Wiki Loves Folklore2∫∞0(x(k−1)∗e(−x/θ))/(Γ(k)θk) dx=22∫0∞(x(k−1)∗e(−x/θ))/(Γ(k)θk) dx=2
This integral is simply the area under a random probability density function (pdf) I chose, but the same applies to any pdf, and since probabilities range from 0 to 1, this integral ranges from 0 to 1 depending on it’s lower and upper bounds. Given the lower and upper bounds are 0 and ∞ respectively, this integral then evaluates to 1. This is simply because when you integrate from 0 to ∞, you’re really taking a summation of the probabilities of each event occurring, and we know that if we add the probabilities of each individual event occurring in a sample space, then the result must equal 1. To illustrate this, I’ll give a simple example. Imagine you flip a coin twice, each flip independent of the other.
Let H represent a flipped Head and T represent a flipped Tail
Your sample space is then (H,H),(H,T),(T,H),(T,T)(H,H),(H,T),(T,H),(T,T)
So in other words, the double coins either both land on head, or both land on tails, or both are opposites of one another.
Summing up these probabilities gives: 1/4+1/4+2/4=4/4=11/4+1/4+2/4=4/4=1
Alright! So if the integral of this pdf (or any other pdf really) from 0 to ∞ always evaluates to 1, then 2 times that integral always evaluates to 2. There you go my dude!
Projecting a sphere to a plane
Four- / other-dimensional