MHB Integration--Finding Principal Value

  • Thread starter Thread starter clickbb08
  • Start date Start date
  • Tags Tags
    Value
Click For Summary
The discussion focuses on calculating the principal value of the integral from negative to positive infinity of Cos(x)/(x^2 + 9) and demonstrating that it equals (π/3e^2). The user expresses confusion regarding the approach to these types of problems, specifically mentioning the isolated singular points at -3i and 3i. A correction is noted, stating that the integral should actually yield (π/3e^3). The method involves using the residue theorem, with emphasis on contributions from the upper half-plane and the real axis. Understanding these concepts is crucial for solving the integral correctly.
clickbb08
Messages
1
Reaction score
0
I'm supposed to show that the P.V. of the integral from (-infinity to infinity) of Cosx/(x^2+9)dx is (pi/3e^2). I don't understand how to go about these kinds of problems. I know that I will have an ISP at -3i and 3i.
 
Physics news on Phys.org
clickbb08 said:
I'm supposed to show that the P.V. of the integral from (-infinity to infinity) of Cosx/(x^2+9)dx is (pi/3e^2). I don't understand how to go about these kinds of problems. I know that I will have an ISP at -3i and 3i.
First it should be
$$
\int_{-\infty}^{\infty}\frac{\cos x}{x^2 + 9}dx = \frac{\pi}{3e^3}
$$

Let $f(z) = \dfrac{e^{iz}}{(z-3i)(z+3i)}$

Then
$$
\int_{-\infty}^{\infty}\frac{\cos x}{x^2 + 9}dx = 2\pi i\sum_{\text{UHP}}\text{Res}_{z=z_j}f(z) + \pi i\sum_{\text{real axis}}\text{Res}_{z=z_j}f(z)
$$

UHP = upper half plane
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
7K