3
$\begingroup$

Evaluate $\lim\limits_{x\to \infty} \frac{\int_0^x \left(\arctan t \right)^2\,dt}{\sqrt{x^2+1}}$

My attempt was to start doing the integral by parts but at some point it just didn't work. Is there a simple way to do it ? Any help will be appreciated ! ( also, this is a highschool problem, so i would like to see some hints at that level) .

$\endgroup$
1
  • 1
    $\begingroup$ Hint: what's the second derivative of the numerator? $\endgroup$ – user217285 Mar 30 '20 at 20:34
2
$\begingroup$

Hints: Use l'Hospital. This is something like $\frac{\infty}{\infty}$. Also use fundamental theorem of calculus that states that $$\frac d{dx}\int_0^xf(t)dt=f(x)$$ Let me know if this is enough.

$\endgroup$
2
$\begingroup$

$\begin{align}\displaystyle\lim_{x\to \infty} \dfrac{\int_0^x \left(\arctan t \right)^2\mathrm{dt}}{\sqrt{x^2+1}}\left(\dfrac{\infty}{\infty}\right)&=\displaystyle\lim_{x\to \infty} \dfrac{\dfrac{\mathrm{d}}{\mathrm{dx}}\displaystyle\int_0^x \left(\arctan t \right)^2\mathrm{dt}}{\dfrac{\mathrm{d}}{\mathrm{dx}}\sqrt{x^2+1}}\text{ (L'Hospital rule)}\\&=\displaystyle\lim_{x\to \infty} \dfrac{\left(\arctan x \right)^2\cdot1-\left(\arctan 0\right)^2\cdot0+\displaystyle\int_0^x \dfrac{\partial \left(\arctan t\right)^2}{\partial x}\mathrm{dt}}{\dfrac{\mathrm{d}}{\mathrm{dx}}\sqrt{x^2+1}}\text{(Leibnitz rule)}\\&=\displaystyle\lim_{x\to \infty} \dfrac{\left(\arctan x \right)^2}{\dfrac{x}{\sqrt{x^2+1}}}\\&= \dfrac{\displaystyle\lim_{x\to \infty}\left(\arctan x \right)^2}{\displaystyle\lim_{x\to \infty}\dfrac{x}{\sqrt{x^2+1}}}\\&=\dfrac{\left(\frac{\pi}{2}\right)^2}{1}\\&=\dfrac{\pi^2}{4}\end{align}$

$\endgroup$
1
$\begingroup$

If you do not want to use L'Hopital's rule, here is my proposition :

Let $ x>0 $, using the substitution : $ \left\lbrace\begin{aligned}u&=\frac{1}{x}\\ \mathrm{d}x&=-\frac{\mathrm{d}u}{u^{2}}\end{aligned}\right. $, we get : \begin{aligned} \int_{0}^{x}{\arctan^{2}{t}\,\mathrm{d}t}&=\int_{\frac{1}{x}}^{+\infty}{\frac{1}{u^{2}}\left(\frac{\pi}{2}-\arctan{u}\right)^{2}\,\mathrm{d}u}\\ &=\frac{\pi^{2}}{4}\int_{\frac{1}{x}}^{+\infty}{\frac{\mathrm{d}u}{u^{2}}}-\pi\int_{\frac{1}{x}}^{+\infty}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}+\int_{\frac{1}{x}}^{+\infty}{\frac{\arctan^{2}{u}}{u^{2}}\,\mathrm{d}u}\\ \int_{0}^{x}{\arctan^{2}{t}\,\mathrm{d}t}&=\frac{\pi^{2}x}{4}-\pi\int_{\frac{1}{x}}^{+\infty}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}+\int_{\frac{1}{x}}^{+\infty}{\frac{\arctan^{2}{u}}{u^{2}}\,\mathrm{d}u} \end{aligned}

