How Does Integration by Expansion Work for Sine Integrals?

Click For Summary
SUMMARY

The discussion focuses on the evaluation of the integral \( I(x) = \frac{1}{\pi} \int^{\pi}_{0} \sin(x \sin t) dt \) and its expansion as \( I(x) = 4 + \frac{2x}{\pi}x + O(x^{3}) \) as \( x \rightarrow 0 \). A user attempted to apply the Maclaurin series for \( f(t) = \sin(x \sin t) \) but encountered difficulties. The conversation emphasizes the necessity of showing prior work in accordance with forum rules, which is crucial for receiving assistance.

PREREQUISITES
  • Understanding of integral calculus, specifically sine integrals.
  • Familiarity with Maclaurin series expansion.
  • Basic knowledge of asymptotic notation, particularly \( O(x^n) \).
  • Experience with differentiation and evaluating derivatives of functions.
NEXT STEPS
  • Study the derivation of the Maclaurin series for \( \sin(x) \) and its applications.
  • Learn about the properties of sine integrals and their expansions.
  • Explore techniques for evaluating integrals involving trigonometric functions.
  • Review asymptotic analysis and how to apply it in mathematical proofs.
USEFUL FOR

Students preparing for exams in calculus, mathematicians interested in integral evaluations, and anyone studying series expansions in mathematical analysis.

wel
Gold Member
Messages
36
Reaction score
0
Consider the integral
\begin{equation}
I(x)= \frac{1}{\pi} \int^{\pi}_{0} sin(xsint) dt
\end{equation}
show that
\begin{equation}
I(x)= 4+ \frac{2x}{\pi}x +O(x^{3})
\end{equation}
as x\rightarrow0.

=> I Have used the expansion of McLaurin series of I(x) but did not work.
please help me.

(Note: It is not my homework or coursework question but it is from past exam paper which i am preparing for my exam)
 
Physics news on Phys.org
Please show your working.

Maclearen series: $$f(t)=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$$

##f(t)=\sin(x\sin t),\; f(0)=0##

##f'(t)=\cdots##

##f''(t)=\cdots##

etc. until you start getting terms in x3
 
wel,
Physics Forums rules require that you show what you have tried. I sent you a PM about this, and am closing this thread. You are welcome to start a new thread for this problem, but you need to show what you have tried.
 

Similar threads

  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K