Discontinutiy of a function

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?

[Torsten]

This function is even holomorphic in the complex plane.

Where do you think are discontinuities ?

Best wishes

Torsten.

[Sergio Manzetti]

Dear Torsten, I wrong the wrong function, the right is:

f(x)= sqrt(1/2) - e^(-2*i*5.344285879^-28*x/1.0545718^-34) (sqrt(1/2)-(1.0545718^-34/2)cos(x) + (5.344285879^-28*x)(2*i*1.0545718^-34))

[Sergio Manzetti]

Dear Torsten, how can I find the discontinuities of this function in the REAL plane?

[Torsten]

No discontinuities - neither of the real or imaginary part on the real axis nor of the function itself in the complex plane.

Best wishes

Torsten.

[Sergio Manzetti]

Dear Torsten, thanks for your feedback, but can you try this function in Matheamtica Wolfram online? It says its discontinuous in R, but it is continuous in "its domain". The command there is simply "Is sqrt(1/2) - e^(-2*i*5.344285879^-28*x/1.0545718^-34) (sqrt(1/2)-(1.0545718^-34/2)cos(x) + (5.344285879^-28*x)(2*i*1.0545718^-34)) continuous? " That shows that it is not continuous in R. So how can two major programs disagree on such a critical point?

[Torsten]

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]

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)