Now, since $ \frac{\arctan^{2}{t}}{t^{2}}=\underset{\overset{t\to +\infty}{}}{\mathcal{O}}\left(\frac{1}{t^{2}}\right) $, and $ t\overset{f}{\mapsto}\frac{\arctan^{2}{t}}{t^{2}} $ is extendable by continuity at $ 0 $, $ f $ is integrable on $ \mathbb{R}^{+} $, meaning $ \lim\limits_{x\to +\infty}{\int\limits_{\frac{1}{x}}^{+\infty}{\frac{\arctan^{2}{u}}{u^{2}}\,\mathrm{d}u}}=\int\limits_{0}^{+\infty}{\frac{\arctan^{2}{u}}{u^{2}}\,\mathrm{d}u}=C \cdot $

Since $ \left(\forall u>0\right), \arctan{u}\leq u $, we get : $$ \left|\frac{1}{x}\int_{\frac{1}{x}}^{1}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}\right|\leq\frac{1}{x}\int_{\frac{1}{x}}^{1}{\frac{\mathrm{d}u}{u}}=\frac{\ln{x}}{x}\underset{x\to +\infty}{\longrightarrow} 0 $$

Thus, $ \frac{1}{x}\int\limits_{\frac{1}{x}}^{+\infty}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}=\frac{1}{x}\int\limits_{\frac{1}{x}}^{1}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}+\frac{1}{x}\int\limits_{1}^{+\infty}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}\underset{x\to +\infty}{\longrightarrow}0 \cdot $

Hence $$ \frac{1}{x}\int_{0}^{x}{\arctan^{2}{u}\,\mathrm{d}u}=\frac{\pi^{2}}{4}-\frac{\pi}{x}\int_{\frac{1}{x}}^{+\infty}{\frac{\arctan{u}}{u^{2}}\,\mathrm{d}u}+\frac{1}{x}\int_{\frac{1}{x}}^{+\infty}{\frac{\arctan^{2}{u}}{u^{2}}\,\mathrm{d}u}\underset{x\to +\infty}{\longrightarrow}\frac{\pi^{2}}{4} $$

Which leads to $$ \lim_{x\to +\infty}{\frac{1}{\sqrt{1+x^{2}}}\int_{0}^{x}{\arctan^{2}{u}\,\mathrm{d}u}}=\lim_{x\to +\infty}{\frac{x}{\sqrt{1+x^{2}}}\times\frac{1}{x}\int_{0}^{x}{\arctan^{2}{u}\,\mathrm{d}u}}=1\times\frac{\pi^{2}}{4} $$

$\endgroup$
1
$\begingroup$

Note that

$${x\over\sqrt{x^2+1}}\to1\quad\text{as }x\to\infty$$

so we can replace the denominator in the limit with simply $x$. Now integration by parts tells us

$$\int_0^x(\arctan t)^2dt=x(\arctan x)^2-\int_0^x{2t\arctan t\over1+t^2}dt$$

so

$${1\over x}\int_0^x(\arctan t)^2dt=(\arctan x)^2-{1\over x}\int_0^x{2t\arctan t\over1+t^2}dt$$

and, since $\arctan t$ is an increasing function with limit $\pi/2$ as $t\to\infty$,

$$0\le{1\over x}\int_0^x{2t\arctan t\over1+t^2}dt\le{\arctan x\over x}\int_0^x{2t\over1+t^2}dt={\arctan x\log(1+x^2)\over x}\le{\pi\over2}{\log(1+x^2)\over x}\to0$$

Thus

$${1\over x}\int_0^x(\arctan t)^2dt\to\left(\pi\over2\right)^2-0={\pi^2\over4}$$

Remark: The limit ${\log(1+x^2)\over x}\to0$ can be found either via L'Hopital or from some cleverly crude integral inequalities:

$$\begin{align} 0\le{\log(1+x^2)\over x}={1\over x}\int_0^x{2t\over1+t^2}dt &={2\over x}\left(\int_0^{x^{1/3}}{t\over1+t^2}dt+\int_{x^{1/3}}^{x^{2/3}}{t\over1+t^2}dt+\int_{x^{2/3}}^x{t\over1+t^2}dt\right)\\ &\le{2\over x}\left({x^{1/3}(x^{1/3}-0)\over1}+{x^{2/3}(x^{2/3}-0)\over(x^{1/3})^2}+{x(x-0)\over(x^{2/3})^2}\right)\\ &={2\over x}\left(x^{2/3}+x^{2/3}+x^{2/3} \right)\\ &={6\over x^{1/3}}\to0 \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.