My Math Forum How to approach this integral?

 Calculus Calculus Math Forum

February 22nd, 2019, 08:44 AM   #1
Newbie

Joined: Feb 2019
From: United states

Posts: 9
Thanks: 0

How to approach this integral?

How approach this integral? I have assumed z=nx and hence
Then derivative gives #dz=ndx#. Now it becomes this
Integral from #\frac{n} {n+1}# to 1 ##\frac{arctan(z)dz} {n arcsin(z)}##. After this I am not able to proceed further.
Attached Images
 IMG-20190222-WA0020.jpg (9.3 KB, 38 views)

 February 22nd, 2019, 10:33 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,969 Thanks: 2219 Where did you get the question from? The fourth answer seems to be correct. Thanks from idontknow
 February 22nd, 2019, 01:09 PM #3 Global Moderator   Joined: May 2007 Posts: 6,822 Thanks: 723 To get the answer, note that for large n the integrand $\approx \frac{tan^{-1}(1)}{sin^{-1}(1)}=\frac{\pi/4}{\pi/2}=\frac{1}{2}$, so $C_n\approx \frac{1}{2n^2}$ Thanks from topsquark and idontknow
February 22nd, 2019, 04:12 PM   #4
Senior Member

Joined: Sep 2016
From: USA

Posts: 645
Thanks: 406

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by mathman To get the answer, note that for large n the integrand $\approx \frac{tan^{-1}(1)}{sin^{-1}(1)}=\frac{\pi/4}{\pi/2}=\frac{1}{2}$, so $C_n\approx \frac{1}{2n^2}$
Interchanging limits like that is not always allowed. For example, your analysis applied to the following sequence would give a wrong answer.
$C_n = \int_0^{n} 1/n \ dx$
In general I would say its a bad idea to think about it this way even if just for "intuition" since the intuition gained is wrong.
Justifying when you can interchange limits or not typically requires the dominated convergence theorem or monotone convergence theorem. Both of which are a bit advanced for a calculus student.

February 23rd, 2019, 01:58 PM   #5
Global Moderator

Joined: May 2007

Posts: 6,822
Thanks: 723

Quote:
 Originally Posted by SDK Interchanging limits like that is not always allowed. For example, your analysis applied to the following sequence would give a wrong answer. $C_n = \int_0^{n} 1/n \ dx$ In general I would say its a bad idea to think about it this way even if just for "intuition" since the intuition gained is wrong. Justifying when you can interchange limits or not typically requires the dominated convergence theorem or monotone convergence theorem. Both of which are a bit advanced for a calculus student.
In your example $C_n=1$ for all $n$. What is the problem?

What is wrong with my answer. It could have been expressed with usual $\epsilon, n$ notation, but the result would be the same.

February 23rd, 2019, 04:04 PM   #6
Senior Member

Joined: Sep 2016
From: USA

Posts: 645
Thanks: 406

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by mathman In your example $C_n=1$ for all $n$. What is the problem? What is wrong with my answer. It could have been expressed with usual $\epsilon, n$ notation, but the result would be the same.
Yes $C_n = 1$ for all $n$. The problem is that
$\int_0^n \lim_{n \to \infty} 1/n \ dx = 0$ for all $n$ so obviously the limit exchange,
$\lim_{n \to \infty} \int_0^n 1/n \ dx = \int_0^n \lim_{n \to \infty} 1/n \ dx$
fails in this example. But this limit exchange is exactly what you implied in your example without justifying why it is allowed in that case.

February 24th, 2019, 01:29 PM   #7
Global Moderator

Joined: May 2007

Posts: 6,822
Thanks: 723

Quote:
 Originally Posted by SDK Yes $C_n = 1$ for all $n$. The problem is that $\int_0^n \lim_{n \to \infty} 1/n \ dx = 0$ for all $n$ so obviously the limit exchange, $\lim_{n \to \infty} \int_0^n 1/n \ dx = \int_0^n \lim_{n \to \infty} 1/n \ dx$ fails in this example. But this limit exchange is exactly what you implied in your example without justifying why it is allowed in that case.
I should have given more detail for the first step. In the range of integration $tan^{-1}(1-\frac{1}{n+1})\le tan^{-1}(nx) \le tan^{-1}(1)$, similarly for $sin^{-1}(nx)$, so that the integrand can be approximated by a constant $k=\frac{tan^{-1}(1)}{sin^{-1}(1)}$ and the integral $\int_{1/(n+1)}^{1/n}dx$ becomes $\approx \frac{k}{n(n+1)}\approx \frac{k}{n^2}$, where $k=1/2$.

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Conway51 Number Theory 31 November 10th, 2017 05:35 PM VisionaryLen Calculus 19 October 28th, 2016 11:29 AM alikim Elementary Math 6 June 10th, 2015 09:35 AM cmmcnamara Advanced Statistics 4 February 10th, 2010 05:49 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top