How Does the Dirac Delta Function Apply to Trigonometric Integrals?

[Pual Black]

Homework Statement

hi
i have to find the result of
$\int_{0}^{\pi}[\delta(cos\theta-1)+ \delta(cos\theta+1)]sin\theta d\theta$

Homework Equations

i know from Dirac Delta Function that
$\int \delta(x-a)dx=1$
if the region includes x=a and zero otherwise

The Attempt at a Solution

first i make the substitution
let $\cos\theta=x$
then $\sin\theta d\theta=-dx$
at $\theta=0 \rightarrow x=1$
at $\theta=\pi \rightarrow x=-1$

$\int_{1}^{-1}[\delta(x-1)+ \delta(x+1)](-dx)$

$\int_{-1}^{1}[\delta(x-1)+ \delta(x+1)](dx)$

$\int_{-1}^{1}\delta(x-1)dx+ \int_{-1}^{1}\delta(x+1)(dx)$

so the first term is equal one but what about the second term. I know that the result must be 2

 

[DrClaude]

In $\delta(x-a) = \delta(x+1)$, what is the value of $a$?

 

[Pual Black]

In $\delta(x-a) = \delta(x+1)$, what is the value of $a$?

In L.H.S
X must be equal a. Else dirac delta will be zero.
R.H.S
X=-1
Or a=-1
Maybe I didn't understand your question.

 

[DrClaude]

a=-1

Correct. So what is the value of the corresponding integral?

 

[George Jones]

$\int_{-1}^{1}\delta(x-1)dx+ \int_{-1}^{1}\delta(x+1)(dx)$

so the first term is equal one

Why? I think that the first term is undefined. Something strange is happening at $x = 1$, so write it as the limit,
$$\int_{-1}^{1} \delta \left( x -1 \right) dx = \lim_{ a \to 1} \int_{-1}^{a} \delta \left( x -1 \right) dx.$$
This two-sided limit, however, does not exist, because
$$\lim_{ a \to 1^-} \int_{-1}^{a} \delta \left( x -1 \right) dx = 0,$$
and
$$\lim_{ a \to 1^+} \int_{-1}^{a} \delta \left( x -1 \right) dx = 1.$$
In the first one-sided limit, $x = 1$ is not in the region of integration, while in the second one-sided limit, $x = 1$ is in the region of integration. Since the two one-sided limits are not equal, the two-sided limit does not exist.

I suppose that you mean the second one-sided limit, i.e., that
$$\int_{-1}^{1} \delta \left( x -1 \right) dx = \lim_{ a \to 1^+} \int_{-1}^{a} \delta \left( x -1 \right) dx.$$

 

[Pual Black]

Correct. So what is the value of the corresponding integral?

Since a=-1 is in the range of integral. The integral is equal one?

 

[Ray Vickson]

Homework Statement

hi
i have to find the result of
$\int_{0}^{\pi}[\delta(cos\theta-1)+ \delta(cos\theta+1)]sin\theta d\theta$

Homework Equations

i know from Dirac Delta Function that
$\int \delta(x-a)dx=1$
if the region includes x=a and zero otherwise

The Attempt at a Solution

first i make the substitution
let $\cos\theta=x$
then $\sin\theta d\theta=-dx$
at $\theta=0 \rightarrow x=1$
at $\theta=\pi \rightarrow x=-1$

$\int_{1}^{-1}[\delta(x-1)+ \delta(x+1)](-dx)$

$\int_{-1}^{1}[\delta(x-1)+ \delta(x+1)](dx)$

$\int_{-1}^{1}\delta(x-1)dx+ \int_{-1}^{1}\delta(x+1)(dx)$

so the first term is equal one but what about the second term. I know that the result must be 2

How do you know the result must be 2? You are integrating through only half of a $\delta$ function. Some people would say that the result is not defined at all, others would say the final result should be 1 (not the 2 that you give). We could also argue that for $a > -1$ we have $\int_{-1}^a \delta(x+1) \, dx = 0.15865$ or $= 0.84134$ (for both of which there exist perfectly good arguments/justifications). However, I would personally take $\int_{-1}^a \delta(x+1) \, dx = 1/2$, for more-or-less good reasons.

 

[Pual Black]

How do you know the result must be 2? You are integrating through only half of a $\delta$ function. Some people would say that the result is not defined at all, others would say the final result should be 1 (not the 2 that you give). We could also argue that for $a > -1$ we have $\int_{-1}^a \delta(x+1) \, dx = 0.15865$ or $= 0.84134$ (for both of which there exist perfectly good arguments/justifications). However, I would personally take $\int_{-1}^a \delta(x+1) \, dx = 1$, for more-or-less good reasons.

The reason why i think the result is 2 because of a problem in Jackson Electrodynamic.
I have now uploaded a pdf file and at page 1 you can see that the integral of dirac delta should be 2.