Net flux in radiative transfer

Question

I encountered the need of computing the net flux of a specific intensity given by $$I_{ex}(\theta,\phi) = I_0 \delta(\theta) \delta(\phi)$$ traversing an horizontal plane whose normal coincides with the $ z $-axis of the coordinate system. The variable $ \theta $ is measured from the $ z $-axis, whereas the variable $ \phi $ is measured from the $ x $-axis as usual. The background medium is vacuum. To compute this net flux, I should solve the integral $$\int_{4 \pi} I_{ex}(\theta^{'},\phi^{'}) \cos\theta^{'}\text{d}\omega^{'} = \int_{0}^{2 \pi} \int_{0}^{\pi} I_{ex}(\theta^{'},\phi^{'}) \cos\theta^{'} sin\theta^{'} \text{d}\theta^{'} \text{d}\phi^{'}$$ where $\omega^{'}$ indicates the solid angle. Substituting the expression of $I_{ex}$ into this last integral, I should get
$$\int_{0}^{2 \pi} \int_{0}^{\pi} I_0 \delta(\theta^{'}) \delta(\phi^{'}) \cos\theta^{'} \sin\theta^{'} \text{d}\theta^{'} \text{d}\phi^{'} = I_0 \int_{0}^{\pi} \delta(\theta^{'}) \cos\theta^{'} \sin\theta^{'} \text{d}\theta^{'} = \cos0 \sin0 = 0$$ In these steps I've exploited the Dirac's delta property $\int_{-a}^{a} f(x) \delta(x) \text{d}x = f(0)$, but on some notes I saw a different approach. Firstly, they set $\mu^{'} = \cos\theta^{'}$, which leads to $\text{d}\mu^{'} = - \sin\theta^{'} \text{d}\theta^{'}$ and so they rewrite the net flux integral as $$\int_{0}^{2 \pi} \int_{-1}^{1} I_{ex}(\mu^{'},\phi^{'}) \mu^{'} \sin\theta^{'} \frac{\text{d}\mu^{'}}{\sin\theta^{'}} d\phi^{'} = \int_{0}^{2 \pi} \int_{-1}^{1} I_{ex}(\mu^{'},\phi^{'}) \mu^{'} \text{d}\mu^{'} \text{d}\phi^{'} = \int_{-1}^{1} I_0 \delta(\mu^{'} - 1) \mu^{'} \text{d}\mu^{'} = I_0$$ I don't understand why the two approaches lead to totally different result. Can someone explain it to me, please? Moreover, could you confirm that in the theory of radiative transfer the following convention is adopted? $$\int_{-a}^{a} f(x) \delta(x - a) \text{d}x = \lim_{\epsilon \to 0^{+}} \int_{-a}^{a + \epsilon} f(x) \delta(x - a) \text{d}x = f(a)$$

Answer 1

Since $\theta^1=\cos^{-1}\mu$,$$\delta(\theta)\to\delta(\cos^{-1}\mu)=\delta(\mu-1). \left|\frac{1}{d(\cos^{-1}\mu)/d\mu}\right|_{\mu=1}= ...\left|\sqrt{\mu^2-1}\right|_{\mu=1}=0$$

where I have used the formula for $\delta(f(x))$ and only shown the relevant term in the second-last part.

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I am not sure about the convention but the equality seems correct.

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$^1$ I dropped the redundant single quote on the vars.