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

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) .

• Hint: what's the second derivative of the numerator? – user217285 Mar 30 '20 at 20:34

## 4 Answers

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.

\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}

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

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}