Solve Complicated Integral: 4+ 8/πx +O(x²) at x→0

  • Thread starter Thread starter wel
  • Start date Start date
  • Tags Tags
    Integral
Click For Summary
SUMMARY

The integral \( I(x) = \int^{2}_{0} (1+t) e^{x \cos[\pi (t-1)/2]} dt \) evaluates to \( I(x) = 4 + \frac{8}{\pi}x + O(x^{2}) \) as \( x \rightarrow 0 \). To simplify the evaluation, one can use a Taylor expansion of the exponential function for small \( x \), which transforms the integral into a more manageable form. Integration by parts is suggested but deemed complicated due to the exponential term. The discussion emphasizes focusing on the behavior of the integral as \( x \) approaches zero.

PREREQUISITES
  • Understanding of Taylor series expansion
  • Familiarity with integration techniques, particularly integration by parts
  • Knowledge of limits and asymptotic notation, specifically \( O(x^2) \)
  • Basic understanding of exponential functions and their properties
NEXT STEPS
  • Study Taylor series expansion for exponential functions
  • Practice integration by parts with complex functions
  • Explore asymptotic analysis and its applications in calculus
  • Learn about the properties of limits and their significance in integral calculus
USEFUL FOR

Mathematicians, calculus students, and anyone interested in advanced integral evaluation techniques, particularly those dealing with asymptotic expansions and Taylor series.

wel
Gold Member
Messages
36
Reaction score
0
Consider the integral
\begin{equation}
I(x)=\int^{2}_{0} (1+t) e^{xcos[\pi (t-1)/2]} dt
\end{equation}
show that
\begin{equation}
I(x)= 4+ \frac{8}{\pi}x +O(x^{2})
\end{equation}
as $$x\rightarrow0.$$

=> Using integration by parts, but its too complicated for me because of huge exponential term.
please help me.
 
Physics news on Phys.org
Notice that you are only to consider the case when x is very small, tending to zero. This means you can make a suitable expansion of the exponential function, leaving a much simpler integral.
 
CAF123 said:
Notice that you are only to consider the case when x is very small, tending to zero. This means you can make a suitable expansion of the exponential function, leaving a much simpler integral.

Or maybe do this:
$$I=I(0)+I'(0)x+O(x^2)$$
:rolleyes:
 

Similar threads

  • · Replies 105 ·
4
Replies
105
Views
11K
Replies
7
Views
2K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
3
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
12
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
2K