 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2015/s1/1/3/solution

Take

g (z) = L (z)^{3} / z^{3}

on the same logarithm branch. At zero,

L (z)^{3} = - z^{3} + O (z^{4})

, so the apparent singularity is removable and the closed contour integral is zero. The bank difference is

(ℓ - iπ)^{3} - (ℓ + iπ)^{3} = - 6 iπ ℓ^{2} + 2 i π^{3} .

The circular contributions vanish by the previous estimates with the third power of the logarithm. Consequently

0 = - 6 iπ \int_{1}^{\infty} \frac{l o g ^{2} ( x - 1 )}{x ^{3}} d x + 2 i π^{3} \int_{1}^{\infty} \frac{d x}{x ^{3}} .

The last integral is

1/2

\int_{1}^{\infty} \frac{lo g ^{2} ( x - 1 )}{x ^{3}} d x = \frac{π ^{2}}{6} .

Increasing the logarithm power has converted the desired real integral into a jump across a branch cut, while removing the pole inside the contour.

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