Differentials of variable substitution

In summary, the conversation discusses the use of variable substitution in the proof of the convolution theorem. The substitution r=s-t is used and the differential relation dr=ds is applied. However, since t is a constant, the "d" of a constant function is 0, which explains why the differential relation dr=ds is used instead of dr=ds-dt.
  • #1
CantorSet
44
0
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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  • #2
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.
 
  • #3
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.
 

FAQ: Differentials of variable substitution

What is a differential of variable substitution?

A differential of variable substitution is a mathematical concept used to simplify complex integrals. It involves substituting a new variable in place of the original variable in the integral, and then using the chain rule to rewrite the integral in terms of the new variable. This can make the integral easier to solve or evaluate.

How does a differential of variable substitution work?

The process of a differential of variable substitution involves substituting a new variable, typically denoted as u, in place of the original variable in the integral. This new variable is chosen in such a way that it simplifies the integral, such as by canceling out terms or making the integrand easier to integrate. This is then followed by using the chain rule to rewrite the integral in terms of the new variable.

What are the benefits of using a differential of variable substitution?

There are several benefits of using a differential of variable substitution. It can simplify complex integrals, making them easier to solve or evaluate. It can also help to identify patterns and relationships between different integrals, which can be useful in solving other integrals. Additionally, it can be used to transform integrals into a more manageable form, allowing for the use of other integration techniques.

What types of integrals can be solved using a differential of variable substitution?

A differential of variable substitution can be used to solve a wide range of integrals, including those involving polynomial, trigonometric, exponential, and logarithmic functions. It can also be used for integrals with multiple variables, as long as one of the variables can be substituted for a new variable using the chain rule.

Are there any limitations to using a differential of variable substitution?

While a differential of variable substitution can be a powerful tool in solving integrals, it may not always be the most efficient or effective method. In some cases, it may not be possible to find a suitable substitution that simplifies the integral. It also may not work for integrals with discontinuities or singularities, as the substitution may not be valid in these cases. Additionally, it may require some creativity and practice to identify the best substitution for a given integral.

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