Take g(z)=L(z)3/z3 on the same logarithm branch. At zero, L(z)3=−z3+O(z4), so the apparent singularity is removable and the closed contour integral is zero. The bank difference isThe circular contributions vanish by the previous estimates with the third power of the logarithm. ConsequentlyThe last integral is 1/2, givingIncreasing the logarithm power has converted the desired real integral into a jump across a branch cut, while removing the pole inside the contour.
(ℓ−iπ)3−(ℓ+iπ)3=−6iπℓ2+2iπ3.
0=−6iπ∫1∞x3log2(x−1)dx+2iπ3∫1∞x3dx.
∫1∞x3log2(x−1)dx=6π2.
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