Proving $\int_0^{\infty} \frac {x^{m-1} - x^{n-1}}{1-x} dx$ and $\int_0^{\infty} x^m (\log x)^m dx$ Prove that:
(a) $\int_0^{\infty} \frac {x^{m-1} - x^{n-1}}{1-x} dx$ is convergent if $0<m<1$ and $0<n<1$;
(b) $\int_0^{\infty} x^m (\log x)^n dx$ is convergent if $m< -1$;
Getting no clue to solve these improper integrals. Thanks for Help!
 A: I began to write this solution, but then realized that the parameters $n$ and $m$ are either not defined or defined incorrectly.

For the second integral, we assume that $n$ is an integer, else $\log^n x$ is undefined on the reals for $x<1$.  We will invoke the standard inequalities for the logarithm function  
$$\frac{x-1}{x}\le \log x\le x-1 \tag 1$$
Thus, for all $\alpha>0$ we have 
$$\frac{x^\alpha -1}{\alpha x^\alpha}\le \log x \le \frac{x^\alpha -1}{\alpha} \tag 2$$
We first note that the integral of interest $\int_0^\infty x^m\log^n x\,dx$, has (possible) singularities at $x=0$, $x=1$, and at $x=\infty$.   Therefore, we will write the integral as the sum
$$\int_0^\infty x^m\log^n x\,dx=\int_0^{1/2} x^m\log^n x\,dx +\int_{1/2}^{e}x^m\log^mx\,dx +\int_{e}^\infty x^m\log^n x\,dx \tag 2$$
For, the first integral on the right-hand side of $(2)$, we see using $(1a)$ that for $n\ge 0$ we have
$$|x^m \log^n x|\ge |x^m|\,|x-1|^n \tag 3$$
while using $(2)$ for $n<0$ we have for all $\alpha >0$
$$|x^m \log^n x|\ge |x^m|\,\frac{|x^{\alpha}-1|^n}{\alpha^n x^{n\alpha}} \tag 4$$
From $(3)$ it is easy to see that for $n\ge 0$, the first integral on the right-hand side of $(2)$ converges for $m>-1$ and diverges for $m,\le -1$ by comparison with the integral $\int_0^{1/2} x^m\,dx$.  Since $(4)$ must be true for all $\alpha>0$, we can see again that the first integral converges for $m>-1$ by taking $\alpha$ so small that $m-n\alpha >-1$; and the integral diverges otherwise.
Similarly, using $(2)$, the second integral on the right-hand side of $(2)$ converges for $n>-1$ by comparison with the convergent integral $\int_{1/2}^{3/2} (x-1)^n\, dx$.
Finally, for the third integral on the right-hand side of $(2)$, we note that $\log x\ge 1$ and increasing with $x$.  Therefore, using $(3)$ we see that 
$$x^m\log^m x=O(x^m)$$
and the third integral on the right-hand side of $(2)$ converges for $m>-1$ and diverges elsewhere by comparison with the integral $$\int_e^{\infty}x^m\,dx$$
Putting it all together, the integral in $(2)$ diverges for all $m$ and $n$.
