Help in understanding the Maple answer

Question

I tried to find the definite integral of an equation using Maple. It returns the answer to me as a limit. Specifically, it is giving me

$$\lim_{r\to \infty} G(r)$$

What does this answer mean? Do I need to put the limit myself to get the final answer?

Regards.

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(incorporating comment into this question)

This is the maple code:

int(2*Pi*lambda*alpha*r*exp(-Pi*r^2*(lambda*alpha-D*ln(Y)))*(1-exp(-Pi*lambda*alpha*(-R^2+r^2))), r = R .. infinity)

Can you please let me know what to do with this answer?

Answer 1

Hint: Using a version with simplified expressions (i.e. straightforward combinations of your parameters) I get with Maple 7 $$G(r) = \int r \exp(-b r^2) \left(1-\exp\left(-c \left(-R^2+r^2\right)\right)\right) d\,r \\ =-\frac{1}{2b\exp(br^2)}+\frac{\exp\left(-(b+c)r^2+cR^2\right) }{2(b+c)} $$ where you can see that the convergence $r\rightarrow \infty$ depends on the sign of $b$ and $b+c-\;$ If both are positive the limit should be 0 and the integral would be $$-G(R) = \frac{ce^{-R^2 b}}{2b(b+c)}$$

Answer 2

As others have suggested, it helps to supply additional assumptions, for example the signs on key subexpressions which affect the limit.

This below was in Maple 18.00,

restart:

int(2*Pi*lambda*alpha*r*exp(-Pi*r^2*(lambda*alpha-DD*ln(Y)))
    *(1-exp(-Pi*lambda*alpha*(-R^2+r^2))), r=R..infinity)   
 assuming lambda*alpha>0, lambda*alpha-DD*ln(Y)>0:         

lprint(%);

   lambda^2*alpha^2*Y^(R^2*DD*Pi)*exp(-Pi*R^2*alpha*lambda)
   /(DD^2*ln(Y)^2-3*DD*ln(Y)*lambda*alpha+2*lambda^2*alpha^2)