Suppose $\alpha$ is a constant, $A$ is a positive real number.
(a) Find the value of $\int_1^A x^{-\alpha} dx$, for $A > 1$.
(b) What is the answer of (a) when $A$ tends to infinity?
For (a), I find out it's $\frac{1}{1-\alpha}( A^{1-\alpha} -1 )$
However, for (b) , I am not sure how would the answer in (a) change when A tends to infinity. Should I state that there are 4 cases ( $\alpha < 0$, $\alpha = 0$, $0 < \alpha < 1$, $\alpha > 1$) and then work out how does the answer in (a) change accordingly?
The following expression: $$\frac{1}{1-\alpha}( A^{1-\alpha} -1 )$$
goes to infinity for $\alpha \in (-\infty, 1)$. It is finite and equal to $\frac{1}{\alpha-1}$ for $\alpha \in (1,\infty)$. For $\alpha = 1$, the answer will be $ln(A)$ which goes to $\infty$ as $A\to\infty$.
Also for the first part, you have to take two cases. $\alpha = 1$ and $\alpha \neq 1$ because the resulting expression will change because of that.
For $\alpha \neq 1$ you get this expression:
$$\frac{1}{1-\alpha}( A^{1-\alpha} -1 )$$
And for $\alpha = 1$, you get
$$ln(A)$$