That outcome is correct given that integral -- if you expected a different output then what you integrate is not correct.
int(dirac(2*pi*f) -((2*pi*m)/3),m,-Inf,Inf)
dirac(2*pi*f) is constant compared to m, so integrating it with respect to m from A to B is going to give you
dirac(2*pi*f)*B - dirac(2*pi*f)*A
dirac(2*pi*f)*(B-A)
but B is inf and A is -inf so that would be dirac(2*pi*f) * (inf-(-inf)) => dirac(2*pi*f) * inf .
dirac(2*pi*f) is 0 whenever f is non-zero. 0 * inf is nan. Therefore the int(dirac*2*pi*f), m,-inf,inf) is going to give nan in all positions except where f == 0.
dirac(2*pi*0) = dirac(0) . dirac is a "distribution", not a function as is typically understood, but the distribution has value infinity except when dirac(expression) is being integrated with respect to a variable involved in the expression. For example int(dirac(2*pi*m), m, -inf, inf) potentially has non-infinite meaning, but dirac(0), m, -inf, inf) is always going to be infinity. So at f == 0, the integral of the dirac(0) is infinity * (infinity - (-infinity)) -> infinity .
Thus, taking into account just the first term, it is NaN in each case where f is non-zero, and infinity in the case where f is 0.
The second term, which is of the form constant*m, integrates to (1/2*constant*m^2) evaluated at infininity minus the evaluation at -infinity. 1/2*constant * infinity^2 is infinity times sign(constant) for non-zero constant (i.e., -2*pi/3) and 1/2*constant*(-infinity)^2 is infinity times sign(constant) . infinity*sign(constant) - infinity*sign(constant) ... is undefined, because it is the subtraction of two infinities.
So the integral of the second term is undefined (nan) for all f values . The first term is nan for all f except 0, and the second term is nan for all f including f == 0.
Therefore, the integral is nan for all f values.
Therefore there is nothing to plot.
... Even if you repair your syntax to stem(f, x) -- since your x is actually your dependent variable and your f is your independent variable.
