Prove the inequality involving multiple integrals

Question

Please, help me to prove that: $$ \int_Q \int \frac{dxdy}{x^{-1} + |\ln y| - 1} \leq 1,$$ where $$ Q = [0; 1] \times [0; 1] $$ Any ideas how to start. Thank you.

Answer 1

If we set $\enspace\displaystyle y:=e^{-t}\enspace$ with $\enspace\displaystyle t\in\mathbb{R}_0^+\enspace$ then because of $\enspace\displaystyle t\geq 1-e^{-t}\enspace$

it’s $\enspace\displaystyle \frac{1}{x}+|\ln y|-1\geq \frac{1}{x}-y\geq 0\enspace$ for $\enspace 0<x\le 1$ .

It follows $$0<\int\limits_0^1 \int\limits_0^1 \frac{dxdy}{\frac{1}{x}+|\ln y|-1} \leq\int\limits_0^1 \int\limits_0^1 \frac{dxdy}{\frac{1}{x} -y}$$ and with $\enspace\displaystyle \int\int \frac{dxdy}{\frac{1}{x} -y}=A+By+\frac{(1-xy)\ln(1-xy)}{y}\enspace$ we get

$$\int\limits_0^1 \int\limits_0^1 \frac{dxdy}{\frac{1}{x} -y}=A+By+\frac{(1-xy)\ln(1-xy)}{y})|_ {x=0}^{x=1}|_{y=0}^{y=1}=\frac{(1-y)\ln(1-y)}{y}|_0^1=1$$ what has to be proofed.

Note: $$\enspace \displaystyle \int\limits_0^1 \int\limits_0^1 \frac{dxdy}{\frac{1}{x}+|\ln y|-1} < \int\limits_0^1 \int\limits_0^1 \frac{dxdy}{ \frac{1}{x} +\frac{1-y^h}{h}-1}|_{0<h<1}<\int\limits_0^1 \int\limits_0^1 \frac{dxdy}{\frac{1}{x} -y}=1 $$