improper integral: exp(ikx) undefined in Matlab?

5 views (last 30 days)
Niklas Kurz on 19 Jun 2021
Edited: Niklas Kurz on 20 Jun 2021
I wanna integrate: With solution: But Matlab gives NAN:
syms x k; assume(k,'integer'); int(exp(1i*k*x),x,-inf,inf)

Steven Lord on 19 Jun 2021
What is the value of your integral when k is equal to 0?
syms x k;
assume(k,'integer');
k = 0;
int(exp(1i*k*x),x,-inf,inf)
ans =
This makes sense, as you're just integrating 1 over the whole real line.
What's the value of your integral when k is 1?
k = 1;
int(exp(1i*k*x),x,-inf,inf)
ans =
NaN
Does this make sense? Let's look at the real and imaginary parts of the function you're integrating.
f = 8*pi;
fplot(real(exp(1i*k*x)), [-f, f], 'k--')
hold on
fplot(imag(exp(1i*k*x)), [-f, f], 'c-') Those oscillations could be problematic. Does this integral exist? Let's look at a series of values of those integrals for gradually increasing limits.
for L = 0:0.5:10
value = int(exp(1i*k*x), -L*pi, L*pi);
fprintf("The value of the integral from %g*pi to %g*pi is %g.\n", -L, L, value);
end
The value of the integral from -0*pi to 0*pi is 0. The value of the integral from -0.5*pi to 0.5*pi is 2. The value of the integral from -1*pi to 1*pi is 0. The value of the integral from -1.5*pi to 1.5*pi is -2. The value of the integral from -2*pi to 2*pi is 0. The value of the integral from -2.5*pi to 2.5*pi is 2. The value of the integral from -3*pi to 3*pi is 0. The value of the integral from -3.5*pi to 3.5*pi is -2. The value of the integral from -4*pi to 4*pi is 0. The value of the integral from -4.5*pi to 4.5*pi is 2. The value of the integral from -5*pi to 5*pi is 0. The value of the integral from -5.5*pi to 5.5*pi is -2. The value of the integral from -6*pi to 6*pi is 0. The value of the integral from -6.5*pi to 6.5*pi is 2. The value of the integral from -7*pi to 7*pi is 0. The value of the integral from -7.5*pi to 7.5*pi is -2. The value of the integral from -8*pi to 8*pi is 0. The value of the integral from -8.5*pi to 8.5*pi is 2. The value of the integral from -9*pi to 9*pi is 0. The value of the integral from -9.5*pi to 9.5*pi is -2. The value of the integral from -10*pi to 10*pi is 0.
So should the value of this integral on the infinite interval be 0, -2, or something inbetween?
Niklas Kurz on 20 Jun 2021
yea, that's the problem aobut the delta-distribution not being a function, so it is hard for matlab to see in infinit domain. Or the variable k is not restricted enough. At least.
syms x; fourier(x^0)
gives the adaped solution (evaluating exactly that integral)