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.)



I wanted to give a special thanks to @robjon for his insightful comments.

We first observe that $\lim_{\epsilon\to 0}e^{-\tan z/\epsilon}=0$ unless $z=\ell \pi$, $\ell$ an integer. Therefore, all of the "action" of the integration will take place over intervals around $\ell \pi$. So, let's first see what is happening for $0<x<\pi/2$.

In the spirit of Laplace's Method, we have for $0<z<\pi/2$, $\tan^z =z^2+O(z^4)$ and thus for $0<x<\pi/2$

$$\begin{align} \epsilon^{-1/2}\int_0^xze^{-\tan^2z/\epsilon}dz&\sim\epsilon^{-1/2}\int_0^xze^{-z^2/\epsilon}dz\\\\ &=\epsilon^{-1/2}\left.\left(-\epsilon^{-z^2/\epsilon}\right)\right|_{z=0}^{z=x}\\\\ &=\epsilon^{1/2}\left(1-e^{-x^2/\epsilon}\right) \end{align}$$

which clearly goes to zero as $\epsilon\to 0$.

Next, we observe that the integration around singularities of the tangent function pose no challenge. Thus, for a general $(L-1)\pi<x<L\pi$, and $\delta >0$ we can write

$$\begin{align} \epsilon^{-1/2}\int_0^x ze^{-\tan^2z/\epsilon}dz&=\epsilon^{-1/2}\sum_{\ell=0}^{L-2}\left(\int_{\ell \pi+\delta}^{(\ell+1)\pi-\delta}ze^{-\tan^2z/\epsilon}dz+\int_{(\ell+1)\pi-\delta}^{(\ell+1)\pi+\delta}ze^{-\tan^2z/\epsilon}dz\right)\\\\ &+\epsilon^{-1/2}\int_{(L-1)\pi+\delta}^{x}ze^{-\tan^2z/\epsilon}dz \tag 1\\\\ \end{align}$$

We observe that in $(1)$ the only integrals that will contribute in the limit as $\epsilon \to 0$ are those around integer multiples of $\pi$. Thus, we have for $(L-1)\pi<x<L\pi$ and $\delta>0$

$$\begin{align} \lim_{\epsilon \to 0}\epsilon^{-1/2}\int_0^x ze^{-\tan^2z/\epsilon}dz&=\lim_{\epsilon \to 0} \epsilon^{-1/2}\sum_{\ell=0}^{L-2}\left(\int_{(\ell+1)\pi-\delta}^{(\ell+1)\pi+\delta}ze^{-\tan^2z/\epsilon}dz\right) \tag 2\\\\ \end{align}$$

We proceed to evaluate the integrals in $(2)$. To that end we have

$$\begin{align} \epsilon^{-1/2}\int_{(\ell+1)\pi-\delta}^{(\ell+1)\pi+\delta}ze^{-\tan^2z/\epsilon}dz &=\epsilon^{-1/2}\left(\int_{-\delta}^{\delta}ze^{-\tan^2z/\epsilon}dz+(\ell +1)\pi\int_{-\delta}^{\delta}e^{-\tan^2z/\epsilon}dz\right)\\\\ &=(\ell +1)\pi\epsilon^{-1/2}\int_{-\delta}^{\delta}e^{-\tan^2z/\epsilon}dz\\\\ &\sim (\ell +1)\pi\epsilon^{-1/2}\int_{-\delta}^{\delta}e^{-z^2/\epsilon}dz\\\\ &= (\ell +1)\pi\int_{-\delta/\epsilon^{1/2}}^{\delta/\epsilon^{1/2}}e^{-z^2}dz\\\\ &\to (\ell +1)\pi^{3/2} \end{align}$$

Summing over $\ell$ we find for $(L-1)\pi<x<L\pi$

$$\lim_{\epsilon \to 0}\epsilon^{-1/2}\int_0^xze^{-\tan^2z/\epsilon}dz=\frac{L(L-1)\pi^{3/2}}{2}$$

One final note concerns the case in which $x=L\pi$. For that case, we see that we need to add one more integral, namely

$$\begin{align} \lim_{\epsilon\to 0}\epsilon^{-1/2}\int_{L\pi-\delta}^{L\pi}ze^{-\tan^z/\epsilon}&=L\pi\int_{-\infty}^0e^{-z^2}dz\\\\ &=\frac12 L\pi^{3/2} \end{align}$$

Thus, for $x=L\pi$ we have

$$\lim_{\epsilon \to 0}\epsilon^{-1/2}\int_0^xze^{-\tan^2z/\epsilon}dz=\frac{L^2\pi^{3/2}}{2}$$

