improper integral: exp(ikx) undefined in Matlab?

조회 수: 7 (최근 30일)
Niklas Kurz
Niklas Kurz 2021년 6월 19일
편집: Niklas Kurz 2021년 6월 20일
I wanna integrate:
With solution:
But Matlab gives NAN:
syms x k; assume(k,'integer'); int(exp(1i*k*x),x,-inf,inf)

채택된 답변

Steven Lord
Steven Lord 2021년 6월 19일
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?
  댓글 수: 1
Niklas Kurz
Niklas Kurz 2021년 6월 20일
편집: Niklas Kurz 2021년 6월 20일
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)

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Dialog Boxes에 대해 자세히 알아보기

태그

Community Treasure Hunt

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

Start Hunting!

Translated by