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Integrating the delta function

  • #1

Homework Statement


By using the substitution u=cosx obtain the value of the integral
[tex]\int[/tex][tex]\delta[/tex](cosx-1/2)dx between 0 and pi


Homework Equations


I have no idea how to go any further with this apart from substituting in for u!?


The Attempt at a Solution

 

Answers and Replies

  • #2
dx
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What's the integral you get after doing the substitution?
 
  • #3
D H
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So, make the substitution! We can't help you if you don't do (or show) your work.
 
  • #4
tiny-tim
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  • #5
Sorry, I realise my post was very specific. To be perfectly honest I don't really know where to start. I know that the delta function is defined as
[tex]\delta[/tex](x-a)=o if x[tex]\neq[/tex]a and tends to infinity is x=a
shifting property
[tex]\int[/tex]dx[tex]\delta[/tex](x-a)f(x) = f(a) where the integral is over all space.

In this case
I=[tex]\int[/tex][tex]\delta[/tex](cosx-0.5)dx where the integral is between zero and pi
=[tex]\int[/tex][tex]\delta[/tex](u-0.5)dx letting u=cosx

not sure how to proceed
 
  • #6
Cyosis
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You're substituting, yet after the substitution you integrate over the original variable,dx? Execute the substitution correctly first. Also put the entire expression between [tex] brackets it saves you time and is a lot neater.
 
  • #7
dx
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You still have dx in your integral. Write everything in terms of u and du. Don't forget to change the limits of the integral too.
 
  • #8
Ok...(my brain is obviously not working right now!)
if u=cosx
du=-sinxdx
dx=-du/sinx
but you need this independent of x? but x=cos^-1(u) ??

changing the limits:
when x=0 u=1 and when x=pi u=-1
 
  • #9
dx
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You can write sin x = √(1 - u2). So the integral becomes

[tex] -\int_{1}^{-1} \delta(u - 1/2) \frac{1}{\sqrt{1-u^2}}du [/tex]
 
  • #10
Cyosis
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To see that [itex]\sin \arccos u =\sqrt{1-u^2}[/itex] use [itex]\cos^2x+\sin^2x=1[/itex] or use a triangle.
 
Last edited:
  • #11
dx
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To see that [itex]\sin \arccos u =\frac{1}{\sqrt{1-u^2}}[/itex] use [itex]\cos^2x+\sin^2x=1[/itex] or use a triangle.
You mean

[tex] \sin \cos^{-1} u = \sqrt{1-u^2} [/tex].
 
  • #12
Cyosis
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Yes I do thanks for pointing it out, it's fixed now.
 
  • #13
thank you that clarifies it alot!
Therefore the solution is going to be -1/(1-1/4)^0.5 = 1/(3/4)^0.5 = 2/sqrt(3) ?
 
  • #14
Cyosis
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Yep, that's correct.
 
  • #15
dx
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Your final answer is correct, but you had a minus sign in your first expression which shouldn't be there.
 
  • #16
thank you !
 

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