# Definite integral+limits question

#### Krushnaraj Pandya

Gold Member
1. The problem statement, all variables and given/known data
The value of $\lim_{t \to \infty} \frac {\int_0^t (\arctan x) ^2}{\sqrt {t^2+1}}$

3. The attempt at a solution
I tried to use $\int_b^a f(x) = \int_b^a f(a+b-x)$ It didn't work out, and I can't see any other way to move ahead. I'd be grateful for your help

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#### Delta2

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Are you allowed to use inequalities like for example $\arctan x<\frac{\pi}{2}$?

#### Krushnaraj Pandya

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Are you allowed to use inequalities like for example $\arctan x<\frac{\pi}{2}$?
Since $\arctan x<\frac{\pi}{2}$ is something basic and known, I guess we can use it. I don't understand how you suggest using it though...

#### Delta2

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well using it you can get an upper bound of $\frac{\pi^2}{4}$ for that limit. I am currently thinking on what we should use to get a lower bound...

#### Krushnaraj Pandya

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well using it you can get an upper bound of $\frac{\pi^2}{4}$ for that limit. I am currently thinking on what we should use to get a lower bound...
I understand that as x tends to infinity- the square of tan inverse will approach (pi)^2/4 ; I do not know how you can change the limit mathematically using this fact...

#### Delta2

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Are you allowed to use inequalities between integrals? Because it is $(\arctan x)^2<\frac{\pi^2}{4}$ integrating both sides of the inequality what do you get?

#### Krushnaraj Pandya

Gold Member
Are you allowed to use inequalities between integrals? Because it is $(\arctan x)^2<\frac{\pi^2}{4}$ integrating both sides of the inequality what do you get?
I have never encountered something like that. The next chapter in my course is differential equations- perhaps this'll be elaborated there?

#### Delta2

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You mean that you haven't done the mini theorem that "if $f(x)<g(x)$ then $\int_a^bf(x)dx<\int_a^bg(x)dx$?. No that theorem has nothing to do with differential equations...

I guess there must be another way to work this, but I suspect $\pi^2/4$ is the answer.

#### Krushnaraj Pandya

Gold Member
You mean that you haven't done the mini theorem that "if $f(x)<g(x)$ then $\int_a^bf(x)dx<\int_a^bg(x)dx$?. No that theorem has nothing to do with differential equations...

I guess there must be another way to work this, but I suspect $\pi^2/4$ is the answer.
Ah, right I have read that theorem (just last night) but it wasn't used anywhere since so I totally forgot about it. Yes, the answer is indeed $\pi^2/4$ should I try using that theorem now to get the result myself or is this still part intuition-part solution?

#### Delta2

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It is not complete solution. Using it you can prove an inequality regarding that limit, namely that $\lim_{t \to \infty} \frac {\int_0^t (\arctan x) ^2dx}{\sqrt {t^2+1}}\leq\frac{\pi^2}{4}$.

#### Krushnaraj Pandya

Gold Member
It is not complete solution. Using it you can prove an inequality regarding that limit, namely that $\lim_{t \to \infty} \frac {\int_0^t (\arctan x) ^2dx}{\sqrt {t^2+1}}\leq\frac{\pi^2}{4}$.
Proved that! cant we use -pi/4<arctan(x) to get a sandwich condition

#### Delta2

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Proved that! cant we use -pi/4<arctan(x) to get a sandwich condition
Nope using $-\frac{\pi}{2}<\arctan x$ wont work... Look in your textbook if there is some section with inequalities for inverse trigonometric functions..

#### Delta2

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I thought of another way, using the mean value theorem for definite integrals.

First we replace t at the limit with a natural number n. The limit as $n\to\infty$ is the same as $t\to\infty$

Then from mean value theorem, we know that for every $n$, there will be $x_n$ such that $\int_0^n(\arctan x)^2dx=(\arctan x_n)^2(n-0)$

So from here just compute the limit $\lim_{n\to\infty}\frac{\int_0^n(\arctan x)^2dx}{\sqrt{n^2+1}}=…$ at the end use the fact that
$\lim_{n\to\infty}(\arctan x_n)^2=\frac{\pi^2}{4}$

#### Krushnaraj Pandya

Gold Member
Nope using $-\frac{\pi}{2}<\arctan x$ wont work... Look in your textbook if there is some section with inequalities for inverse trigonometric functions..
sorry, I mistyped; I meant $-\frac{\pi}{2}<\arctan x$ but you got it anyway. Why won't that work, is it because the lower limit is 0 and thus not a limiting condition on the lower side of arctan?

