Divergence of integral $\int_{a}^{t} \frac{1}{g-px^{b}} dx$

Question

How can I show that $$\lim \limits_{t \rightarrow (\frac{g}{p})^{1/b}}{\int_{a}^{t}} \frac{1}{g-px^{b}} dx ,$$ $g = $gravitational acceleration, $p,b >0$ diverges to $\infty$ ?

Answer 1

Let $x_0 = (\frac{g}{p})^{1/b}$. Note that $$ \lim_{x\rightarrow x_0} \frac{x_0 - x}{g - px^b} =^H \lim_{x\rightarrow x_0} \frac{-1}{ - pbx^{b-1}} = \frac{1}{pb}x_0^{-\frac{1}{b}+1} = \frac{1}{gb}x_0^{-\frac{1}{b}} > 0$$ is finite and not equal $0$. That means that if you pick any $C\in(0,\frac{1}{gb}x_0^{-\frac{1}{b}})$, then for $x$ close enough to $x_0$ you have: $$ C\frac{1}{x_0-x} \le \frac{1}{g - px^b} $$ Since integral $ \int_a^{x_0} \frac{1}{x_0-x} dx $ is divergent this means that also $ \int_a^{x_0} \frac{1}{g-px^b} dx$ is divergent.