Should maxima be able to do this integral? (A.K.A. what am I missing?)

The way to approach is not to look at forms of the the indefinite 
integral, but look squarely at the definite integral itself. Near zero 
the integrand behaves like 1/y, which is divergent. Near infinity it 
behaves like 1/y^2, which converges. So, the behaviour at zero makes 
the integral divergent.


On Wednesday, October 23, 2002, at 05:53  PM, C Y wrote:

> I'll readily concede I could be making a mistake here, but just in case
> I'm not...  When I try the following integral:
>
> (C1) integrate(1/(y*(y+1)),y);
>
> (D1) 			      LOG(y) - LOG(y + 1)
>
> which is fine and equals LOG(y/(y+1)).  But when I do a definite
> integral from zero to inf, this happens:
>
> (C9) integrate(1/(y*(y+1)),y,0,inf);
>
> Integral is divergent
>  -- an error.  Quitting.  To debug this try DEBUGMODE(TRUE);)
>
> which initially looks like it makes sense, since log(y/(y+1)) doesn't
> immediately look like it can be evaulated at the infinity limit,
> although it can be evaluated at y=0.  But if we do a manipulation of
> the results from the indefinite integral using Laws of Logarithms, we
> get the form:
>
> LOG(y) - LOG(y + 1)=-(LOG(y+1) - LOG(y))=-LOG(1+y/y)=-LOG(1+1/y)
>
> which can be evaluated at inf but not at zero.  If we use the latter
> form at inf and the first form at zero, which should be legal since the
> two forms are equal? evaluating at the limits and subtracting gives us
> for the integral:
>
> -LOG(1)-LOG(0)=-1
>
> I suppose I messed up and/or did something illegal somewhere, but can
> someone enlighten me why this is wrong?  Sorry if this is an obvious
> mistake on my part.
>
> CY
>
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