Attached is page 34 of "Advanced Global Illumination", 2nd edition, by Dutre et al. I don't understand how (2.22) is derived from (2.21) and the given incident radiance distribution $L(x\leftarrow\Psi)=L_{in}\delta(\Psi-\Theta)$. Could you please give me a detailed derivation? Thanks a lot!
The only thing you really need to know in order to derive (2.22) from (2.21) is that the $\delta$-distribution satisfies
$$ \int \delta(x) f(x)\,\text{d}x = f(0). $$
(The $\delta$-distribution is not a function in the common sense, so the above integral is merely a useful notation and not to be read as e.g. a Lebesgue or Riemann integral.)
Unfortunately, the notation that the authors use for the substitution of the incident radiance distribution is misleading. The $\Theta$ in $L(x\leftarrow\Psi) = L_{in}\delta(\Psi-\Theta)$ is completely unrelated to the integration variable $\omega_\Theta$, so I chose to denote it by $\tilde\Psi$ instead.
Here is the derivation of (2.22) from (2.21):
$$\frac{\int_{\Omega_x} \int_{\Omega_x} f_r(x, \Psi \rightarrow \Theta) L_{in}\delta(\Psi-\tilde\Psi)\cos(N_x,\Theta)\cos(N_x,\Psi)\,\text{d}\omega_\Psi\,\text{d}\omega_\Theta}{\int_{\Omega_x}L_{in}\delta(\Psi-\tilde\Psi)\cos(N_x,\Psi)\,\text{d}\omega_\Psi}$$
$$\frac{\int_{\Omega_x} \int_{\Omega_x-\tilde\Psi} f_r(x, \Psi+\tilde\Psi \rightarrow \Theta) L_{in}\delta(\Psi)\cos(N_x,\Theta)\cos(N_x,\Psi+\tilde\Psi)\,\text{d}\omega_\Psi\,\text{d}\omega_\Theta}{\int_{\Omega_x-\tilde\Psi}L_{in}\delta(\Psi)\cos(N_x,\Psi+\tilde\Psi)\,\text{d}\omega_\Psi}$$
$$\frac{\int_{\Omega_x} f_r(x, \tilde\Psi \rightarrow \Theta) L_{in}\cos(N_x,\Theta)\cos(N_x,\tilde\Psi)\,\text{d}\omega_\Theta}{L_{in}\cos(N_x,\tilde\Psi)}$$
$$\int_{\Omega_x} f_r(x, \tilde\Psi \rightarrow \Theta) \cos(N_x,\Theta)\,\text{d}\omega_\Theta$$
Now you only need to notice that we want equation (2.21) to be true for all such "test functions" $L_{in}\delta(\Psi-\tilde\Psi)$ for the incident radiance, i.e. for all possible values of $\tilde\Psi$.