Does this integral have a closed form expression?

Question

I have been working with two related integrals. The first one yields a simple expression, but I can't seem to find a simple expression for the second.

Integral 1

$$ \begin{align*} I_x &= \int_z^{1-y+z} \sqrt{\frac{1}{1-x-y+z} + \frac{1}{x-z}} \, dx\\ &= \int_z^{1-y+z} \sqrt{\frac{1-y}{(1-x-y+z)(x-z)}} \, dx\\ &= \sqrt{1-y} \int_z^{1-y+z} \frac{1}{\sqrt{(1-x-y+z)(x-z)}} \, dx\\ \end{align*} $$

We can substitute $u=\frac{x-z}{1-y}$, so $x=u(1-y)+z$, and $dx=(1-y) \, du$. Then we have:

$$ \begin{align*} I_x &= \sqrt{1-y} \int_0^1 u^{\frac{1}{2}-1}(1-u)^{\frac{1}{2}-1} \, du\\ &= \sqrt{1-y} \, \operatorname{B} \left( \tfrac{1}{2}, \tfrac{1}{2} \right)\\ &= \sqrt{1-y} \, \pi \end{align*} $$

So this integral has a simple expression because the substitution led to the exact form of the beta function.

Integral 2

$$ I_z = \int_{\max(0,x+y-1)}^{\min(x,y)} \sqrt{\frac{1}{1-x-y+z} + \frac{1}{x-z} + \frac{1}{y-z} + \frac{1}{z}} \, dz $$

This seems related to the first integral, but I can't seem to simplify it. Even if we assume one of the four "cases" of integration limits ($\int_0^x$, $\int_0^y$, $\int_{x+y-1}^x$, or $\int_{x+y-1}^y$) and try to proceed, it doesn't seem to help.

Am I missing something simple? Is there an obscure integration technique that would help? Or is there just no way to simplify this?

Update (6-26-14): From Jacquet & Szpankowski, 2003, "Markov Types and Minimax Redundancy for Markov Sources" (link), p.10, we have the following, which may (or may not) help here:

$$ \begin{align*} I_{JS} &= 4 \int_0^1 \frac{1}{\sqrt{(1{-}x)x}} \, dx \int_0^{\min(x,1{-}x)} \frac{1}{(1{-}x{-}y)(x{-}y)} \, dy\\ &= 8 \int_0^{\frac{1}{2}} \frac{\log(1{-}2x) - \log\left(1{-}2\sqrt{(1{-}x)x}\right)}{\sqrt{(1{-}x)x}} \, dx\\ &= 16 \int_0^{\frac{\pi}{4}} \log \left( \frac{\cos(2\theta)}{1{-}\sin(2\theta)} \right) \, d\theta\\ &= 16 \, G\\ \end{align*} $$

where $G \approx 0.915965594$ is Catalan's constant.

Answer 1

First of all, regarding the limits of integration $a=\max{(0,x+y-1)}$ and $b=\min{(x,y)}$, you do not have to break up your analysis into four separate cases to accommodate the two possible values taken by the min/max functions. These functions can be represented in terms of elementary functions as follows:

$$\max{(a,b)}=\frac{a+b}{2}+\frac12|a-b|=\frac{a+b}{2}+\frac12\sqrt{(a-b)^2},\\ \min{(a,b)}=\frac{a+b}{2}-\frac12|a-b|=\frac{a+b}{2}-\frac12\sqrt{(a-b)^2}.$$

With that in mind, and after making the substitutions $u=2z-x-y$ followed by $w=u+1$, a little algebra reduces the integral to:

$$\begin{align} \mathcal{I}{\left(x,y\right)} &=\int_{\max{(0,x+y-1)}}^{\min{(x,y)}}\sqrt{\frac{1}{z}+\frac{1}{z-(x+y-1)}+\frac{1}{x-z}+\frac{1}{y-z}}\,\mathrm{d}z\\ &=\int_{\frac{x+y-1+|1-x-y|}{2}}^{\frac{x+y-|x-y|}{2}}\sqrt{\frac{1}{z}+\frac{1}{z-(x+y-1)}+\frac{1}{x-z}+\frac{1}{y-z}}\,\mathrm{d}z\\ &=\frac12\int_{-1+|1-x-y|}^{-|x-y|}\sqrt{\frac{1}{\frac{u+x+y}{2}}+\frac{1}{\frac{u+x+y}{2}-(x+y-1)}+\frac{1}{x-\frac{u+x+y}{2}}+\frac{1}{y-\frac{u+x+y}{2}}}\,\mathrm{d}u\\ &=\frac12\int_{-1+|1-x-y|}^{-|x-y|}\sqrt{\frac{2}{u+x+y}+\frac{2}{u-x-y+2}+\frac{2}{-u+x-y}+\frac{2}{-u-x+y}}\,\mathrm{d}u\\ &=\frac{\sqrt{2}}{2}\int_{-1+|1-x-y|}^{-|x-y|}\sqrt{\frac{1}{u+x+y}+\frac{1}{u-x-y+2}+\frac{1}{-u+x-y}+\frac{1}{-u-x+y}}\,\mathrm{d}u\\ &=\frac{\sqrt{2}}{2}\int_{|1-x-y|}^{1-|x-y|}\sqrt{\frac{1}{w+x+y-1}+\frac{1}{w-x-y+1}+\frac{1}{1-w+x-y}+\frac{1}{1-w-x+y}}\,\mathrm{d}w\\ &=\frac{\sqrt{2}}{2}\int_{|1-x-y|}^{1-|x-y|}\sqrt{\frac{2w}{w^2-(x+y-1)^2}+\frac{2(1-w)}{(1-w)^2-(x-y)^2}}\,\mathrm{d}w\\ &=\int_{|1-x-y|}^{1-|x-y|}\sqrt{\frac{w}{w^2-|x+y-1|^2}+\frac{(1-w)}{(1-w)^2-|x-y|^2}}\,\mathrm{d}w. \end{align}$$

At this point I'm not exactly sure how best to proceed. My gut is starting to tell me that this integral is going to wind up being some ghastly thing in terms of elliptic integrals.