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Numerical integral where the bounds change with each evaluation

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I'm trying to take the following integral:
I tried this symbolically, but I think that was throwing me off. I am attaching some numerical data for the function A(t) and the time vector. How would you do this numerically? I imagine it would be something like the following, but I can't quite get it to work.
omega = logspace(-1,1,1000);
fun = @(t,omega) A*sin(omega*t);
for i = 1:1000;
w=omega(i);
f(i) = integral(@(t) fun(t,w),0,2*pi/w);
end
  3 Comments
Walter Roberson
Walter Roberson on 16 May 2023
A(t) just means that A is a function of t.
It is the expression for some kind of tranform of A(t), but I am not sure what the name of this transform is.

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Answers (1)

Walter Roberson
Walter Roberson on 16 May 2023
A = rand
A = 0.8252
omega = logspace(-1,1,1000);
fun = @(t,omega) A*sin(omega*t);
for i = 1:1000;
w=omega(i);
f(i) = integral(@(t) fun(t,w),0,2*pi/w);
end
plot(omega, f)
All of the values are within round-off error of 0.
However... there is a difference between what your mathematical expression shows here, compared to the integral you were calculating there and here.
In previous discussion, A was a constant. In the expression here, A is a function of t. That makes a big difference.
For example,
A = @(t) t.^2 - t + 1;
omega = logspace(-1,1,1000);
fun = @(t,omega) A(t).*sin(omega*t);
for i = 1:1000;
w=omega(i);
FUN = @(t) fun(t,w);
f(i) = integral(FUN,0,2*pi/w);
end
plot(omega, f)
syms t Omega
A = @(t) t.^2 - t + 1;
omega = logspace(-1,1,1000);
fun = @(t,omega) A(t).*sin(omega*t);
F = int(fun(t,Omega), t, 0, 2*pi/Omega)
F = 
f = subs(F, Omega, omega);
plot(omega, f)
  4 Comments
Walter Roberson
Walter Roberson on 16 May 2023
"how would you modify it to use a vector of type double which is also a function of t"
Vectors are not functions.
If you have the value of A(t) sampled at particular t, then calculate A(t).*sin(omega*t) at those t, and use trapz() or similar to do numeric integration, making sure to pass in the appropriate t values to trapz()
Walter Roberson
Walter Roberson on 16 May 2023
Hmmm, first, could you confirm for me that you want omega to range from -1 to 1, and the bounds of integration is 0 to 2*pi/omega -- and since omega ranges from -1 to +1, that means that for negative omega you want negative upper bound, and you want to go right through to infinite upper bound as omega passes through 0 ?

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