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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. 
February 22nd, 2019, 10:33 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,759 Thanks: 2138 
Where did you get the question from? The fourth answer seems to be correct.

February 22nd, 2019, 01:09 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,770 Thanks: 700 
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 $C_n\approx \frac{1}{2n^2}$

February 22nd, 2019, 04:12 PM  #4  
Senior Member Joined: Sep 2016 From: USA Posts: 628 Thanks: 398 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
\[ C_n = \int_0^{n} 1/n \ dx \] 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: 6,770 Thanks: 700  Quote:
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: 628 Thanks: 398 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
$\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: 6,770 Thanks: 700  Quote:
 

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