Definite integral+limits question

  • Thread starter Krushnaraj Pandya
  • Start date
In summary, the conversation discusses finding the value of a limit using various methods such as using inequalities, the mean value theorem for definite integrals, and the limit definition. The final answer is suspected to be ##\pi^2/4## and is proven using the mean value theorem for definite integrals. The conversation also touches upon the use of inequalities for inverse trigonometric functions and the mini theorem for integrals.
  • #1
Krushnaraj Pandya
Gold Member
697
73

Homework Statement


The value of ## \lim_{t \to \infty} \frac {\int_0^t (\arctan x) ^2}{\sqrt {t^2+1}}##

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
 
Physics news on Phys.org
  • #2
Are you allowed to use inequalities like for example ##\arctan x<\frac{\pi}{2}##?
 
  • #3
Delta² said:
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...
 
  • #4
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...
 
  • #5
Delta² said:
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...
 
  • #6
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?
 
  • #7
Delta² said:
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?
 
  • #8
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.
 
  • #9
Delta² said:
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?
 
  • #10
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}##.
 
  • #11
Delta² said:
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! can't we use -pi/4<arctan(x) to get a sandwich condition
 
  • #12
Krushnaraj Pandya said:
Proved that! can't we use -pi/4<arctan(x) to get a sandwich condition

Nope using ##-\frac{\pi}{2}<\arctan x## won't work... Look in your textbook if there is some section with inequalities for inverse trigonometric functions..
 
  • #13
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}##
 
  • #14
Delta² said:
Nope using ##-\frac{\pi}{2}<\arctan x## won't 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?
 
  • #15
Delta² said:
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?
 
  • #16
Krushnaraj Pandya said:
\
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..
 
  • #17
Delta² said:
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?
 
  • #18
Krushnaraj Pandya said:
\
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.
 
Last edited:
  • Like
Likes StoneTemplePython
  • #19
Delta² said:
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?
 
  • #20
Ray Vickson said:
How can we know for sure that the ##x_n## goes to infinity?
Well , right now I can't think something rigorous. The intuition behind it , is that for large n the value of the integral is going to 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 going to be similar to yours as to why ##c_n/n\to 0##
 
  • #21
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.
 
  • #22
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
 
Last edited:
  • #23
Delta² said:
Well , right now I can't think something rigorous. The intuition behind it , is that for large n the value of the integral is going to 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 going to 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.##
 
Last edited:
  • Like
Likes Delta2
  • #24
Delta² said:
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?
 
  • Like
Likes Delta2
  • #25
StoneTemplePython said:
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
 
  • #26
Delta² said:
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.
umm, ok since this is the first time I've ever used the Newton-leibniz rule (its not in our syllabus but I learned it for kicks and an advantage in my entrance exams) I think I might be making a mistake- would you please confirm my work?
the question is in infinity/infinity form, derivative of the numerator should be ## f(t) \frac {dt}{dt} = {arctan {t}}^2## right? the denominator follows simple differentiation and then another application of Hospital rule. Am I correct?
 
  • #27
Krushnaraj Pandya said:
umm, ok since this is the first time I've ever used the Newton-leibniz rule (its not in our syllabus but I learned it for kicks and an advantage in my entrance exams) I think I might be making a mistake- would you please confirm my work?
the question is in infinity/infinity form, derivative of the numerator should be ## f(t) \frac {dt}{dt} = {arctan {t}}^2## right? the denominator follows simple differentiation and then another application of Hospital rule. Am I correct?
The derivative of the numerator is ##(\arctan t)^2##. You don't have to apply Hospital rule again, cause the limit of the numerator becomes finite that is ##\frac{\pi^2}{4}##
 
  • Like
Likes Krushnaraj Pandya
  • #28
Delta² said:
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.
This is my first time using the Newton leibniz theorem, (not in the syllabus, learned it for kicks and an advantage in e so just want to be sure I'm right. the derivative of the numerator should be ## f(t) \frac {dt}{dt} = {arctant}^2## the denominator follows simple differentiation then a successive application of Hospital rule. Am I correct?
Delta² said:
The derivative of the numerator is ##(\arctan t)^2##. You don't have to apply Hospital rule again, cause the limit of the numerator becomes finite that is ##\frac{\pi^2}{4}##
Ah! yes right. This is the best method, amazingly simple! Thank you
 
  • Like
Likes Delta2

What is a definite integral?

A definite integral is a mathematical concept used to find the area under a curve between two points (limits) on a graph. It is represented by the symbol ∫ and is a fundamental part of calculus.

What are limits in a definite integral?

Limits in a definite integral are the two points on a graph between which the area is being calculated. The lower limit indicates the starting point and the upper limit indicates the end point. These limits are used to define the boundaries of the area being calculated.

How do you solve a definite integral?

To solve a definite integral, you first need to determine the limits and the function being integrated. Then, you can use various techniques such as the Fundamental Theorem of Calculus, integration by substitution, or integration by parts. Once you have integrated the function, you can evaluate it at the limits to find the area under the curve.

What is the difference between definite and indefinite integrals?

The main difference between definite and indefinite integrals is the presence of limits. Definite integrals have specific limits, while indefinite integrals do not. In other words, definite integrals give a specific numerical value, while indefinite integrals give a function as the result.

Why are definite integrals important in science?

Definite integrals are important in science because they allow us to calculate a wide range of physical and mathematical quantities such as displacement, velocity, acceleration, and area under a curve. They are also essential in solving differential equations, which are used to model many natural phenomena in various scientific fields.

Similar threads

  • Calculus and Beyond Homework Help
Replies
8
Views
588
Replies
9
Views
959
  • Calculus and Beyond Homework Help
Replies
8
Views
719
  • Calculus and Beyond Homework Help
Replies
20
Views
1K
  • Calculus and Beyond Homework Help
Replies
19
Views
688
  • Calculus and Beyond Homework Help
Replies
16
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
934
  • Calculus and Beyond Homework Help
Replies
3
Views
929
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
Back
Top