From: //noreply@axelvogt.de   
      
   On 15.03.2019 16:37, acer wrote:   
   > On Friday, March 15, 2019 at 3:41:59 AM UTC-4, nma%12...@gtempaccount.com   
   wrote:   
   >> Maple 2019 gives   
   >>   
   >> int(exp(I*w*t)/(c+I*w),w=-infinity..infinity) assuming c>0   
   >>   
   >>          0   
   >>   
   >> Next, when I tell it that t>0, it then gives   
   >>   
   >> int(exp(I*w*t)/(c+I*w),w=-infinity..infinity) assuming c>0,t>0   
   >>   
   >>         2*Pi*exp(-c*t)   
   >>   
   >> Which is correct.   
   >>   
   >> Should it give zero for the first case?  With Mathematica 11.3   
   >>   
   >> sol = Integrate[ConditionalExpression[Exp[I w t]/(c + I*w), c > 0],   
   >>          {w, -Infinity,Infinity}]   
   >>   
   >> gives   
   >>   
   >> ConditionalExpression[(Pi*(1 + Sign[t]))/E^((c*t)/Sign[t]),   
   >>            Element[t, Reals] && c > 0]   
   >>   
   >> Which is zero, only when t<0 and  2*Pi*exp(-c*t) when t>0   
   >>   
   >> So Maple and Mathematica agree when told t>0. But why Maple gives zero when   
   one does not tell it anything about if t is positive or not? Do you think this   
   is an OK result by Maple?   
   >>   
   >> Thanks   
   >> --Nasser   
   >   
   > The problem is that the `meijerg` method is producing the 0 result and,   
   without the assumption on t, all other methods are producing FAIL.   
   >   
   > But with an assumption such as t>0 or t::real the `ftoc` and `ftocms`   
   methods produce a non-FAIL result and that is what `int` then returns.   
   >   
   > Yes, there is a bug in how the `int` is handling this example with its   
   `meijerg` method. With that method forced it produces 0 even for t=2,   
   >   
   >     int(exp(I*w*(2))/(c+I*w),w=-infinity..infinity,   
   >         method=meijerg) assuming c>0;   
   >   
   >                        0   
   >   
   > The `ftoc` method will allow this, hoewever,   
   >   
   >     int((exp(I*w*(t))/(c+I*w)),w=-infinity..infinity)   
   >       assuming t::real, c>0;   
   >   
   >                 2   
   >        signum(t)  exp(-c t) Pi + signum(t) exp(-c t) Pi   
   >   
   >   
   > Nasser, I will submit a bug report.   
   >   
      
      
   One can look at it as Fourier integral (any t) or use evalc (thus t real)   
   Then it works for me. Have not looked closer for t=0.   
      
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