Power Series of Integral xarctan(3x)

Click For Summary
SUMMARY

The integral of the function xarctan(3x) from 0 to 0.1 can be evaluated using power series expansion. The discussion reveals that the power series for arctan(3x) and x/(1+9x^2) were derived, leading to the expression: [(-1)^(n-1)*9^(n-1)*x^(2n+1)]/[(2n+1)(2n)(2n-1)] + [(-1)^(n-1)*9^(n-1)*x^(2n)]/[(2n)(2n-1)]. The calculated result of 0.0050927 was identified as incorrect, prompting a request for assistance in identifying potential errors in differentiation or integration processes.

PREREQUISITES
  • Understanding of power series expansions, specifically for functions like arctan.
  • Familiarity with differentiation and integration techniques in calculus.
  • Knowledge of Taylor series and their convergence criteria.
  • Experience with evaluating definite integrals.
NEXT STEPS
  • Study the derivation of the Taylor series for arctan(u) and its applications.
  • Learn about the integration techniques for rational functions, particularly using logarithmic forms.
  • Explore convergence criteria for power series and their implications in calculus.
  • Practice evaluating integrals involving power series to solidify understanding.
USEFUL FOR

Students and educators in calculus, particularly those focusing on integral calculus and power series methods. This discussion is beneficial for anyone looking to deepen their understanding of series expansions and integration techniques.

EstimatedEyes
Messages
31
Reaction score
0

Homework Statement



Evaluate the integral of xarctan(3x) from 0 to 0.1 by expressing the integral in terms of a power series.

Homework Equations





The Attempt at a Solution



I differentiated xarctan(3x) until I got two functions that I could turn into power series (arctan(3x) and x(1/(1+9x^2)).

After I found the power series of these two functions I integrated them as many times as was necessary to arrive back at the integral of the original function and got (both as sums from one to infinity):
[(-1)^(n-1)*9^(n-1)*x^(2n+1)]/[(2n+1)(2n)(2n-1)] + [(-1)^(n-1)*9^(n-1)*x^(2n)]/[(2n)(2n-1)]

When I evaluated it from 0 to 0.1, I ended up with 0.0050927 which is not the correct answer, but I don't see where I would have done it wrong. Any help would be appreciated.
 
Physics news on Phys.org
A faster way to find the power series would be knowing \arctan{u}=\sum_{n=0}^{\infty}(-1)^{n}\frac{u^{2n+1}}{2n+1} convergent for |x|\leq{1}. x\arctan{3x}=\sum_{n=0}^{\infty}(-1)^{n}\frac{3^{2n+1}}{2n+1}x^{2(n+1)}. I haven't checked your work but maybe you made a mistake while differentiating.

On a separate note, \int \frac{x}{1+9x^{2}}\,dx=\frac{1}{18}\ln{|1+9x^{2}|}+C, not the arctan you were looking for.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
889
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
6
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K