How to find $\lim_{x\to\infty} \frac{ \int_x^1 \arctan(t^2)\, dt}{x} $

Question

Any tips on how to find this limit:

$$\lim_{x\to\infty} \frac{ \displaystyle\int_x^1 \arctan(t^2)\, dt}{x} $$

Answer 1

Let $L$ be given by the limit

$$L=\lim_{x\to \infty}\frac{1}{x}\int_x^1 \arctan(t^2)dt$$

Note that the limit is in the form for L'Hospital's Rule. Thus, we have

$$L=\lim_{x\to \infty}\frac{1}{x}\int_x^1 \arctan(t^2)dt=\lim_{x\to \infty}\left(- \arctan(x^2)\right)=-\pi/2$$

Answer 2

Do you really mean $\int_x^1$? Seems a little curious, $\int_1^x$ would make more sense, given that $x$ is tending to infinity.

Yes, you could get this from L'Hopital as suggested. Imo using L'Hopital is very often a bad thing, because one doesn't learn anything. A simple direct proof is better here because it works in cases where the Fundamental Theorem of Calculus doesn't quite apply, it allows you to prove similar things for series, where L'Hopital makes no sense, etc. So:

Theorem. If $f$ is continuous on $[1,\infty)$ and $\lim_{x\to\infty}f(x)=L$ then $$\lim_{x\to\infty}\frac1x\int_1^xf(t)\,dt=L.$$ Just like a well known result for series.

Lemma. Suppose $f$ is continuous on $[a,b]$ and $|L-f(t)|<\epsilon$ everywhere on $[a,b]$. Then $|L-\frac1{b-a}\int_a^bf(t)\,dt|<\epsilon$.

Proof $$\left|L-\frac1{b-a}\int_a^bf\right| =\left|\frac1{b-a}\int_a^b(L-f)\right| \le\frac1{b-a}\int_a^b|L-f| <\frac1{b-a}\int_a^b\epsilon=\epsilon.$$

Proof of the theorem: Let $\epsilon>0$. Choose $A>1$ so $|L-f(t)|<\epsilon$ for all $t\ge A$. Suppose $x>A$. Then $$L-\frac1x\int_1^xf(t)\,dt =-\frac1x\int_1^Af(t)\,dt+\frac AxL+\frac{x-A}x\left(L-\frac1{x-A}\int_A^xf(t)\,dt\right)=I+II+III.$$It's clear that $I$ and $II$ tend to $0$ as $x\to\infty$, and the lemma shows that $|III|<\epsilon$ for all $x>A$. So the whole thing is less than $3\epsilon$ if $x$ is large enough.