TITLE: Evaluating the limit of a certain definite integral
QUESTION [10 upvotes]: Let $\displaystyle f(x)= \lim_{\epsilon \to 0} \frac{1}{\sqrt{\epsilon}}\int_0^x ze^{-(\epsilon)^{-1}\tan^2z}dz$ for $x\in[0,\infty)$.
Evaluate $f(x)$ in closed form for all $x\in[0,\infty)$ and sketch a graph of this function.

Hints, as well as solutions are welcome for this question :-)
Edit:  So far, I have, from substituting $\sqrt{\epsilon}u$ = z,
$\displaystyle f(x)= \lim_{\epsilon \to 0} \int_0^{\sqrt{\epsilon}u}  \sqrt{\epsilon}ue^{-(\epsilon)^{-1}\tan^2\sqrt{\epsilon}u}du$
But we can split the integral into two terms, with the first integral equal to zero, by dominated convergence theorem.  I think we only need to look at:
$\displaystyle f(x)= \lim_{\epsilon \to 0} \int_0^{a}  \sqrt{\epsilon}ue^{-(\epsilon)^{-1}\tan^2\sqrt{\epsilon}u}du$ +
$\displaystyle \lim_{\epsilon \to 0} \int_a^{\sqrt{\epsilon}u}  \sqrt{\epsilon}ue^{-(\epsilon)^{-1}\tan^2\sqrt{\epsilon}u}du$
= $$0+\displaystyle \lim_{\epsilon \to 0} \int_a^{\sqrt{\epsilon}u}  \sqrt{\epsilon}ue^{-(\epsilon)^{-1}\tan^2\sqrt{\epsilon}u}du$$
(I'm not sure if integrating away from the origin helps much, to be honest.)

REPLY [6 votes]: Consider what happens on an interval $[n\pi-\frac\pi2,n\pi+\frac\pi2]$. Let $x=\tan(z)$ and $u=x/\sqrt{\epsilon}$, then
$$
\begin{align}
\lim_{\epsilon\to0}\frac1{\sqrt{\epsilon}}\int_{n\pi-\frac\pi2}^{n\pi+\frac\pi2}e^{-\tan^2(z)/\epsilon}\,\mathrm{d}z
&=\lim_{\epsilon\to0}\frac1{\sqrt{\epsilon}}\int_{-\infty}^\infty e^{-x^2/\epsilon}\frac{\mathrm{d}x}{1+x^2}\\
&=\lim_{\epsilon\to0}\int_{-\infty}^\infty e^{-u^2}\frac{\mathrm{d}u}{1+\epsilon u^2}\\
&=\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}u\\[6pt]
&=\sqrt\pi\tag{1}
\end{align}
$$
For all $\epsilon\gt0$, we have
$$
\frac1{\sqrt\epsilon}e^{-\tan^2(z)/\epsilon}\le\frac{|\cot(z)|}{\sqrt{2e}}\tag{2}
$$
Thus, for any $\lambda\gt0$, Dominated Convergence says
$$
\lim_{\epsilon\to0}\frac1{\sqrt{\epsilon}}\int_{n\pi-\frac\pi2}^{n\pi+\frac\pi2}\big[|z-n\pi|\ge\lambda\big]e^{-\tan^2(z)/\epsilon}\,\mathrm{d}z=0\tag{3}
$$
where $[\cdot]$ are Iverson Brackets. Combining $(1)$ and $(3)$ gives
$$
\lim_{\epsilon\to0}\frac1{\sqrt{\epsilon}}\int_{n\pi-\frac\pi2}^{n\pi+\frac\pi2}\big[|z-n\pi|\lt\lambda\big]e^{-\tan^2(z)/\epsilon}\,\mathrm{d}z=\sqrt\pi\tag{4}
$$
Limits $(3)$ and $(4)$ tell us that $\frac1{\sqrt\epsilon}e^{-\tan^2(z)/\epsilon}$ is an approximation of
$$
\sqrt\pi\sum_{n\in\mathbb{Z}}\delta(z-n\pi)\tag{5}
$$
where $\delta(z)$ is the Dirac delta function.
Thus,
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{\epsilon\to0}\frac1{\sqrt{\epsilon}}\int_0^xz\,e^{-\tan^2(z)/\epsilon}\,\mathrm{d}z
=\left\{\begin{array}{}
\displaystyle\pi^{3/2}\,\frac{\lfloor x/\pi\rfloor^2+\lfloor x/\pi\rfloor}2&\text{if }x\not\in\pi\mathbb{Z}\\
\displaystyle\pi^{-1/2}\,\frac{x^2}2&\text{if }x\in\pi\mathbb{Z}
\end{array}\right.}\tag{6}
$$
The plot would look something like