清空筛选条件
清空筛选条件

I am trying to get a numerical answer of following improper integral but not getting the correct value

2 次查看（过去 30 天）
Tarun Agarwal
Tarun Agarwal 2021-7-25
评论： John D'Errico 2021-7-25
syms x;
f =(exp(1/cos(x)));
A=vpa(int(f,-11/7,11/7));
A
Please correct me where I am going wrong and how to evaluate this value using some approximations or other functions.

请先登录，再回答此问题。

回答（1 个）

Chunru
Chunru 2021-7-25
Ran in:
Your integration interval [-11/7 11/7] include two points with infinity.
Try the following:
syms x;
f =(exp(1/cos(x)));
A=vpa(int(f, x, -10/7,10/7));
A
A = 
78.865629794567528914520336604775
  2 个评论
Walter Roberson
Walter Roberson 2021-7-25
In particular, 11/7 is an approximation of pi/2, and is slightly larger than pi/2 . cos(pi/2) is 0, so 1/cos(pi/2) would pass through infinity.
The situation would have been different if it had been exp(-1/cos(x))
John D'Errico
John D'Errico 2021-7-25
Ran in:
To expand on the answer here...
syms x
f =(exp(1/cos(x)));
fplot(f,[-11/7,-10/7])
fplot(f,[10/7,11/7])
The point being that cos(pi/2) == 0, and that pi/2 is just slightly less than 11/7.
pi/2 < 11/7
ans = logical
1
But just slightly so. That means your integrand has a singularity at +/- pi/2. And that begs the question of whether the integral is well posed, having a finite value at all.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Calculus 的更多信息

标签

产品


版本

R2020a

Community Treasure Hunt

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

Start Hunting!

Translated by