[arcadi7]

the funny thing is that the calculation in this article misses the point, especially in the Feynman context. First, beyond all trickery, the log(alpha) answer might suggest that something bad happens at alpha=0 . What makes this integral interesting is that it is equal to zero identically for alpha<1 .

The reason, of course, is that this integral is not randomly chosen -- it represents the two-dimensional coulomb potential (log(r)) of the sphere (circle) of radius 1 at distance alpha from the center. By when point alpha is inside the circle , the potential is constant (or zero -- no force) . When alpha is outside, the potential is log(r) as if all the mass of a circle is at its center. The expression under the log in the integral is just (square of ) the distance between the point alpha and point on a unit circle.

beyond tricks -- the physical reason for the singular behavior of this integral is gauss theorem for coulomb potential . so no magic.