## UnkleRhaukus 4 years ago Show that $\mathcal{L} \{{1 \over x}f(x)\} =\int _s^∞F(s)ds$

1. JamesJ

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

2. UnkleRhaukus

how do we swap order of integration again

3. JamesJ

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.

4. JamesJ

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

5. JamesJ

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

6. UnkleRhaukus

|dw:1328840642309:dw|

7. JamesJ

Look at the region in x,s-space.

8. JamesJ

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 $\int_s^\infty F(s') \ ds'$

9. JamesJ

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

10. UnkleRhaukus

so x,s' space yeah

11. JamesJ

Got it?

12. UnkleRhaukus

is this the right region s' from s to ∞ x form 0 to ∞|dw:1328841005870:dw|

13. JamesJ

Yes, hence when you change the order of integration, do the limits change?

14. UnkleRhaukus

well now x gos from 0 to ∞ and s' goes from s to ∞. the limits have not changed

15. UnkleRhaukus

ok i think i have it

16. UnkleRhaukus

I got there thank you James

17. JamesJ

It's a nice result, and what you might hypothesize, given the Laplace transform of xf(x)