How to evaluate the following integral which is improper I want to evaluate 
$$ \int_0^1 \frac{ x^n - x^m }{\ln x } d x ,\;\;\;\;\;n>0,m>0$$
My books says the answer is $\ln \left( \frac{n+1}{m+1} \right)$. however, I think this integral diverges. Notice if we apply integration by parts, then 
we will have 
$$ \left( \frac{ x^{n+1} }{n+1 } - \frac{ x^{m+1} }{m+1} \right)\frac{1}{\ln x} \bigg\rvert_{0^-}^{1^+} - \int ..... $$
but the first part I think diverges, so the integral diverges.
 A: We assume $n>0,m>0$. 


*

*The integrand being continuous over $(0,b]$, $0<b<1$, with a finite limit as $x \to 0^+$, we deduce that the only potential issue of convergence is near $1^-$.  Let's see how
$\dfrac{x^n-x^m}{\ln x} $ behaves near $x=1$, $x<1$. One may write,
by a Taylor series expansion, as $x \to 1^-$, $$ x^{n}=e^{n\ln
   x}=e^{n\ln(1-(1-
   x))}=e^{n\left[(x-1)+O((x-1)^2)\right]}=1+n(x-1)+O((x-1)^2) $$ then  $\displaystyle \lim_{x \to 1^-}\frac{x-1}{\ln x}=1$ gives
$$ \frac{x^n-x^m}{\ln x}=\frac{n(x-1)-m(x-1)+O((x-1)^2)}{\ln
   x}=(n-m)+O(x-1) $$ thus the given integral is convergent.

*A closed form may be obtained by observing that $$\frac{x^n-1}{\ln
   x}=n\int_{0}^{1}x^{ny} dy$$ yielding $$ \begin{align}
   \int_0^1\frac{x^n-x^m}{\ln
   x}\:dx&= \int_0^1\frac{(x^n-1)-(x^m-1)}{\ln
   x}\:dx
\\\\&=\int_{0}^{1}dy\int_0^1(nx^{ny}-mx^{my})\:dx \\\\&=
   \int_0^1\left(\frac{n}{ny+1}-\frac{m}{my+1}\right)dy \\\\&=\ln
   \left(\frac{n+1}{m+1} \right) \end{align} $$ as expected.

A: $$
\begin{align}
\int_0^1\frac{x^n-x^m}{\log(x)}\,\mathrm{d}x
&=\lim_{a\to0}\int_a^\infty\frac{e^{-(m+1)t}-e^{-(n+1)t}}{t}\,\mathrm{d}t\tag{1}\\
&=\lim_{a\to0}\left(\int_a^\infty\frac{e^{-(m+1)t}}{t}\,\mathrm{d}t-\int_a^\infty\frac{e^{-(n+1)t}}{t}\,\mathrm{d}t\right)\tag{2}\\
&=\lim_{a\to0}\left(\int_{(m+1)a}^\infty\frac{e^{-t}}{t}\,\mathrm{d}t-\int_{(n+1)a}^\infty\frac{e^{-t}}{t}\,\mathrm{d}t\right)\tag{3}\\
&=\lim_{a\to0}\int_{(m+1)a}^{(n+1)a}\frac{e^{-t}}{t}\,\mathrm{d}t\tag{4}\\
&=\lim_{a\to0}\int_{(m+1)a}^{(n+1)a}\frac{1+O(t)}{t}\,\mathrm{d}t\tag{5}\\
&=\lim_{a\to0}\left(\log\left(\frac{n+1}{m+1}\right)+O((n-m)a)\right)\tag{6}\\[3pt]
&=\log\left(\frac{n+1}{m+1}\right)\tag{7}
\end{align}
$$
Explanation:
$(1)$: $x=e^{-t}$ and convert the bottom limit of integration to a limit
$(2)$: break the integral into two
$(3)$: substitute $t\mapsto t/(m+1)$ and $t\mapsto t/(n+1)$
$(4)$: combine the integrals
$(5)$: $e^{-t}=1+O(t)$
$(6)$: integrate
$(7)$: evaluate the limit
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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$\ds{\int_{0}^{1}{x^{n} - x^{m} \over \ln\pars{x}}\,\dd x:\ {\large ?}
\,,\qquad n > 0\,,\,\,\,m>0}$.

\begin{align}
\int_{0}^{1}{x^{n} - x^{m} \over \ln\pars{x}}\,\dd x & =
\int_{0}^{1}\pars{x^{n} - x^{m}}\,\,\,
\overbrace{\pars{-\int_{0}^{\infty}x^{t}\,\dd t}}
^{\ds{1 \over \ln\pars{x}}}\,\,\,\dd x =
\int_{0}^{\infty}\int_{0}^{1}\pars{x^{m + t} - x^{n + t}}\dd x\,\dd t
\\[5mm] & =
\int_{0}^{\infty}\pars{{1 \over t + m + 1} - {1 \over t + n + 1}}\dd t =
\left.\ln\pars{t + m + 1 \over t + n + 1}
\right\vert_{\ t\ =\ 0}^{\ t\ \to\ \infty}
\\[5mm] & =
\bbx{\ds{\ln\pars{n + 1 \over m + 1}}}
\end{align}
