6.72 Assume(16.2.01)

6.72.1 John Trapp
6.72.2 Carl DeVore (19.2.01)
6.72.3 E. Elbraechter (20.2.01)
6.72.4 Adri van der Meer (20.2.01)

6.72.1 John Trapp

I am sure that this has been reported before.

> assume(-3<t);additionally(t<3); 
> about(t); 
Originally t, renamed t~: 
  is assumed to be: RealRange(Open(-3),Open(3)) 
 
> int(2/(t^2-9),t); 
                                 
> int(-2/(9-t^2),t); 
 
> -int(2/(9-t^2),t);
 

In each case the solution is 1/3ln(t~-3)-1/3ln(t~+3). Maple does not recognize the domain of the integrand at all and we have the log of a negative number. Ugly.

6.72.2 Carl DeVore (19.2.01)

If you allow for the constant of integration to be complex, then the result makes sense. If you evaluate this anitiderivative between real limits, you will get a real answer.

I do realize, however, that this is difficult to explain to a beginning calculus student when you’re trying to teach them Maple.

6.72.3 E. Elbraechter (20.2.01)

From the online help of ‘int’ the statement:

Note that no constant of integration appears in the result.

Therefore the integral of an real valued integrand can have a constant imaginary part;

> restart; 
> assume(-3<t, t<3); 
>   int(2/(t^2-9),t): 
>   evalc(%): 
> J  := %; 
            J := 1/3 ln(3 - t) - 1/3 ln(t + 3) + 1/3 I Pi 
 
> Jr := Re(J); 
               Jr := 1/3 ln(3 - t) - 1/3 ln(t + 3)
 

For -3 < t < 3 this is a real valued expression.

6.72.4 Adri van der Meer (20.2.01)

Try:

> f := int(2/(x^2-9),x=0..t);
 

and Maple will recognize the assumption made on t.