# Why the BRDF of specular reflection is infinite in the reflection direction?

I know the BRDF of specular reflection is nonzero only in the reflection direction.

But why it is infinite?

A paragraph on page 36 of Advanced Global Illumination:

• Where are you getting this infinity from? Is this a formula you're using, or something you've read, or something about a particular shader? – Dan Hulme Feb 24 '17 at 12:58
• @DanHulme Question updated. – chaosink Feb 24 '17 at 13:39
• What's your maths background? Are you familiar with the Dirac delta distribution? – Dan Hulme Feb 24 '17 at 13:40
• @DanHulme Haven't heard of that... Learning CG always makes me worry about my math. – chaosink Feb 24 '17 at 13:44

In the BRDF curve, the total area under the curve is the albedo: what fraction of incident light is reflected in total (as opposed to being absorbed or transmitted). For a perfect reflector, the area under the curve sums to 1, because it reflects all of the incident light. This area limits how high the curve can be. For example, a perfect diffuse reflector has a BRDF that's a flat line at $\frac{1}{2\pi}$, because it reflects the light evenly across the whole hemisphere, the solid angle of a hemisphere is $2\pi$, and the area has to be 1.
The remark about "$\delta$-functions" is because of a function (properly speaking, it's not a function but a distribution) called the Dirac $\delta$ (delta). This distribution is an infinitely thin, infinitely tall spike at $x = 0$, whose area is 1, just like our perfect specular BRDF.
• Thanks for your time and patience! I've understand what you mean. However, I doubt that the total area under the BRDF curve(you mean the surface in the hemisphere coordinate system?) is the albedo. In my another question Energy conservation of BRDF, the last equation $\forall\Phi:\int_{\Omega_x}f_r(x,\Phi\to\Theta)cos(N_x,\Theta)d\omega_\Theta\le1$ is the necessary and sufficient condition for energy conservation.(To be continued) – chaosink Feb 24 '17 at 16:37
• So I think albedo maybe equal $\int_{\Omega_x}f_r(x,\Phi\to\Theta)cos(N_x,\Theta)d\omega_\Theta$ (I'm not sure since I haven't found any relative document for the time). An additional factor $cos(N_x,\Theta)$ is in the formula. From this formula, if we consider the situation that it equals 1 and the BRDF is constant, we can get the BRDF equals $\frac{1}{\pi}$ but not $\frac{1}{2\pi}$ since $\int_{\Omega_x}cos(N_x,\Theta)d\omega_\Theta=\pi$. – chaosink Feb 24 '17 at 16:37