#### Krushnaraj Pandya

Gold Member
I thought of another way, using the mean value theorem for definite integrals.

First we replace t at the limit with a natural number n. The limit as $n\to\infty$ is the same as $t\to\infty$

Then from mean value theorem, we know that for every $n$, there will be $x_n$ such that $\int_0^n(\arctan x)^2dx=(\arctan x_n)^2(n-0)$

So from here just compute the limit $\lim_{n\to\infty}\frac{\int_0^n(\arctan x)^2dx}{\sqrt{n^2+1}}=…$ at the end use the fact that
$\lim_{n\to\infty}(\arctan x_n)^2=\frac{\pi^2}{4}$
\
Got it! That was extremely clever, how do you manage to gain such intuition?

#### Delta2

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\
Got it! That was extremely clever, how do you manage to gain such intuition?
It was a moment of divine inspiration :D. Except that I was just thinking what other properties and theorems for definite integrals we could use..

#### Krushnaraj Pandya

Gold Member
It was a moment of divine inspiration :D. Except that I was just thinking what other properties and theorems for definite integrals we could use..
wow, hope I'll get there someday...Thanks a lot.
I have 2 more doubts posted as 2 threads...one I've already solved and want to know if there's a shorter way. The second one I have no idea about how to proceed. would you mind taking a look?

#### Ray Vickson

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Got it! That was extremely clever, how do you manage to gain such intuition?
Following the intuition by Delta2, a possibly easier way is the followig. First, plot the integrand $\arctan^2(x)$ over an interval $0 \leq x \leq M$ for some moderate-to-large $M$. You will notice that the function is almost constant over the majority of the interval, so we have
$$\arctan^2(x) \doteq \lim_{w \to \infty} \arctan^2(w) = \frac{\pi^2}{4}$$
over almost the whole interval. Thus, $\int_0^n \arctan^2(x) \, dx = n \pi^2/4 + c_n$, where $c_n \to \text{some constant}$ as $n \to \infty.$ Since $c_n/\sqrt{n^2+1} \to 0$, the required limit is just $\pi^2/4.$

Note added in edit: we have that $g(x) = \pi^24 - \arctan^2(x) \geq 0$ approaches zero as $x \to \infty$. If $C = \lim_{n \to \infty} \int_0^n g(x) \, dx$ exists, then we have $0 < c_n < C$ for all $n > 0$, so it is, indeed, true that $c_n/n \to 0.$ While visual examination of a plot suggests that $C$ is finite, that is not a proof. We need to look at how near $\arctan(x)$ is to $\pi/2$ for large $x$, so we can tell whether $\arctan^2(x) - \pi^2/4$ goes to zero quickly enough to have a finite integral on $[0, \infty).$ You can verify that everything does, in fact, go through as needed.

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• StoneTemplePython

#### Ray Vickson

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I thought of another way, using the mean value theorem for definite integrals.

First we replace t at the limit with a natural number n. The limit as $n\to\infty$ is the same as $t\to\infty$

Then from mean value theorem, we know that for every $n$, there will be $x_n$ such that $\int_0^n(\arctan x)^2dx=(\arctan x_n)^2(n-0)$

So from here just compute the limit $\lim_{n\to\infty}\frac{\int_0^n(\arctan x)^2dx}{\sqrt{n^2+1}}=…$ at the end use the fact that
$\lim_{n\to\infty}(\arctan x_n)^2=\frac{\pi^2}{4}$
How can we know for sure that the $x_n$ goes to infinity?

#### Delta2

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How can we know for sure that the $x_n$ goes to infinity?
Well , right now I cant think something rigorous. The intuition behind it , is that for large n the value of the integral is gonna be $n\frac{\pi^2}{4}+\epsilon(n)$ so comparing this with $(\arctan x_n)^2(n-0)$ we get that $\lim_{n\to\infty}(\arctan x_n)^2=\frac{\pi^2}{4}+\lim_{n\to\infty}\frac{\epsilon(n)}{n}$. Well as to why $\lim_{n\to\infty}\frac{\epsilon(n)}{n}=0$ I guess the argument is gonna be similar to yours as to why $c_n/n\to 0$

#### Delta2

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After thinking of this over and over again I think the most rigorous approach is to use the good old Del Hospital rule, it is very straightforward as well if we first work the limit $lim_{t\to\infty}\frac{\int_0^t(\arctan x)^2 dx}{t}$ with Hospital rule.

#### StoneTemplePython

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a couple small points: if OP wants to clean up the denominator, setting up some inequalities like the below should sandwich the result.

