Absolute convergence of Hook Integral I am read Dan Romik’s book “Surprising Math of Longest Increasing subsequence.”
For a non-increasing, compactly supported function $f:[0,\infty)\to[0,\infty)$ such that $\int f=1,$ he defines the hook-length, naturally, as $$h_f(x,y)=f(x)-y+f^{-1}(y)-x,$$ where $f^{-1}(y)=\inf\{x: f(x)\le y\}.$
The hook integral is defined as $$I(f)=\int_0^\infty\int_0^{f(x)} \log(h_f(x,y)) \, dy \, dx.$$
He says that (it is easy to see that) the hook integral is absolutely convergent. But I could not prove the absolute convergence of the hook-integral.
Any help would be appreciated.
 A: I'm looking through the textbook myself and encountered the same difficulty. Here's my answer after about half a day's thought (the 1 coffee cup difficulty was totally a lie).
Let $A$ be the region we're integrating over, and let $A_1 = \{(x,y) \in A: h_f(x,y) < 1\}$ and $A_2 = \{(x,y) \in A: h_f(x,y) \geqslant 1\}$. Then it is sufficient to show that both $\int_{A_1} -\log(h_f(x,y)) \,dy \,dx$ and $\int_{A_2} \log(h_f(x,y)) \,dy \,dx$ are finite.
But now we note that, by $f$ being compactly supported, $A_2$ is a measurable region with finite area, and $\log(h_f(x,y))$ is non-negative and bounded above, so the integral over $A_2$ is certainly finite. For the other integral, to bound it from above, we need to bound $h_f$ from below (since $|\log|$ is decreasing between $0$ and $1$). In this region, the obvious way would be to note that $f(x) - y + f^{-1}(y) - x \geqslant f(x) - y \geqslant 0$. Let $K$ be a finite constant such that $f(x) = 0$ for all $x>K$ and let $B_1 = \{(x,y) \in \mathbb{R}^2: 0 \leqslant f(x)-y \leqslant 1 \}$ and note that $A_1 \subset B_1$. So
\begin{align}
\int_{A_1}-\log(h_f(x,y))\,dy\,dx &\leqslant \int_{A_1} -\log(f(x)-y)\,dy\,dx \\
& \leqslant \int_{B_1}-\log(f(x) - y)\,dy\,dx \\
& \leqslant \int_{x=0}^{K} \int_{y=f(x)-1}^{f(x)}-\log(f(x) - y)\,dy\,dx \\
& = \int_{x=0}^{K} 1 \,dx = K < \infty
\end{align}
and we're done.