Putting it all together we have

$$\lim_{\epsilon \to 0}\epsilon^{-1/2}\int_0^xze^{-\tan^2z/\epsilon}dz= \begin{cases} \frac{L(L-1)\pi^{3/2}}{2},&(L-1)\pi<x<L\pi\\\\ \frac{L^2\pi^{3/2}}{2},&x=L\pi \end{cases} $$

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    $\begingroup$ This only works for $x\lt\pi$. The interesting part comes when $x\ge\pi$. $\endgroup$ – robjohn Jul 8 '15 at 9:23
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    $\begingroup$ The part of your answer where you get $2z^2+O(z^6)$ is garbled. Are you claiming that $\tan^2(z)=2z^2+O(z^6)$? If so, that seems to be problematic as $\tan^2(z)=z^2+O(z^4)$. $\endgroup$ – robjohn Jul 8 '15 at 10:43
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    $\begingroup$ You note that $\lim\limits_{\epsilon\to0} e^{-\tan^2(x)/\epsilon}=0$. However, what is needed is $\lim\limits_{\epsilon\to0} \frac1{\sqrt\epsilon}e^{-\tan^2(x)/\epsilon}=0$ for $x\ne0$, and that $\frac1{\sqrt\epsilon}e^{-\tan^2(x)/\epsilon}$ is uniformly bounded in $\epsilon$ for $x$ away from $0$ (the last is so that we can use Dominated Convergence). $\endgroup$ – robjohn Jul 9 '15 at 0:35
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    $\begingroup$ @robjon I did note that and for purpose of gaining an understanding of where the "action" is in the integral. I don't see that this note is inappropriate. $\endgroup$ – Mark Viola Jul 9 '15 at 0:52
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    $\begingroup$ @LebronJames You're quite welcome. And special thanks goes to Robjon for showing the way forward! $\endgroup$ – Mark Viola Jul 9 '15 at 2:50

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

enter image description here

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    $\begingroup$ @LebronJames: I am out getting dinner. I will address your questions as soon as I can. $\endgroup$ – robjohn Jul 9 '15 at 1:54
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    $\begingroup$ @LebronJames: on $\big[-\frac\pi2,\frac\pi2\big]$, $\frac1{\sqrt\epsilon}e^{-\tan^2(x)/\epsilon}$ is a family of approximations to $\sqrt\pi$ times the Dirac delta function. Since $\tan^2(x)$ has a period of $\pi$, on $\mathbb{R}$, $\frac1{\sqrt\epsilon}e^{-\tan^2(x)/\epsilon}$ is an approximation to $\sqrt\pi$ times the sum of Dirac delta functions translated to $x=n\pi$. Thus, on an interval containing a neighborhood of $n\pi$, the integral of $\frac1{\sqrt\epsilon}f(x)e^{-\tan^2(x)/\epsilon}$ tends to $\sqrt\pi\,f(n\pi)$. $\endgroup$ – robjohn Jul 9 '15 at 4:58
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    $\begingroup$ @LebronJames: on an interval ending at $n\pi$, the integral of $\frac1{\sqrt\epsilon}f(x)e^{-\tan^2(x)/\epsilon}$ tends to $\frac{\sqrt\pi}2\,f(n\pi)$. In the case at hand, $f(x)=x$. $\endgroup$ – robjohn Jul 9 '15 at 5:01
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    $\begingroup$ @robjohn It doesn't change the thrust of your development, but the bound you want to establish dominated convergence is $$\frac{1}{\sqrt{\epsilon}}e^{-\tan^2x/\epsilon}\le \frac{1}{\sqrt{2e\tan^2x}}$$However, doesn't the DC theorem require the dominating function be integrable? Neither $\cot^2x$ nor $|\cot x|$ are integrable on any interval containing a singularity. So, does that not invalidate the reasoning here? $\endgroup$ – Mark Viola Jul 9 '15 at 6:21
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    $\begingroup$ @Dr.MV: Oops... that is the bound, but that works as well. All we are trying to show with dominated convergence is $(3)$ and for that $|x|\ge\lambda$. Since $\frac1{\sqrt\epsilon}e^{-\tan^2(x)/\epsilon}\le\frac{\cot(\lambda)}{\sqrt{2e}}$, the integrands are bounded by $\frac{\cot(\lambda)}{\sqrt{2e}}$ on $|x|\ge\lambda$ and $|x|\lt\frac\pi2$ so the dominating function is integrable (bounded on a bounded set). $\endgroup$ – robjohn Jul 9 '15 at 8:58

For small $\epsilon$, $\tan^2(\sqrt{\epsilon} u ) \approx \epsilon u^2$, so the integral is of approximately $\sqrt{\epsilon}u e^{-u^2}$ which can be integrated.

  • $\begingroup$ Thanks for the cool observation, @martycohen! :-) $\endgroup$ – User001 Jul 8 '15 at 5:57

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