$\int_0^1 \frac{x^p}{1-x^q}\; dx＝∞$

Question

When $p>-1, \;q>0$, I want to prove $$\int_0^1 \frac{x^p}{1-x^q}\; dx = \infty.$$

Any help would be appreciated. I observed by graph soft this is true, but I cannot prove. I want to find some function $\varphi(x)$ which satisfies

$$\int_0^1 \varphi(x)dx = \infty,$$ such that $ \frac{x^p}{1-x^q}>\varphi(x)$. Thank you.

Answer 1

Hint: It is enough to show that $\int_{1/2}^{1} \frac 1 {1-x^{q}}dx=\infty$ (because $\inf \{x^{p}: \frac 1 2 \leq x \leq 1\} >0$). Expand $\frac 1 {1-x^{q}}$ as $\sum_n x^{nq}$ and integrate term by term. Compare with $\sum \frac 1 n$.

Answer 2

It diverges about $x = 1$. To see this, substitute $y = 1-x$. This transforms the integral to $ \int_0^1 f(y) \, dy$ where $$ f(y) = \frac{(1-y)^p}{1-(1-y)^q} $$ Note that $$ (1-y)^q = 1 - q y + O(y^2) $$ and hence $$ f(y) \sim \frac{1}{q y} $$ as $y \to 0$. But $\int_0^1 dy/y = \infty$.

The $x^p = (1-y)^p$ in the numerator is a red-herring.