 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2008/d/2/a/solution

There is no singularity at the lower endpoint one. Put

t = lo g (1 + x)

, so

d x = e^{t} d t

. The nonnegative integral becomes

\int_{l o g 2}^{\infty} \frac{e ^{(1 - α) t}}{t ^{2}} d t .

For

α = 1

, it converges as a reciprocal-square integral. For

α > 1

, exponential decay makes it converge. For

α < 1

, its integrand eventually grows without bound, so its integral diverges. Thus

α \geq 1.

This is the logarithmic borderline in a power-tail integrability test.

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