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The integration of

$$\int^\pi_0 \sin(\theta) \delta(\cos\theta - 1)\, d \theta$$

can be derived by replacing $d\theta$ with $d\cos\theta$

$$\begin{align*} \int^\pi_0 \sin(\theta) \delta(\cos\theta - 1)\, d \theta &= -\int^{-1}_1 \delta(x - 1)\, dx \quad (\text{let } x = \cos\theta, dx = -\sin(\theta)\,d\theta) \\ &= \int^1_{-1} \delta(x - 1)\, dx \\ &= 1 \end{align*}$$

However, if I don't replace the base and apply the definition of Dirac delta ($\int f(x)\delta(x -a)\,dx = f(a)$)directly, the result become $0$.

$$\begin{align*} \int^\pi_0 \sin(\theta) \delta(\cos\theta - 1)\, d \theta &= \sin(0) \\ &= 0 \end{align*}$$

What is wrong with the second approach?

I have seen the other solution that replace $\delta(\cos\theta - 1)$ at Dirac delta integral of cosx

$$\begin{align*} \int^\pi_0 \sin(\theta) \delta(\cos\theta - 1)\, d \theta &= \int^\pi_0 \sin(\theta) \frac{\delta(\theta - \frac{\pi}{2})}{|-\sin\frac{\pi}{2}|}\, d \theta \\ &= \int^\pi_0 \sin(\theta) \delta(\theta - \frac{\pi}{2})\, d \theta \\ &= \sin(\frac{\pi}{2}) \\ &= 1 \end{align*}$$

Is the replacement of the Dirac delta function necessary? Why it's required and how is the replacement derived?

gyro
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    You can't "apply the definition of Dirac delta directly" because $\delta(\cos\theta-1),{\rm d}\theta$ is not of the form $\delta(\theta-a),{\rm d}\theta$ for any constant $a$. You don't have to use the rule to replace $\delta$ of a function, but it is the general method and you don't always have nice substitutions like you have here. – anon Oct 21 '22 at 06:10
  • Where can I find the general equation of replacing the Dirac delta function? – gyro Oct 21 '22 at 06:42
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    Use the composition with a function property of $\delta$. There is no unambiguous answer to your integral because the delta contribution occurs exactly at an endpoint of the integration $\theta=0$ or $x=1$. Formally, the result is $H(0)$, that is, whichever number you assign to the step function at zero argument – Sal Oct 21 '22 at 08:32
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    It is interesting that $H(0)$ is assigned to the different values on Dirac delta function - Wikipedia and Heaviside step function - Wikipedia. One is $1$, and the other is $0$. – gyro Oct 22 '22 at 13:19

1 Answers1

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What if the second integral method produces the correct result ?

Notice that the third method (with the link) uses the result of $\delta(\cos \theta)$ rather than finding the root of $\delta(\cos \theta - 1)$. The zero should be found from the equation $\cos \theta_{0} = 1$ which gives $0 + 2 \, n \, \pi$ and not the $\frac{\pi}{2}$ given in the example (which is the root of the equation $\cos \theta_{1} = 0)$.

The given integral $$ \int \sin(a \theta) \, \delta(\cos(a \theta) - b) \, d\theta$$ can be integrated directly by making use of the Heaviside step function as seen by $$ \int \sin(a \theta) \, \delta(\cos(a \theta) - b) \, d\theta = - \frac{1}{a} \, H(\cos(a \theta) - b) + c_{0}.$$ Evaluating this integral for the limits desired gives $$ \int_{0}^{\pi} \sin(a \theta) \, \delta(\cos(a \theta) - b) \, d\theta = \frac{H(1 - b) - H(\cos(a \pi) - b)}{a}. $$ If $a = 1$ then $$ \int_{0}^{\pi} \sin(\theta) \, \delta(\cos(\theta) - b) \, d\theta = H(1 - b) - H(-1 - b) $$ and if $b=1$ the result reduces to $$ \int_{0}^{\pi} \sin(\theta) \, \delta(\cos(\theta) - 1) \, d\theta = 0. $$

A related integral is given in this question and solution Integrating Dirac delta functions with trigonometric arguments

Leucippus
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  • Wow! I need some time to understand Heaviside step function. However, where did I make a mistake in the first method? – gyro Oct 21 '22 at 12:33
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    @leucippus Hello my friend! It's been a long, long while. I hope all is going well. As for this result, there is an issue. For $b=1$, $H(1-b)-H(-1-b)=H(0)-H(-2)=H(0)$. And $H(0)$ is not uniquely defined. To that extent, the object $\int_0^\pi \sin(\theta)\delta(\cos(\theta)-1),d\theta$ is not well-defined. – Mark Viola Oct 21 '22 at 17:01