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Definite integral+limits question

  • #1
Krushnaraj Pandya
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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
 

Answers and Replies

  • #2
Delta2
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Are you allowed to use inequalities like for example ##\arctan x<\frac{\pi}{2}##?
 
  • #3
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...
 
  • #4
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...
 
  • #5
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...
 
  • #6
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?
 
  • #7
Krushnaraj Pandya
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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?
I have never encountered something like that. The next chapter in my course is differential equations- perhaps this'll be elaborated there?
 
  • #8
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.
 
  • #9
Krushnaraj Pandya
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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.
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
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}##.
 
  • #11
Krushnaraj Pandya
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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}##.
Proved that! cant we use -pi/4<arctan(x) to get a sandwich condition
 
  • #12
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..
 
  • #13
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}##
 
  • #14
Krushnaraj Pandya
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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?
 
  • #15
Krushnaraj Pandya
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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}##
\
Got it! That was extremely clever, how do you manage to gain such intuition?
 
  • #16
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..
 
  • #17
Krushnaraj Pandya
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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
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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  • #19
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?
 
  • #20
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##
 
  • #21
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.
 
  • #22
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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  • #23
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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  • #24
Krushnaraj Pandya
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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.
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?
 
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  • #25
Krushnaraj Pandya
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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
interesting, i'll delve into it further as soon as I'm done with my midterms
 

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