Evaluate the integral from 0 to infinity of log(x)/(1 + x^2) dx, and explain why the same argument fails for the integral of log(x)/(1 + x^3).

Evaluate the integral from 0 to infinity of log(x)/(1 + x^2) dx, and explain why the same argument fails for the integral of log(x)/(1 + x^3).

Approach: Split the range at 1 and substitute x by 1/x on the outer piece, then compare the two halves after checking that each converges on its own.

0. Split the integral at x = 1. On the piece from 1 to infinity apply the reciprocal substitution x = 1/u, which gives dx = -du/u^2 and 1 + x^2 = (u^2 + 1)/u^2, so that piece becomes the integral from 0 to 1 of log(1/u)/(1 + u^2) du = minus the integral from 0 to 1 of log(u)/(1 + u^2) du. The two halves are therefore negatives of each other and the total is 0. Each half converges separately, since log(u) is integrable at 0 and log(x)/x^2 is integrable at infinity, so the cancellation is legitimate rather than a difference of two infinities. The odd symmetry comes from 1 + x^2 being invariant under x to 1/x up to the factor x^2 that the Jacobian supplies. For 1 + x^3 that invariance is lost, since the substitution produces (1 + u^3)/u^3 against a Jacobian of u^{-2}, leaving an extra factor of u, and the integral is a nonzero value involving pi^2.

Follow-up: What is the integral from 0 to infinity of log(x)/(1 + x^2)^2 dx, and does the same splitting argument still give zero?

Key concepts: reciprocal substitution, odd symmetry, convergence check, definite integral.