I am trying to get a numerical answer of following improper integral but not getting the correct value
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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.
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Answers (1)
Chunru
on 25 Jul 2021
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
2 Comments
Walter Roberson
on 25 Jul 2021
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
on 25 Jul 2021
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
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.
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