Why $\lim_{\varepsilon \to 0} \varepsilon \int_0^\infty \frac{z\,dz}{(z-1)^2 + \varepsilon^2z^3} = \pi$? I need help evaluating the following limit:
$$
\lim_{\varepsilon \to 0}\left[2\varepsilon\int_{0}^{\infty} \frac{x^{3}\,\mathrm{d}x}
{\left(x^{2} - \varepsilon^{2}\right)^{2} + x^{6}}\right]
$$
Making the substitutions $y = x^{2}$ and $z = y/\varepsilon^{2}$ I was able to put the integral in a form that seems more tractable:
$$
\lim_{\varepsilon \to 0}\left[\varepsilon
\int_{0}^{\infty}\frac{z\,\mathrm{d}z}{\left(z - 1\right)^{2} + \varepsilon^{2}z^{3}}\right]
$$
I have two reasons to think that the limit evaluates to $\pi$:

*

*The first one is that it's necessary for the result I'm trying to reach in a physics problem, which obviously isn't a very good justification.

*The second one ist that I made a numerical calculation that gave me a result very close to $\pi$: I restricted the integral to the interval $[0,10]$, because the integrand has a very sharp peak at $z=1$.


*

*Then, I divided that interval in $2 \cdot 10^{6}$ subintervals and used Simpson's method to calculate the integral. With
$\varepsilon = 10^{-4}$ I got the result $3.1417$.

*I tried to calculate it using the residues method, but I couldn't find the roots of the third degree polynomial in the denominator.

Does anyone have an idea on how to evaluate the limit analytically? Any help is appreciated.
 A: 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 without loss of generality,  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 $\varepsilon\to 0$.
A: 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 $\varepsilon \to 0.$
A: 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.

A: By substituting $x=\varepsilon y^{-1/2}$, we get
$$ \int_{0}^{\infty} \frac{2\varepsilon x^3 \, \mathrm{d}x}{(x^2-\varepsilon^2)^2 + x^6} = \int_{0}^{\infty} \frac{\varepsilon \, \mathrm{d}y}{y(y-1)^2 + \varepsilon^2} = \operatorname{Im}\int_{0}^{\infty} \frac{\mathrm{d}y}{\sqrt{y}(y-1) - i\varepsilon}. $$
Proof with Gap: The above limit is essentially the same as the imaginary part of
$$ \lim_{\epsilon \to 0^+} \int_{0-i\varepsilon}^{\infty-i\varepsilon} \frac{\mathrm{d}z}{\sqrt{z}(z-1)}
= \operatorname{PV}\!\!\int_{0}^{\infty} \frac{\mathrm{d}x}{\sqrt{x}(x-1)} + i\pi \, \underset{z=1}{\mathrm{Res}}\, \frac{1}{\sqrt{z}(z-1)}. $$
Since the Cauchy PV on the right-hand side is a real number, only the residue term contributes to the imaginary part with the value of $\pi$.
Actual Proof: Let $\mathcal{I} = (1-\frac{1}{2020},1+\frac{1}{2020})$ and consider the following regularized version of the integrand:
$$ f_{\varepsilon}(y) := \begin{cases}
\frac{1}{\sqrt{y}(y-1) - i\varepsilon} - \frac{1}{(y-1) - i\varepsilon} = \frac{(y-1)(1-\sqrt{y})}{(\sqrt{y}(y-1) - i\varepsilon)((y-1) - i\varepsilon)} , & y \in \mathcal{I}; \\
\frac{1}{\sqrt{y}(y-1) - i\varepsilon}, & y \notin \mathcal{I}; 
\end{cases}$$
Then by using the simple inequality $\left| \frac{1}{z} \right| \leq \frac{1}{\left| \operatorname{Re}(z) \right|}$, we obtain the bound:
$$ \left| f_{\varepsilon}(y) \right| \leq \begin{cases}
\frac{1}{\sqrt{y}(\sqrt{y}+1)}, & y \in \mathcal{I}; \\
\frac{1}{\sqrt{y}\left|y-1\right|}, & y \notin \mathcal{I};
\end{cases} $$
It is easy to check that this bound is integrable over $[0, \infty)$. Moreover, $f_{\varepsilon}$ converges pointwise to the real-valued function $f_0$ as $\varepsilon \to 0^+$. So, by the Dominated Convergence Theorem,
$$ \lim_{\varepsilon \to 0^+} \operatorname{Im}\int_{0}^{\infty} f_{\varepsilon}(x) \, \mathrm{d}y
= \operatorname{Im} \underbrace{ \int_{0}^{\infty} f_0(x) \, \mathrm{d}y }_{\in \mathbb{R}} = 0. $$
Therefore it follows that
$$
\lim_{\varepsilon \to 0^+} \operatorname{Im}\int_{0}^{\infty} \frac{\mathrm{d}y}{\sqrt{y}(y-1) - i\varepsilon}
= \lim_{\varepsilon \to 0^+} \operatorname{Im}\int_{1-\frac{1}{2020}}^{1+\frac{1}{2020}} \frac{\mathrm{d}y}{y-1 - i\varepsilon}
= \pi
$$
