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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.

3 Answers3

1

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.
Olivier Oloa
  • 120,989
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$$ \begin{align} \int_0^1\frac{x^n-x^m}{\log(x)}\,\mathrm{d}x &=\lim_{a\to0}\int_a^\infty\frac{e^{-(m+1)t}-e^{-(n+1)t}}{t}\,\mathrm{d}t\tag{1}\\ &=\lim_{a\to0}\left(\int_a^\infty\frac{e^{-(m+1)t}}{t}\,\mathrm{d}t-\int_a^\infty\frac{e^{-(n+1)t}}{t}\,\mathrm{d}t\right)\tag{2}\\ &=\lim_{a\to0}\left(\int_{(m+1)a}^\infty\frac{e^{-t}}{t}\,\mathrm{d}t-\int_{(n+1)a}^\infty\frac{e^{-t}}{t}\,\mathrm{d}t\right)\tag{3}\\ &=\lim_{a\to0}\int_{(m+1)a}^{(n+1)a}\frac{e^{-t}}{t}\,\mathrm{d}t\tag{4}\\ &=\lim_{a\to0}\int_{(m+1)a}^{(n+1)a}\frac{1+O(t)}{t}\,\mathrm{d}t\tag{5}\\ &=\lim_{a\to0}\left(\log\left(\frac{n+1}{m+1}\right)+O((n-m)a)\right)\tag{6}\\[3pt] &=\log\left(\frac{n+1}{m+1}\right)\tag{7} \end{align} $$ 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

robjohn
  • 345,667
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\int_{0}^{1}{x^{n} - x^{m} \over \ln\pars{x}}\,\dd x:\ {\large ?} \,,\qquad n > 0\,,\,\,\,m>0}$.

\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}

Felix Marin
  • 89,464