Solution

ID: past-exam-of-the-perturbation-methods-and-applied-complex-variables-course-of-the-university-of-oxford/2008/a/1/vi/solution

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

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