Discontinutiy of a function

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Sergio Manzetti 2017-12-6

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/371438-discontinutiy-of-a-function

编辑： Sergio Manzetti 2017-12-8

Hi, I have the following command:

if true
syms h g x C 
h = 1
g = 4
C = 1/(2.*pi);
f = (C - (exp(-2.*g.*1i.*x./h)).*((g.*x)/2.*h.*1i));
disc = feval(symengine, 'discont', f, x);

However I get the result:

disc =

Empty sym: 1-by-0

Is there any way to get the discontinuity of this function in the real and imaginary plane?

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Torsten 2017-12-8

编辑：Torsten 2017-12-8

I don't understand why "Mathematica" treats this function as a mapping from R to R.

Regarding it as a mapping from R to C or from C to C, it is continous (and even has much higher smoothness properties).

Best wishes

Torsten.

Sergio Manzetti 2017-12-8

编辑：Sergio Manzetti 2017-12-8

Thanks Torsten, so I can safely say that this function is continuous in R and C? You see, the integral of it and its hermitian counterpart in MATLAB gives single valued numbers only when using double precision, but when using regular precision I get a result composite of several functions. So I though that composite result was actually the integral over several fragments of the function across R, and thus not a continuous integral. (which one would expect if it was continuous in R)

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