How to calculate $\lim_{\varepsilon\rightarrow 1}\int_0^\varepsilon \int_0^z \int_0^y\frac 1 {1-x^3} \, dx \, dy \, dz$? 
How to calculate the integral:
  $$\lim_{\varepsilon\rightarrow 1}\int_0^\varepsilon \int_0^z \int_0^y\frac 1 {1-x^3} \, dx \, dy \, dz\quad?$$

One solution is about infinite series, but I don't fully understand that solution.  Any other approaches?
 A: \begin{align}
& \int_0^\varepsilon \left( \int_0^z \left( \int_0^y\frac 1 {1-x^3} \, dx\right) \, dy\right) \, dz \\[12pt]
= {} & \iiint\limits_{0 \,\le\,x\,\le\,y\,\le\,z\,\le \, \varepsilon } \frac 1 {1-x^3} \,d(x,y,z).
\end{align}
Since the innermost function depends on $(x,y,z)$ only through $x$, putting $\displaystyle \int \cdots\,dx$ on the outside will mean on the inside we're just integrating constant functions, so that might be simpler.
From the expression
$$
0 \le x\le y\le z\le\varepsilon \tag 1
$$
we get $0\le x\le \varepsilon,$ so we have
$$
\int_0^\varepsilon \cdots\,dx.
$$
Then from $(1)$ we get $x\le y \le\varepsilon,$ so we get
$$
\int_0^\varepsilon \left( \int_x^\varepsilon \cdots\,dy \right) dx.
$$
Then from $(1)$ we have $y\le z\le\varepsilon,$ so we have
$$
\int_0^\varepsilon \left( \int_x^\varepsilon \left( \int_y^\varepsilon \cdots \,dz \right) dy \right) dx.
$$
So we have
\begin{align}
& \int_0^\varepsilon \left( \int_x^\varepsilon \left( \int_y^\varepsilon \frac 1 {1-x^3} \,dz \right) dy \right) dx \\[10pt]
= {} & \int_0^\varepsilon \int_x^\varepsilon \frac{\varepsilon-y}{1-x^3}\,dy\,dx \\[10pt]
= {} & \frac 1 2 \int_0^\varepsilon \frac{(\varepsilon-x)^2}{1-x^3} \, dx 
\end{align}
If $\varepsilon=1,$ then
\begin{align}
& \frac{(\varepsilon-x)^2}{1-x^3} = \frac{(1-x)(1+x)}{(1-x)(1+x+x^2)} \\[10pt]
= {} & \frac{1+x}{1+x+x^2} = \underbrace{\frac{1/2}{1+x+x^2}}_{\Large\text{complete the square, etc.}} + \underbrace{\frac{(1/2) + x}{1+x+x^2}}_{\Large\text{routine substitution}}
\end{align}
At this point I would ponder the question of why it was expressed as $\lim\limits_{\varepsilon\to1}.$
