Integrating the delta function

[captainjack2000]

Homework Statement

By using the substitution u=cosx obtain the value of the integral
$$\int$$$$\delta$$(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

 

[dx]

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

 

[D H]

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

 

[tiny-tim]

Hi captainjack2000!

(have a delta: δ )

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

Hint: what is the definition of the Dirac delta function?

(See also http://en.wikipedia.org/wiki/Dirac_delta_function#Delta_function_of_more_complicated_arguments)

 

[captainjack2000]

Sorry, I realize 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

 

[Cyosis]

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.

 

[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.

 

[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

 

[dx]

You can write sin x = √(1 - u²). So the integral becomes

$$-\int_{1}^{-1} \delta(u - 1/2) \frac{1}{\sqrt{1-u^2}}du$$

 

[Cyosis]

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

 

[dx]

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

You mean

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

 

[Cyosis]

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

 

[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) ?

 

[Cyosis]

Yep, that's correct.

 

[dx]

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

 

[captainjack2000]

thank you !