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I am a bit stuck on solving this integral:

$$g(\omega)=\left(\frac{L}{2\pi}\right)^d\int({d^dq})\delta(\omega-cq^\alpha)$$

The range of values of $q$ is $[0,\infty).$

I have tried referencing this post $n$-dimensional integral of delta function with no success, as I do not understand the portion of rewriting the delta function in a more convenient expression.

This is what I have thus far... $$g(\omega)=\left(\frac{L}{2\pi}\right)^dVol_{d}\int({dq})q^{d-1}\delta(\omega-cq^\alpha)$$ Where $Vol_{d}=\frac{\pi^{d/2}}{(d/2)!}$

Then replace the $q^{d-1}$ in the integral with $(\frac{\omega}{c})^{1/\alpha}$

Resulting in $\left(\frac{L}{2\pi}\right)^d\frac{\pi^{d/2}}{(d/2)!}(\frac{\omega}{c})^{(d-1)/\alpha}$

My only hesitation is replacing the $q^{d-1}$ term with $(\frac{\omega}{c})^{(d-1)/\alpha}$ due to the delta function substitution.

Hosam Hajeer
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  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Apr 10 '22 at 18:50
  • I think you want to use the surface area of the $d$-dimensional sphere, not the volume. Is $\omega$ the same as $w$? – John Barber Apr 10 '22 at 19:06
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    Also, there's a $\delta$ function identity you may find useful. If the function $f(x)$ has a simple root at $x = y$, then $\delta\bigl(f(x)\bigr) = \delta(x-y)/|f'(y)|$. – John Barber Apr 10 '22 at 19:08

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