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Calculate $$\lim_{a\to 1}\int_{0}^{a}x\ln(1-x)dx, a\in (0,1)$$

I calculate the integral but when I calculate the limit I get $\ln(0)$ and the limit should be $-\frac{3}{4}$.

How to approach the exercise in other way?

Zacky
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DaniVaja
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  • How are you approaching the integral? – JacobCheverie Apr 24 '19 at 17:34
  • By parts.I take f(x)=ln(1-x) so f'(x)=1/(1-x) and g'(x)=x so g(x)=x^2/2 – DaniVaja Apr 24 '19 at 17:35
  • @DaniVaja You should have $f'(x)=-\frac{1}{1-x}$. – kccu Apr 24 '19 at 17:36
  • That will yield $$\int_0^a \dfrac{x^2}{2(1-x)}$$

    How do you handle that?

     – JacobCheverie Apr 24 '19 at 17:40
  • I get $\frac{-a^{2}-2a-ln(a-1)^{2}}{4}$ – DaniVaja Apr 24 '19 at 17:43
  • I use substitution of $u=1-x, du=-dx, x=1-u$ to get

    $$\int_0^a \dfrac{(1-u)^2}{2u} du = \int_0^a \Bigg( \dfrac{1}{2u} - 1 + \dfrac{u}{2} \Bigg) du$$

    This will result in $\dfrac{1}{2} ln|2(1-x)| -1 +x +\dfrac{1}{4}(1-x)^2$

     – JacobCheverie Apr 24 '19 at 17:47
  • @DaniVaja Write it instead as $-\frac{1}{2}\ln(1-a) - \frac{a^2}{4}-\frac{a}{2}$. Now you have to combine the $\ln(1-a)$ term with the rest (i.e., $f(x)g(x)$ evaluated at $x=a$) before taking the limit. – kccu Apr 24 '19 at 17:53
  • My whole expression is

    $$\lim_{a\rightarrow 1} \dfrac{a^2}{2} \ln|1-a| - \dfrac{1}{2} \ln|2(1-a)| -a -\dfrac{1}{4}(1-a)^2 + \dfrac{1}{2}\ln|2| +\dfrac{1}{4}$$

     – JacobCheverie Apr 24 '19 at 17:57

2 Answers2

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Here’s another approach:

$$\lim_{a\to 1}\int_{0}^{a}x\log(1-x)dx=-\lim_{a\to 1}\int_{0}^{a}\sum_{n\ge 1}{\frac{x^{n+1}}{n}}dx=-\lim_{a\to 1}\sum_{n\ge1}{\int_{0}^{a}{\frac{x^{n+1}}{n}}}dx=-\lim_{a\to 1}\sum_{n\ge 1}{\frac{a^{n+2}}{n(n+2)}}$$ $$=-\sum_{n\ge 1}{\frac{1}{n(n+2)}}=-\frac{1}{2}\sum_{n\ge 1}{\left(\frac{1}{n}-\frac{1}{n+2}\right)}= -\frac{1}{2}\sum_{n\ge 1}{\left(\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2}\right)}$$ $$ =-\frac{1}{2}{\overbrace{\sum_{n\ge 1}{\left(\frac{1}{n}-\frac{1}{n+1}\right)}}}^{=1}-\underbrace{\frac{1}{2}\sum_{n\ge 1}{\left(\frac{1}{n+1}-\frac{1}{n+2}\right)}}_{\text{this is 1/2}}$$ $$=-\frac{1}{2}\times 1-\frac{1}{2}\times\frac{1}{2}=-\frac{3}{4}$$

DINEDINE
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1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\left.\lim_{a \to 1^{\Large -}} \int_{0}^{a}x\ln\pars{1 - x}\,\dd x\,\right\vert_{\ a\ >\ 0}} \\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\,& \lim_{a \to 1^{\Large -}}\bracks{% {1 \over 2}\,a^{2}\ln\pars{1 - a} - \int_{0}^{a}\pars{{1 \over 2}\,x^{2}}{-1 \over 1 - x}\,\dd x} \\[5mm] = &\ \lim_{a \to 1^{\Large -}}\bracks{% {1 \over 2}\,a^{2}\ln\pars{1 - a} - {1 \over 2}\int_{0}^{a}\pars{1 + x - {1 \over 1 - x}}\,\dd x} \\[5mm] = &\ \lim_{a \to 1^{\Large -}}\bracks{% {1 \over 2}\,a^{2}\ln\pars{1 - a} - {1 \over 2}\,a - {1 \over 4}\,a^{2} - {1 \over 2}\,\ln\pars{1 - a}} \\[5mm] = &\ \bbx{-\,{3 \over 4}} \end{align}

Felix Marin
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