Why $\lim_{\varepsilon \to 0} \varepsilon \int_0^\infty \frac{z\,dz}{(z-1)^2 + \varepsilon^2z^3} = \pi$?
We'll break the solution into several steps.
- Evaluate the integral.
Note that the integrand is strictly positive for $z\in[0,\infty)$. Further, as Oliver Diaz noted, the integrand is improperly integrable at $\infty$.
Let $p_{\varepsilon}(z) = (z-1)^2 +\varepsilon^2 z^3=\varepsilon^2z^3+z^2-2z+1$, or $p(z)$ for short. $p$ has a positive minimum at $z=\frac{\sqrt{6 \epsilon ^2+1}-1}{3 \epsilon ^2}$; then $p$ has three roots, one real and two complex. Denote them $\xi_i,$ $i=1,2,3$ (note these depend on $\varepsilon$), where WLOG we take $\xi_1$ to be the real root.
Claim 1: $$ \int \frac{z}{p_{\varepsilon}(z)}\,dz = \sum_{i=1}^3 \frac{\log(z-\xi_i)\xi_i}{p_{\varepsilon}'(\xi_i)} $$Indeed, by partial fractions we have $$ \int \frac{z}{p_{\varepsilon}(z)}\,dz = \int \frac{C_1}{z-\xi_1}+\frac{C_2}{z-\xi_2}+\frac{C_3}{z-\xi_3}\,dz $$for some $C_1,C_2,C_3$; then well-known facts about the partial fraction decomposition give $C_i = \frac{\xi_i}{p_{\varepsilon}'(\xi_i)}$.
- Tidy things up.
Ok, we found an antiderivative. Now let's look at the limits of integration: $$ \left.\sum_{i=1}^3 \frac{\log(z-\xi_i)\xi_i}{p_{\varepsilon}'(\xi_i)}\right|_{0}^{\infty} $$
Claim 2: The upper limit is 0. We could use Oliver Diaz's observation, or we could note that, reusing our notation above, $C_1+C_2+C_3=0$, so all of the logs will vanish as $z\to\infty$.
So we have $$ \varepsilon \int_0^{\infty} \frac{z}{p_{\varepsilon}(z)}\,dz = -\varepsilon \sum_{i=1}^3 \frac{\log(-\xi_i)\xi_i}{p_{\varepsilon}'(\xi_i)} $$ $$ = \epsilon \left(\frac{\xi _1 \log \left(-\xi _1\right)}{-2 \xi _1-3 \xi _1^2 \epsilon ^2+2}+\frac{\xi _2 \log \left(-\xi _2\right)}{-2 \xi _2-3 \xi _2^2 \epsilon ^2+2}+\frac{\xi _3 \log \left(-\xi _3\right)}{-2 \xi _3-3 \xi _3^2 \epsilon ^2+2}\right) $$
- Write the Maclaurin series expansion and deduce the limit.
Several hideous derivatives later (who am I kidding, I used Mathematica's Series command), we have $$ \varepsilon \int_{0}^{\infty} \frac{z}{p_{\varepsilon}(z)}\,dz=\epsilon \left(\frac{\xi _1 \log \left(-\xi _1\right)}{2-2 \xi _1}+\frac{\xi _2 \log \left(-\xi _2\right)}{2-2 \xi _2}+\frac{\xi _3 \log \left(-\xi _3\right)}{2-2 \xi _3}\right)+O\left(\epsilon ^3\right) $$In particular, this boils down to $$ =\pi +\epsilon (-2 \log (\epsilon )-1)-\frac{15 \pi \epsilon ^2}{8}+O\left(\epsilon ^3\right), $$which approaches $\pi$ as $\varepsilon\to 0$.
