Matlab not computing integral

I am new to Matlab and cannot figure out why Matlab will not calculate this integral. Any help would be greatly appreciated. Thanks!
clc;
%Symbolic Variables
syms lambda_x x x_prime y y_prime pi k
%Constants
k = 10;
x = 5;
y = 6;
x_prime = -1;
y_prime = 2;
g1 = exp(1i*lambda_x*(x-x_prime))*exp(-1i*abs(y-y_prime)*sqrt((k^2)-(lambda_x^2)));
g2 = sqrt((k^2)-(lambda_x^2));
g3 = g1/g2;
g3integral = int(g3,lambda_x,-inf,inf, 'PrincipalValue', true);
pretty(g3integral)

4 件のコメント

Aquatris
Aquatris 2018 年 9 月 13 日
編集済み: Aquatris 2018 年 9 月 13 日
There might not be a closed form solution to the integral. Where does this equation come from? Did you try any online integral calculator?
Brandon Peter
Brandon Peter 2018 年 9 月 13 日
That is what I am attempting to integrate
Jan
Jan 2018 年 9 月 14 日
What about integrating it numerically and not symbolically?
Brandon Peter
Brandon Peter 2018 年 9 月 14 日
How would you go about doing that?

サインインしてコメントする。

回答 (1 件)

Walter Roberson
Walter Roberson 2018 年 9 月 14 日

0 投票

Numerically it is undefined because the denominator passes through 0, giving a division by 0.
The cauchy principle value does not appear to exist. Take the limit of g3 as lambda_x approaches -infinity or +infinity and you will see it being infinite. Take the limit as lambda_x approaches -10 or +10 and you will see complex infinities with different sign on the two sides of the discontinuity.

カテゴリ

タグ

質問済み:

2018 年 9 月 13 日

回答済み:

2018 年 9 月 14 日

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by