$\frac {\int_0^t (\arctan x) ^2dx}{t}\frac{t}{t+1}= \frac {\int_0^t (\arctan x) ^2 dx}{(t+1)}= \frac {\int_0^t (\arctan x) ^2 dx}{\sqrt {(t+1)^2}}= \frac {\int_0^t (\arctan x) ^2 dx}{\sqrt {t^2 + 2t +1}} \leq \frac {\int_0^t (\arctan x) ^2 dx}{\sqrt {t^2+1}} \leq \frac {\int_0^t (\arctan x) ^2 dx}{\sqrt {t^2}} = \frac {\int_0^t (\arctan x) ^2 dx}{t}$

also the fact that $\arctan^2 x$ has a limit implies that the Cesaro mean evaluates to said limit. We are dealing with integrals -- specifically $\frac{1}{t} \int_0^t (\arctan x) ^2 dx$

not series here, but it's a closely related point -- and we could work with

$\frac{1}{t} \sum_{x=1}^t (\arctan x) ^2$

if we were so inclined. Normally its the case that people use integrals to estimate and bound series, but amusingly, I think the above integrals can be sandwiched by two cesaro means, to give the result. Given monotonicity / negative convexity for the function which exists for all $x \geq 1$, it shouldn't be too hard

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#### Ray Vickson

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Well , right now I cant think something rigorous. The intuition behind it , is that for large n the value of the integral is gonna be $n\frac{\pi^2}{4}+\epsilon(n)$ so comparing this with $(\arctan x_n)^2(n-0)$ we get that $\lim_{n\to\infty}(\arctan x_n)^2=\frac{\pi^2}{4}+\lim_{n\to\infty}\frac{\epsilon(n)}{n}$. Well as to why $\lim_{n\to\infty}\frac{\epsilon(n)}{n}=0$ I guess the argument is gonna be similar to yours as to why $c_n/n\to 0$
I made an error in my edit of post #18, but it is repairable. We can write
$$\int_0^n \arctan^2 x \, dx = \frac{\pi^2}{4} n - c_n,$$
where $c_n > 0.$ In post #18 I mistakenly said that $c_n \to \text{constant}$ as $n \to \infty$, and that is false. In fact, if we look at $\delta(x) = \pi^2/4 - \arctan^2(x)$ we have that $\delta(x) \sim \pi/x$ for large $x>0$. Thus, $c_n ~ K + \pi \ln n$ for a constant $K > 0$ Therefore, $c_n \to +\infty$ as $n \to \infty.$ However, we are really interested in the average instead of the total, so we need only note that $c_n/n \to 0$ because $\ln n / n \to 0.$

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• Delta2

#### Krushnaraj Pandya

Gold Member
After thinking of this over and over again I think the most rigorous approach is to use the good old Del Hospital rule, it is very straightforward as well if we first work the limit $lim_{t\to\infty}\frac{\int_0^t(\arctan x)^2 dx}{t}$ with Hospital rule.
Ah! why didn't I think of that, so I suppose we can use the newton-leibnitz theorem on the numerator and same old differentiation in the denominator-right?

• Delta2

#### Krushnaraj Pandya

Gold Member
a couple small points: if OP wants to clean up the denominator, setting up some inequalities like the below should sandwich the result.

$\frac {\int_0^t (\arctan x) ^2dx}{t}\frac{t}{t+1}= \frac {\int_0^t (\arctan x) ^2 dx}{(t+1)}= \frac {\int_0^t (\arctan x) ^2 dx}{\sqrt {(t+1)^2}}= \frac {\int_0^t (\arctan x) ^2 dx}{\sqrt {t^2 + 2t +1}} \leq \frac {\int_0^t (\arctan x) ^2 dx}{\sqrt {t^2+1}} \leq \frac {\int_0^t (\arctan x) ^2 dx}{\sqrt {t^2}} = \frac {\int_0^t (\arctan x) ^2 dx}{t}$

also the fact that $\arctan^2 x$ has a limit implies that the Cesaro mean evaluates to said limit. We are dealing with integrals -- specifically $\frac{1}{t} \int_0^t (\arctan x) ^2 dx$

not series here, but it's a closely related point -- and we could work with

$\frac{1}{t} \sum_{x=1}^t (\arctan x) ^2$

if we were so inclined. Normally its the case that people use integrals to estimate and bound series, but amusingly, I think the above integrals can be sandwiched by two cesaro means, to give the result. Given monotonicity / negative convexity for the function which exists for all $x \geq 1$, it shouldn't be too hard
interesting, i'll delve into it further as soon as I'm done with my midterms

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