@jschatzman Agreed for the boundary case. My mistake. However, there are still a multitude of cases where Maple gives the wrong answer. I am wondering what approach I should take with problems like this to assure that Maple produces correct results.

For example, Wkipedia gives the formula

where rect is defined to be 1 for -1/2<f<1/2, 1/2 for f=+-1/2, and 0 for |f|>1/2, which I believe is correct. Their sinc function is the definition with pis.  However, as I indicated above, Maple sometimes gives different answers for equivalent expression written slightly differently.

I am not trying to get hyper-critical about Maple. I am just trying to get it to work on more complicated integrals for me, am concerned that the results are not always correct, and I was looking for guidance on how to be certain of correctness.

@jschatzman (int(sin(x)*exp(-2*I*Pi*f*x)/x, x = -infinity .. infinity) assuming (Im(f) = 0))

gives a yet different result, also wrong. It is correct except at the points f=-1/(2Pi) and +1/(2Pi), where the solution given has the value pi/2 (but should be Pi).

However

int(sin(x)*exp(-I*x)/x, x = -infinity .. infinity) gives Pi  (correct)

int(sin(x)*exp(x*I)/x, x = -infinity .. infinity)  gives undefined (incorrect)

It seems that depending on how the integral is written, you can get from Maple:

a) The correct answer.

b) 0 when it should be Pi

c) Pi when it should be  conditioned on f

d) Pi/2 when it should be Pi

e) undefined when it should be Pi

@jschatzman These results are both correct

(int(sin(x)*exp(-2*I*Pi*f*x)/x, x = -infinity .. infinity) assuming (1/(2*Pi) < abs(f)))  gives 0

(int(sin(x)*exp(-2*I*Pi*f*x)/x, x = -infinity .. infinity) assuming (abs(f) <= 1/(2*Pi))) gives pi (greek)

But without the condition on f, the result is incorrect.

@Mariusz Iwaniuk  Wow that is very interesting. Thank you. However, I am still confused.

A) pi and Pi

both are displayed as the greek letter pi, so I would guess that they are equivalent in Maple.

However

int(sin(x)*exp(-2*I*pi*x)/x, x = -infinity .. infinity)   gives  pi(greek) as a result - wrong answer

and

int(sin(x)*exp(-2*I*Pi*x)/x, x = -infinity .. infinity)   gives  0 as a result - correct answer

What is going on?

B)

(int(sin(x)*exp(-2*I*Pi*f*x)/x, x = -infinity .. infinity) assuming (0 < f))  gives the correct answer

even though this is an even function of f, and therefore it shouldn't care about the sign.

By contrast,

int(sin(x)*exp(-2*I*Pi*f*x)/x, x = -infinity .. infinity)   gives pi (greek) - wrong answer

Again, what is going on?

c) How can I get assuredly correct results?