Differentials of variable substitution

[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 ?

 

[Fredrik]

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.

 

[CantorSet]

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.

I see. Thanks for the explanation, Fredrik.