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?
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John
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If you don't want to use infinite series, you need to decompose the integrand into partial fractions. – Matthew Leingang Jun 26 '17 at 23:55
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1If you change the order of integration so integration with respect to $x$ is outermost, then the inner integrals with respect to $y$ and $z$ are of a function only of $x$ so they should be easy to do. – Daniel Schepler Jun 26 '17 at 23:58
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Also, do you mean $\lim_{\epsilon \to 1^-}$? Otherwise, for $\epsilon > 1$, the integral isn't strictly defined. (Though possibly, using a Cauchy principal value might make it defined - but I'm not at all sure.) – Daniel Schepler Jun 27 '17 at 00:03
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\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 $\lim\limits_{\varepsilon\to1}.$
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1I had $\int_0^\epsilon\int_x^\epsilon\int_0^y \frac {1}{1-x^3}\ dz\ dy\ dx$ – Doug M Jun 27 '17 at 00:44
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"At this point I would ponder the question of why it was expressed as limε→1".The function is not defined for x=1,so I thought this was a problem... – John Jun 27 '17 at 15:25