Differentials of variable substitution

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SUMMARY

This discussion focuses on the differential of variable substitution in the context of the convolution theorem. The substitution of variables, specifically r = s - t, leads to the differential relation dr = ds, simplifying the integral expression. The conversation clarifies why the alternative differential relation dr = ds - dt is not applicable, as the differential of a constant function is zero. This understanding is crucial for correctly applying variable substitutions in integral calculus.

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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
[tex]\int f(t) \int g(s - t) ds dt.[/tex]

If we use the substitution

[tex]r = s - t[/tex]

we get the differential relation as

[tex]dr = ds[/tex]

so the above equation becomes

[tex]= \int f(t) \int g(r) dr dt = \int f(t) dt \int g(r) dr[/tex]

But why didn't we use the differential relation

[tex]dr = ds - dt[/tex] ?
 
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You did, but the "d" of a constant function is =0. Note that you did the variable substitution r=s-t in the integral

[tex]\int g(s-t)ds[/tex]

where t is just a constant; it represents the constant function h defined by h(s)=t for all s.
 
Fredrik said:
You did, but the "d" of a constant function is =0. Note that you did the variable substitution r=s-t in the integral

[tex]\int g(s-t)ds[/tex]

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.
 

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