IMC 2004 · Day 2 · P11

First we use the inequality $$x^{-1} - 1 \ge |\ln x|, \quad x \in (0, 1],$$ which follows from $$\left. (x^{-1} - 1) \right|_{x=1} = \left. |\ln x| \right|_{x=1} = 0,$$ $$(x^{-1} - 1)' = -\frac{1}{x^2} \le -\frac{1}{x} = |\ln x|', \quad x \in (0, 1].$$ Therefore $$\int_0^1 \int_0^1 \frac{dx\,dy}{x^{-1} + |\ln y| - 1} \le \int_0^1 \int_0^1 \frac{dx\,dy}{|\ln x| + |\ln y|} = \int_0^1 \int_0^1 \frac{dx\,dy}{|\ln(x \cdot y)|}.$$ Substituting $y = u/x$, we obtain $$\int_0^1 \int_0^1 \frac{dx\,dy}{|\ln(x \cdot y)|} = \int_0^1 \left( \int_u^1 \frac{dx}{x} \right) \frac{du}{|\ln u|} = \int_0^1 |\ln u| \cdot \frac{du}{|\ln u|} = 1.$$

Substituting $s = x^{-1} - 1$ and $u = s - \ln y$, $$\int_0^1 \int_0^1 \frac{dx\,dy}{x^{-1} + |\ln y| - 1} = \int_0^{\infty} \int_s^{\infty} \frac{e^{s-u}}{(s+1)^2 u}\,du\,ds = \int_0^{\infty} \left( \int_0^u \frac{e^s}{(s+1)^2}\,ds \right) \frac{e^{-u}}{u}\,du.$$ Since the function $\dfrac{e^s}{(s+1)^2}$ is convex, $$\int_0^u \frac{e^s}{(s+1)^2}\,ds \le \frac{u}{2} \left( \frac{e^u}{(u+1)^2} + 1 \right)$$ so $$\int_0^1 \int_0^1 \frac{dx\,dy}{x^{-1} + |\ln y| - 1} \le \int_0^{\infty} \frac{u}{2} \left( \frac{e^u}{(u+1)^2} + 1 \right) \frac{e^{-u}}{u}\,du = \frac{1}{2} \left( \int_0^{\infty} \frac{du}{(u+1)^2} + \int_0^{\infty} e^{-u}\,du \right) = 1.$$