# Integrating the delta function

1. May 19, 2009

### captainjack2000

1. The problem statement, all variables and given/known data
By using the substitution u=cosx obtain the value of the integral
$$\int$$$$\delta$$(cosx-1/2)dx between 0 and pi

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

3. The attempt at a solution

2. May 19, 2009

### dx

What's the integral you get after doing the substitution?

3. May 19, 2009

### D H

Staff Emeritus
So, make the substitution! We can't help you if you don't do (or show) your work.

4. May 19, 2009

### tiny-tim

Hi captainjack2000!

(have a delta: δ )
Hint: what is the definition of the Dirac delta function?

5. May 19, 2009

### captainjack2000

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
$$\delta$$(x-a)=o if x$$\neq$$a and tends to infinity is x=a
shifting property
$$\int$$dx$$\delta$$(x-a)f(x) = f(a) where the integral is over all space.

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

not sure how to proceed

6. May 19, 2009

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 $$brackets it saves you time and is a lot neater. 7. May 19, 2009 ### dx You still have dx in your integral. Write everything in terms of u and du. Don't forget to change the limits of the integral too. 8. May 19, 2009 ### captainjack2000 Ok...(my brain is obviously not working right now!) if u=cosx du=-sinxdx dx=-du/sinx but you need this independent of x? but x=cos^-1(u) ?? changing the limits: when x=0 u=1 and when x=pi u=-1 9. May 19, 2009 ### dx You can write sin x = √(1 - u2). So the integral becomes [tex] -\int_{1}^{-1} \delta(u - 1/2) \frac{1}{\sqrt{1-u^2}}du$$

10. May 19, 2009

### Cyosis

To see that $\sin \arccos u =\sqrt{1-u^2}$ use $\cos^2x+\sin^2x=1$ or use a triangle.

Last edited: May 19, 2009
11. May 19, 2009

### dx

You mean

$$\sin \cos^{-1} u = \sqrt{1-u^2}$$.

12. May 19, 2009

### Cyosis

Yes I do thanks for pointing it out, it's fixed now.

13. May 19, 2009

### captainjack2000

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. May 19, 2009

### Cyosis

Yep, that's correct.

15. May 19, 2009

### dx

Your final answer is correct, but you had a minus sign in your first expression which shouldn't be there.

16. May 19, 2009

### captainjack2000

thank you !

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