The function is strictly decreasing from infinity to zero on , so there is a unique positive root, which tends to infinity. Let . Taking logarithms givesFirst . Put with , expand, and cancel :The leading balance gives . Substitution leaves a residual ; the derivative of is of order , so the root error is . ThusThe second term is logarithmic rather than a constant multiple of . This is logarithmic inversion of a Gaussian divided by its argument.