# Differentials of variable substitution

1. Sep 22, 2010

### CantorSet

Hi everyone,

This question is a bit involved but it pertains to calculating the differential of a variable substitution used in the proof of the convolution theorem (http://en.wikipedia.org/wiki/Convolution_theorem)

Consider
$$\int f(t) \int g(s - t) ds dt.$$

If we use the substitution

$$r = s - t$$

we get the differential relation as

$$dr = ds$$

so the above equation becomes

$$= \int f(t) \int g(r) dr dt = \int f(t) dt \int g(r) dr$$

But why didn't we use the differential relation

$$dr = ds - dt$$ ?

2. Sep 22, 2010

### Fredrik

Staff Emeritus
You did, but the "d" of a constant function is =0. Note that you did the variable substitution r=s-t in the integral

$$\int g(s-t)ds$$

where t is just a constant; it represents the constant function h defined by h(s)=t for all s.

3. Sep 28, 2010

### CantorSet

I see. Thanks for the explanation, Fredrik.