Numerical integral where the bounds change with each evaluation

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

[FannoFlow]

Wolfram doesn't like your input either

https://www.wolframalpha.com/input?i=integral+A%28t%29+*+sin%28omega+*+t%29+dt+from+0+to+2*pi%2Fomega

syms omega t A(t)


f = int(A(t) * sin(omega * t), t, 0, 2*pi/omega)

f = 

simplify(f)

ans = 

[FannoFlow]

Does A(t) have a special meaning in this context?

[Walter Roberson]

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.

Answer 1

[Walter Roberson]

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)

[L'O.G.]

Thank you so much @Walter Roberson. Just one question: in the numerical example, how would you modify it to use a vector of type double which is also a function of t rather than the anonymous function A(t) that you used?

[Walter Roberson]

syms t Omega
A = @(t) sqrt(1-t.^2);
omega = logspace(-1,1,20);
fun = @(t,omega) A(t).*sin(omega*t);
F = simplify(int(fun(t,Omega), t, 0, 2*pi/Omega))

F = 

f = double(subs(F, Omega, omega));
subplot(2,1,1);
plot(omega, real(f), 'b-');
subplot(2,1,2);
plot(omega, imag(f), 'r-.')

[Walter Roberson]

"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]

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 ?