Quick question about integration

  • Thread starter dand5
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  • #1
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Main Question or Discussion Point

I have a quick question about integration after a change of variables has been made.

Suppose there is a function [tex] R(t_{1},t_{2}) [/tex] that actually just
depends on the difference [tex] t_{1} - t_{2} [/tex]. The goal is then to
simplify the following integral:

[tex]
\frac{1}{T^{2}}\int^{T}_{0}\int^{T}_{0} R(t_{1},t_{2}) dt_{1}dt_{2}
[/tex]

by using the substitution [tex] t_{1}' = t_{1} [/tex] and [tex] t_{2}'= t_{1} - \tau [/tex].

A straight substitution yields:
[tex]
\frac{1}{T^{2}}\int \int^{T}_{0} R(\tau) dt_{1}'(dt_{1}' - d\tau)
[/tex]

I am uncertain about two things:

1) the integration bounds on the outer integral after the substitution has been made
2) whether or not [tex]dt_{1}'[/tex] in the outer integral is zero since
[tex]dt_{1}[/tex] is held constant when integrating over [tex]dt_{2}[/tex] before the substitution was made.

As a heads up the final result is supposed to be:
[tex]
\frac{1}{T^{2}}\int^{T}_{-T}\left(T-\left|\tau\right|\right)R(\tau) d\tau
[/tex]

Thanks in advance for any responses.
 

Answers and Replies

  • #2
mathman
Science Advisor
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To simplify writing I will use s instead of tau.

Let s=t1-t2, t=t2. Then the differential looks like R(s)dsdt, with s limits -t,T-t and the t limits 0,T.

Next reverse the order of integration. The t limit is split into 2 parts. For s<0, the limits are -s and T, while for s>0, the limits are 0 and T-s. The s limits are -T and T.

When you integrate with respect to t, you get T-|s|.
 
  • #3
28
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ok I see it now, Thanks. Also, the determinant of the Jacobian of the coordinate transformation is 1, so I guess it is correct to say that dt1dt2=dsdt.
 

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