<s>[INST] Showing that $\int_{0}^{\infty}\frac{x^{m-1}-x^{n-1}}{1-x}dx$ is convergent if $ 0
Showing that $\int_{0}^{\infty}\frac{x^{m-1}-x^{n-1}}{1-x}dx$ is convergent if $ 0<m<1 , 0<n <1$
Let $I = \int_{0}^{\infty}\frac{x^{m-1}-x^{n-1}}{1-x}dx$ then $I = \int_{0}^{\infty}\frac{x^{m-1}}{1-x}dx -  \int_{0}^{\infty}\frac{x^{n-1}}{1-x}dx$
We consider $I_1 =  \int_{0}^{\infty}\frac{x^{m-1}}{1-x}dx$
Let $f_1(x) = \frac{x^{m-1}}{1-x}$ then $f_1(x)$ has three points of discontinuity at $x = 0,1 ,\infty$ if $0<m<1$
Let us consider $g_1(x) =  \frac{1}{x^{1-m}}$ then  $\lim_{x \to 0+}\frac{f_1(x)}{g_1(x)} = 1$
Now the integral of the function $g_1(x)$ from $0$ to $a(<1)$ is convergent so the $\int_{0}^{a}f_1(x)dx$ is convergent.
Now we consider the function $f_1(x)$ at the point $x = 1$ then we consider the function $g_1(x) = 1-x$ then $\lim_{x \to 1+}\frac{f_1(x)}{g_1(x)} = \frac{1-x}{x^{m-1}(1-x)} = 1$
Now we see that the function $\int_{a}^{1}g(x) = \int_{a}^{1}\frac{1}{1-x}dx $ is not convergent so the integral $\int_{a}^{1}f_1(x)dx$ is not convergent.
Where am I going wrong in this step?
 [/INST] A simple way to show that the integral is convergent is by cutting the integral into three parts
\begin{align}
\int^{1/2}_0 + \int^{3/2}_{1/2} +  \int^\infty_{3/2}\frac{x^{m-1}-x^{n-1}}{1-x}d x = I_1+I_2+I_3.
\end{align}
It is clear that $I_3$ is convergent since
\begin{align}
I_3 \le C\int^\infty_{3/2}\left(\frac{1}{x^{2-m}}+\frac{1}{x^{2-n}}\right)dx<\infty
\end{align}
for some constant $C>0$.
It is also clear that $I_1$ is convergent since
\begin{align}
I_1 \le C\int^{1/2}_0 \left(\frac{1}{x^{1-m}}+\frac{1}{x^{1-n}}\right)dx.
\end{align}
Lastly, let us deal with $I_2$. Assume $n\ge m$. Notice by the mean value theorem, we have that
\begin{align}
I_2 = \int^{3/2}_{1/2} x^{m-1}\frac{1-x^{n-m}}{1-x} \le C\int^{3/2}_{1/2} x^{m-1} d x
\end{align}
</s>