A simple application of dirac delta shift theorem help

[Seda]

A "simple" application of dirac delta "shift theorem"...help

Homework Statement

show that for a, b, c, d positive:

δ(a/b-c/d) = bdδ(ad-bc)

Homework Equations

∫f(x)δ(x-a)dx = f(a)

The Attempt at a Solution

Ok so I start with

∫δ(a/b-c/d)f(x)dx

But I am not sure how to apply the shift theorem. It seems I need to somehow relate a/b and x so that I can get it in the form of the shift theorem. But trying integration by substitution I always get tangled up. If I let u=a/b, then I can't relate dx to du to intergrate.

If I just say let's call a/b as "x". then dx = what?

ugh, this is a simple problem too. Seems like its an easy canditate for one of the first proofs shown after learning about the shift theorem so I feel pretty dumb that I'm not sure even where to start...

 

[lanedance]

thats not quite the shift theorem but a integral prepty of the delta function, though maybe it is...

I'm thinking you may want to consider a substitution, now first can you solve
\int f(x)\delta(mx-c)dx

 

[lanedance]

then how about associating x with a, and then the something around the substitution ybd = xd-cd. I haven't totally nailed it but it feels close

 

[Seda]

thats not quite the shift theorem but a integral prepty of the delta function,

I'm thinking you may want to consider a substitution, now first can you solve
\int f(x)\delta(mx-c)dx

Let u= mx ; du = mdx

(1/m)∫f(u/m)δ(u-c)du

Zero everywhere except where u=c therefore ..

sorry, I'm stuck

 

[lanedance]

\int f(x)\delta(mx-c)dx

you pretty much has it, but I'd do as follows, let u = mx-c, then x=(u+c)/m, dx=du/m
\int f(\frac{u+c}{m})\delta(u)\frac{1}{m}du

which gives
\int f(x)\delta(mx-c)dx =\frac{1}{m} \int f(\frac{x+c}{m})\delta(u) du= \frac{1}{m} f(\frac{c}{m})

so you can see there are a few similarities in transformation to the original problem, it just working up the substitution correctly

 

[Seda]

But my equation doesn't even have an x. So that's where I get tangled up.

I'm missing something obvious.

What I mean is, no matter what I substitute, I'm not doing to be able to determine what to replace for dx.

 

[lanedance]

or how about letting x(a) = a/b-c/d, then y(a) = bdx(a) = ad-bc, and dy = bd.dx

then yo uhave sometihing like
\int \delta(bdy) dy= \int \delta(x)bd.dx<br />