improper integral: exp(ikx) undefined in Matlab?

6 views (last 30 days)
Niklas Kurz
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)

Accepted Answer

Steven Lord
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?
  1 Comment
Niklas Kurz
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)

Sign in to comment.

More Answers (0)

Tags

Community Treasure Hunt

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

Start Hunting!

Translated by