space.

Answer 7

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
\[ \int_s^\infty F(s') \ ds' \]

Answer 8

So look at the region in x,s'-space. It's very regular.

Answer 9

so x,s' space yeah

Answer 10

Got it?

Answer 11

is this the right region s' from s to ∞
x form 0 to ∞