Simple almost solution
Starting from $$ \int_0^\infty \frac{\varepsilon \, z\,dz}{(z-1)^2 + \varepsilon^2z^3} $$ the variable change $z=1/y$ results in $$ \int_0^\infty \frac{\varepsilon \, dy}{y(1-y)^2 + \varepsilon^2} $$ Then the variable change $y = \varepsilon t$ results in $$ \int_0^\infty \frac{dt}{\varepsilon t (\frac{1}{\varepsilon}-t)^2 + 1} $$ Finally, the variable change $s=t-\frac{1}{\varepsilon}$ results in $$ \int_{-\frac{1}{\varepsilon}}^\infty \frac{ds}{\varepsilon (s+\frac{1}{\varepsilon}) s^2 + 1} = \int_{-\frac{1}{\varepsilon}}^\infty \frac{ds}{s^2 + \varepsilon s^3 + 1} \to \int_{-\infty}^{\infty} \frac{ds}{s^2 + 1} = \pi. $$
Problem
The denominator has a zero close to $s=-\frac{1}{\varepsilon}$ so some extra work is needed to make sure that this doesn't give a contribution which won't vanish when $\varepsilon \to 0.$
Here I another solution based only on Calculus and a little complex analysis.
(i). Define
$$ I(\varepsilon):= \int_0^\infty \frac{x^3\,dx}{(x^2-\epsilon^2)^2 + x^6}\,dx\tag{1}\label{one} $$ The change of variables ($ z =\tfrac{z^2}{\varepsilon^2}$) that Marcos Gil used and obtain $$ I(\varepsilon)=\varepsilon\int^\infty_0\frac{z}{(z-1)^2 +\varepsilon^2z^3}\,dz\tag{2}\label{two} $$
(ii). Analysis of the roots of the denominator of the integrand in $\eqref{two}$. Using Integrand's notation, let $p_\varepsilon(z)=(z-1)^2+\varepsilon^2z^3$. For any $0<r<1$, there is $\eta>0$ such that for all $0<\varepsilon<\eta$, $$ |p_\varepsilon(z)-(z-1)^2|=\varepsilon^2|z|^3<|p_\varepsilon(z)|$$ for all $z$ along the curve $\gamma_r(\theta)=1+re^{i\theta}$. By Rouché's theorem, $p_\varepsilon$ has two solutions inside the ball $B(1;r)$. These solutions are fully complex (and thus pairwise conjugate) since $p_\varepsilon(x)>0$ for all $x\geq0$.
Since $p_\varepsilon(1)=\varepsilon^2>0$, $p_\varepsilon(x)\xrightarrow{x\rightarrow-\infty}-\infty$, and $p_\varepsilon(z)\rightarrow (z-1)^2$ uniformly in compact sets, we have that $p_\varepsilon$ has negative root. Furthermore, if $z^*_\varepsilon=a_\varepsilon+i b_\varepsilon$, ($b_\varepsilon>0$) is one of the complex roots of $p_\varepsilon$, and $c_\varepsilon$ is the negative root of $p_\varepsilon$, then
(a) $\lim_{\varepsilon\rightarrow0}a_\varepsilon + i b_\varepsilon=1$.
(b) $\lim_{\varepsilon\rightarrow0}c_\varepsilon=-\infty$.
To ease notation, let us drop the $\varepsilon$ subscript in what follows. Well know relations between the roots of polynomials and their coefficients give \begin{aligned} c(a^2+b^2)=-\varepsilon^{-2},\qquad c+2a = -\varepsilon^{-2},\qquad 2ac +a^2+b^2=-2\varepsilon^{-2} \end{aligned} Any of this equations imply that $$\lim_{\varepsilon\rightarrow0}c_\varepsilon\varepsilon^2=-1\tag{3}\label{three}$$ On the other hand, direct substitution of $z^*_\varepsilon$ into $p_\varepsilon$ gives \begin{aligned} (a-1)^2+b^2 &=\varepsilon^2\big(a^2+b^2\big)^{3/2}\\ a^3-3ab^2+\frac{a^2-b^2-2a+1}{\varepsilon^2}&=0\\ 3a^2b-b^3+\frac{2b(a-1)}{\varepsilon^2}&=0 \end{aligned} where the fist equation follows from $|p_\varepsilon(z^*_\varepsilon)-(z^*_\varepsilon-1)^2|=\varepsilon^2|z^*_\varepsilon|^3$ and the last two equations follows by equating real and imaginary parts of $p_\varepsilon(z^*_\varepsilon)=0$. As $a_\varepsilon\xrightarrow{\varepsilon\rightarrow0}1$ and $b_\varepsilon\xrightarrow{\varepsilon\rightarrow0}0$, we get $$ \lim_{\varepsilon\rightarrow0}\frac{a_\varepsilon -1}{\varepsilon}=0,\qquad \lim_{\varepsilon\rightarrow0}\frac{b_\varepsilon}{\varepsilon}=1\tag{4}\label{four} $$
(iii). Partial fractions decomposition of the integrand in $\eqref{two}$ gives $$ \frac{z}{p_\varepsilon(z)}=\frac{1}{\varepsilon^2}\Big(\frac{A}{z-c} +\frac{Bz+C}{(z-a)^2+b^2}\Big) $$ where $$ A=-B=\frac{c}{c^2-2ac+a^2+b^2},\qquad C= \frac{a^2+b^2}{c^2-2ac+a^2+b^2}\tag{5}\label{five} $$ Putting things together leads to \begin{aligned} I(\varepsilon)&=\frac{1}{\varepsilon}\Big(\int^\infty_0 \frac{A}{z-c} + \frac{B(z-a)}{(z-a)^2+b^2} \,dz\Big) +\frac{Ba +C}{\varepsilon}\Big(\int^\infty_0 \frac{dz}{(z-a)^2+b^2}\Big)\\ &=\frac{A}{\varepsilon}\log\Big(\frac{z-c}{\sqrt{(z-a)^2+b^2}}\Big)\Big|^\infty_0 + \frac{Ba +C}{\varepsilon b}\arctan\big(\frac{z-a}{b}\big)\Big|^\infty_0\\ &=-\frac{A}\varepsilon\log\Big(-\frac{c}{\sqrt{a^2+b^2}}\Big) + \frac{Ba +C}{\varepsilon b}\Big(\frac{\pi}{2} -\arctan\big(-\frac{a}{b}\big)\Big) \end{aligned} It follows from $\eqref{five}$ and $\eqref{three}$ that $$ \lim_{\varepsilon\rightarrow0}\frac{-A_\varepsilon}{\varepsilon^2}=1=\lim_{\varepsilon\rightarrow0}\frac{C_\varepsilon}{\varepsilon^4} $$ Consequently, $\lim_{\varepsilon\rightarrow0}\frac{-A_\varepsilon}\varepsilon\log\Big(-\frac{c_\varepsilon}{\sqrt{a^2_\varepsilon+b^2_\varepsilon}}\Big)=0$. Similarly, from $\eqref{five}$, $\eqref{four}$ and $\eqref{three}$ \begin{aligned} \lim_{\varepsilon\rightarrow0}\frac{B_\varepsilon a_\varepsilon +C_\varepsilon}{\varepsilon b_\varepsilon}&=\lim_{\varepsilon\rightarrow0}\frac{-a_\varepsilon c_\varepsilon +a^2_\varepsilon+b^2_\varepsilon}{b_\varepsilon\varepsilon(c^2_\varepsilon-2a_\varepsilon c_\varepsilon+a^2_\varepsilon+b^2_\varepsilon)}\\ &=\lim_{\varepsilon\rightarrow0}\frac{\varepsilon}{b_\varepsilon}\frac{-c_\varepsilon\varepsilon^2 a_\varepsilon +\varepsilon^2(a^2_\varepsilon+b^2_\varepsilon)}{(c_\varepsilon\varepsilon^2)^2 - 2\varepsilon^2(c_\varepsilon\varepsilon^2) a_\varepsilon +\varepsilon^2(a^2_\varepsilon + b^2_\varepsilon)} = 1 \end{aligned} Hence, $\lim_{\varepsilon\rightarrow0}\frac{B_\varepsilon a_\varepsilon +C_\varepsilon}{\varepsilon b}\Big(\frac{\pi}{2} -\arctan\big(-\frac{a_\varepsilon}{b_\varepsilon}\big)\Big)=\pi$.
Comments:
- These statements presented and proven in (ii) may be obtained using Vieta's formulas for the cubic equation however, these may be unpalatable to write and analyze on paper (in the 18th and 19th centuries that is what one would have done). Here any symbolic language (Maxima, Ada, Mathematica, Maple etc) may be handy as others have mentioned and used.
- With a little more effort, we could have also obtained some asymptotics for $I(\varepsilon)-\pi$ without resorting to computer aid estimates.